Analytical Chemistry Complete Guide

1 UNIT 1. INTRODUCTION 1.1 Definition of Analytical Chemistry  It is a branch of chemistry which deals with:  theoretical foundation; and  method of chemical analysis.  Steps to follow in analytical chemistry:  Separation;  identification; and  determination of the relative amounts of the component(s) in a sample. 2 1.2 Role of Analytical Chemistry  Analytical chemistry has played a vital role in the development of science.  It plays a vital role in many research areas in  Pharmacy  Veterinary pharmacy  Biochemistry  Biology  Geology  Environmental sciences  Medical science and other sciences.  Many chemists and biochemists devote a significant part of their time in the laboratory to gathering quantitative information about systems of interest to them.  In this regard analytical chemistry serves as an important tool for the scholarly efforts of such investigators. 3 4 1.3 Classification of Analytical Chemistry  Hence, analytical chemistry is divided in to two categories: Qualitative analysis: reveals the chemical identity of the species in the sample- identification. It also includes separation. Example: NaCl (aq) sample gives yellow color by flame test or up on ignition. 2. Quantitative analysis: deals with the relative amount of one or more of these species, or analytes, in numerical terms, i.e., separation and determination. N.B Qualitative information is required before a quantitative chemical analysis 1. 5 Classifying Quantitative analytical methods • Gravimetric Methods: Determine the mass of the analyte or some compound chemically related to it. • Volumetric Methods: The volume of a solution containing sufficient reagent to react completely with the analyte is measured. • Spectroscopic Methods: Based on measurement of the interaction between electromagnetic radiation and analyte atoms or molecules. • Chromatographic Methods: Separation techniques and quantitation is based on calibration curve. • Electro-analytical Methods: Involve the measurement of electrical properties such as voltage, current, resistance and quantity of electrical charge. 6 1.4 Methods of Chemical Analysis  There are three methods, namely: 1. 1. 2. Chemical method Physical method Physico-chemical method Classical Method Instrumental methods Classical method  It is a chemical method, and involves chemical change- Analy. Rxn. Analyte /unknown solution + Reagent Compound  The element or ion to be determined is converted to a compound having a specific property. 7 Classical Method cont’d…  Examples: CaCl2 is a salt which is soluble in water CaCl2 (aq) + H2SO4 CaSO4 (ppt ) NH4+ (aq) + OHNH3 (gas ) Pb2+ + CrO42- PbCrO4 yellow ppt  Generally, the following are changes to easily recognize  Formation of precipitate (ppt)  Changes in color  Evolution of gas  Generation of sound  The substance responsible for the change is called reagent. 8 Instrumental Methods  It includes both: Physical method 2. Physico-chemical method  The analysis is done with the help of instruments.  It is a chemical method, and involves chemical reaction 1. Leaf Sample Chromatographic Separation Green chlorophyll Red Carotid Yellow Xanthophyll  In this case, classical method (column chromatography) takes 3 hrs; however, instrumental method (HPLC) takes about 20 minutes to accomplish the separation. 9 Analytical Methods • Classical Methods: Wet chemical methods such as precipitation, extraction, distillation, boiling or melting points, gravimetric and titrimetric measurements. • Instrumental Methods: Analytical measurements (conductivity, electrode potential, light absorption or emission, mass-to-charge ratio, fluorescence etc.) are made using instrumentation. Methods of Analysis cont’d…  Limitations of classical method  Sensitivity  Speed of analysis  Scale of operation o There are four scales of operation such as macro, micro, semimicro and ultra-micro analysis. o Macro analysis   large amount of sample is used Reaction is carried out using classical tools (test tubes, beaker, etc…) and procedures (filtration, decantation, etc…). o Micro analysis  It is 1/100th of the quantity involved in macro analysis  Reaction is carried out by micro crystalloscopic or by drop method. 11 Scale of operation cont’d… o Semi-micro analysis    It uses 1/20th - 1/25th of the quantity involved in macro analysis Reaction is carried out using classical tools. It retains the principle of macro analysis, but uses small amount of substance. o Ultra-micro analysis   In this case, ‹ 1 mg of sample is used Carried out under the microscope- preferably handled by instruments  So, which one of the above analysis is highly sensitive?  Ultra-microscopic analysis is highly sensitive as it requires very small amount of sample.  The most advantageous analysis is semi-micro analysis because it:  uses classical tools and procedures;  reduces consumption of chemicals;  uses cheap instruments** (main reason); and  Saves time (greater speed of analysis). 12 Methods involved in analytical reaction  Analytical reactions can be carried out into two ways:  Dry method and Wet method o Dry Method     The unknown substance and reagent are in solid state-no formation of solution. Reaction is performed by heating, e.g., flame test The time required to carry out the experiment is less- ‹ 1 minute o Wet method  Both the analyte and reagent have to be in a solution form (liquid state)  Involves conversion of a solid sample into solution form Solid sample   solvent Solution Solvent used can be water, acid, base, etc… Example: CuO is insoluble in water 13 Wet Method cont’d… Fe(OH)3 Redish brown CuO HCl FeCl3 H2SO4 CuSO4 (aq) o Mostly we use the wet method in laboratory 14 Features of wet Method Wet analysis uses electrolytes 2. Wet analysis is carried out using classical tools 3. Sample composition is determined in the form of ions not in the form of elements Example: 1. HCl KClO3 o o o H+ K+ + Cl- + ClO3- Both of the samples contain the element chlorine, but HCl contains chloride ion and KClO3 contains ClO3- ion. HCl + AgNO3 AgCl (white ppt) indicates the presence of ClKClO3 + AgNO3 No reaction 15 Features of wet Method cont’d… 4. If an element forms different types of oxidation state, each ions has its own characteristic properties Example: Fe2+ Fe2+ Acidic sol/H+ Basic sol/OH- Fe(OH)2 Dark green ppt Fe(OH)3 Reddish brown ppt  Reaction conditions: three major reaction conditions o o o o o Suitable media Temperature Concentration Sensitivity of a reaction Interfering effects 16 Types of Analysis Fractional analysis: direct detection of the intended ion in the presence of others 2. Systematic analysis: definite sequence of series for the analysis of the intended ion. Step-by-step detection. Example: 1. Na+, K+, NH4+ NH3 OH- ppt Hg2+ G1 solution ppt Pb2+ Clppt heat solution NH3 ppt solution ppt Ag+ 17 1.5 Steps in Quantitative Chemical Analysis 1. Selecting a Method of Analysis  A vital first step in quantitative analysis is the selection of a method.  factors that need to be considered in the selection process are:  the level of accuracy required.  Cost of analysis (total investment)  number of samples to be analyzed.  Time required for analysis  Skill required  complexity of the sample as well as the number of components in the sample. 18 2. Sampling  The sample whose composition represents that of the bulk of material from which it was taken.  Where the bulk is large and heterogeneous, great effort is required to get a representative sample.  Proper sampling is the most difficult step and the source of greatest error 3. Preparing a Laboratory Sample A solid laboratory sample is  ground to decrease particle size,  Mixed to assure homogeneity, and  stored for various lengths of time before the analysis begins. If water is absorbed or adsorbed during each of these steps, depending upon the humidity of the environment, any loss or gain of water changes the chemical composition of solids, :. it is a good idea to dry samples just before starting an analysis. Alternatively, the moisture content of the sample can be determined at the time of analysis in a separate analytical procedure. 19 4. Defining Replicate Samples  Most chemical analysis are performed on replicate samples whose masses or volumes have been determined by careful measurements with an analytical balance or with a precise volumetric device. Replication improves the quality of results and provides a measure of their reliability. 5. Preparing Solutions of the Samples  Most analyses are performed on solution of the sample.  the solvent should dissolve the entire sample rapidly and completely.  The conditions of dissolutions should be sufficiently mild that loss of     the analyte cannot occur. Unfortunately, many materials that must be analyzed are insoluble in common solvent. Examples include silicate minerals, high-molecular-weight polymers, or specimens of an animal tissue. Conversion of the analyte in such materials into a soluble form can be difficult and time-consuming task. It involves heating the sample with aqueous solutions of strong acids, strong bases, oxidizing agents, reducing agents, or some combination of such reagents; ignition of the sample in air or oxygen; or hightemperature fusion of the sample in the presence of various fluxes. 20 6. Eliminating Interferences  Species other than the analyte that affect the final measurement are called interferences.  the analyte should be eliminated from interferences before the final measurement is made. 7. Calibration and Measurement • All analytical results depend on a final measurement X of a physical property of the analyte. This property must vary in a known and reproducible way with the concentration CA of the analyte. Ideally, the measurement of the physical property is directly proportional to the concentration. That is, CA = k X  The process of determining k is thus an important step in most analyses and is termed a calibration. 21 8. Calculating Results • Computing analyte concentrations from experimental data is ordinarily a simple and straightforward task, particularly with modern calculators or computers.  Such computations are based on the raw experimental data collected in the measurement step, the stoichiometry of the chemical reaction up on which the analysis is based, and instrumental factors. 9. Evaluating Results from Estimating Their Reliability • Analytical results are incomplete without an estimate of their reliability. • The experimenter must provide some measure of the uncertainties associated with computed results if the data are to have any value. 22 23 UNIT 2 IONIC EQUILIBRIA 2.1 Chemical Equilibrium  Chemical reactions are of two types: irreversible and reversible reactions.  A reversible reaction proceed in two opposite directions, and eventually reaches a point of balance, equilibrium state.  Equilibrium principles: law of mass action, law of chemical equilibria and Le Chatelier’s principle.  Chemical equilibrium is reaction at equilibrium state.  It is a dynamic state of a reversible chemical reaction.  At equilibrium, rate of forward reaction is equal to rate of backward reaction. 24 Cont’d… 2.2 Factors influencing chemical equilibrium  The value of the equilibrium constant is independent of concentration.  If the concentration of the reactants/products in an equilibrium mixture changes, the position of the equilibrium may change, keeping the equilibrium constant value constant.  However, it is affected by temperature changes.  The shift in equilibrium positions by external factors such as T, P, or C can be described using Le Chatelier’s principle. 25 Cont’d… 2.3 Acid-Base Equilibrium 2.3.1 Theories of Acids and Bases a/ The Arrhenius Theory • An acid is a substance that contains ionizable hydrogen atom/s and produces H+ in aqueous solution. HA + H2O  A- + H3O+  A base is a substance that contains the OH (hydroxyl) group and produces hydroxide ions, OH- in aqueous solution. BOH  B+ + OHHCN + H2O  H3O+ + CN NaOH  Na+ + OH26 Acid-Base Equilibria Usefulness of Arrhenius Concept  It can explain a number of phenomena like neutralization, salt hydrolysis, strength of acids and bases, etc…  However, NH3 is known to be basic, but does not contain any hydroxide. Similarly, CO2 is known to be acidic, but does not contain a proton. Limitations of Arrhenius Concept  However, it was found to have the following shortcomings. 1. 2. Inability to explain acidic and basic character of certain substances, but this limitation can be overcome if water is supposed to play a role. Inability to explain the reaction between an acid and base in absence of water. NH3 (g) + HCl (g) NH4Cl (s) 27 b/ The Bronsted–Lowry Theory • Acids are proton (H+) donors, or any hydrogen- containing molecule or ion capable of releasing a proton, H+ Ex. HCl, HNO3, H2SO4, HCOOH, HCN  Bases are proton(H+) acceptors or any molecule or ion that can accept a proton Acid  H+ + Base Base + H+  Acid Ex. CH3COOH(aq) + NH3(aq)  CH3COO–(aq) + NH4+(aq) acid base 3. HCO3- + H2O  CO32- + H3O+ Acid Base 28 Conjugate acid-base pairs Conjugate base is a substance formed when an acid loses a proton conjugate acid is the one that is formed when the base accepts proton. Acid1 + Base2  C. Base1 + C. Acid2 Example. HNO2+ H2O  NO2- + H3O+ Acid1 base2 base1 acid2 HNO2 (acid1) and NO2- (base1) are one conjugate acid–base pair, H2O (base2) and H3O+ (acid2) are the other conjugate acid– base pair. The stronger the acid, the weaker is its conjugate base; the weaker the acid, the stronger is its conjugate base. 29 Amphoteric/Amphiprotic substances are substances which possess both acidic and basic properties, i.e., act as acid in the presence of base and base in the presence of acid. Example. H2O, NH3, CH3OH, HCO3-, HS-, H2PO4-, HPO42-, etc. HCl + H2O  Cl- + H3O+ Acid Base NH3 + H2O  NH4+ + OHBase Acid NH3 + OH- NH2- + H2O acid base 30 The Bronsted–Lowry Theory Note that Advantages of Bronsted–Lowry concept:  Is not restricted to molecules but includes even ionic species to act as acids/bases  Explains the basic character of the substances like Na2CO3 and NH3 which do not have hydroxide ion.  Explain acid-base reactions in the non-aqueous medium even in the absence of a solvent HCl and NH3 Limitations of Bronsted–Lowry concept:  It can not explain the reactions between acidic oxides and basic oxides  Substances like BF3, AlCl3 do not have any hydrogen and hence can not give a proton but are known to behave as acids 31 c/ Lewis Theory Acid is a substance that can accept an electron pair Abase is a substance that can donate an electron pair Lewis Base can be  Molecules containing an atom with lone pair of electron(s)  All atoms attain stable electronic configuration  It may contain substances which contain electronegative atoms- O, N, etc.  It includes anions Lewis acid can be Molecules having a central atom with an incomplete octet or containing multiple bonds. 32  AlCl3 + NH3  H3N-AlCl3 L. acid L. base SO3 + O2-  SO42L. acid L. base Molecules with a central atom capable of expanding its outer octet by using empty d-orbitals. SnCl4 + 2Cl-  SnCl62L. acid L. base Simple cations Fe3+ + 6H2O  Fe(H2O)63+ Acid Base 33 2.3.2. Dissociation of acids and bases ACIDS: 1/ monoprotic acid 2/ diprotic or 3/ polyprotic acids Monoprotic acids contain only one ionizable hydrogen atom per molecule. HA  H+ + AExample. HCl, HNO3, HClO4, CH3COOH, HF, HCN, etc Diprotic Acids contain two ionizable hydrogen atoms per mole of an acid. H2A  2H+ + AExample. H2S, H2SO4, H2CO3, etc H2S  2H+ + S2Polyprotic contain more than two ionizable hydrogen atoms per mole of an acid. Ex. H3PO4 + H2O  3H+ + PO43i/ Strong acids are acids which dissociate completely in a particular solvent without leaving undissociated acids in a solution. The most common strong acids are HClO4, HCl, HNO3 and H2SO4. HA(aq) + H2O(l)  H3O+(aq) + A–(aq) HCl(aq) + H2O(l)  H3O+(aq) + Cl–(aq) ii/ Weak acids are acids which dissociate partially into their respective ions in solution. A solution of weak acids contains ionized molecules (ions) and unionized molecules 34 HA(l) + H2O(l)  H3O+(aq) + A–(aq) Example CH3COOH, HF, HCN, HNO2, HOOCH, etc HF + H2O F- + H3O+ Strong bases are substances that dissociate completely in a particular solvent, or it has a strong tendency to accept proton. For instance, LiOH, NaOH, Ca(OH)2, CsOH, and Sr(OH)2 B(aq) + H2O(l)  BH+(aq) + OH–(aq) Base(s) acid(s) acid(w) Base(w) Strong bases have weak conjugate acids. Weak Bases are bases which dissociate partially in to their respective ions in aqueous solution. B(aq) + H2O(l)  BH+(aq) + OH–(aq) 35  Dissociation constant, Ka or Kb is used to describe the equilibrium that exists between the weak acid/ base and its conjugate base/ acid in a dilute solution. HA(l) + H2O(l)  H3O+(aq) + A–(aq) The equilibrium constant for the weak acid is called an acid dissociation constant, Ka, and is written as: Ka = [H3O+] [A–] [HA] Polyprotic acids are described by a series of acid dissociation steps, each characterized by its own acid dissociation constant. 36 H3PO4 + H2O  H3O+ + H2PO4Ka1 = [H3O+] [H2PO4-] = 7.11x10-3 [H3PO4] 2/ H2PO4- + H2O  H3O+ + HPO42Ka2 = [H3O+] [HPO42-] = 6.32x10-8 [H2PO4-] 3/ HPO42- + H2O  H3O++ PO43Ka3 = [H3O+] [PO43-] = 4.5x10-13 [HPO42-]  The overall dissociation constant (Ka) is the product of the individual constants. Ka = Ka1 Ka2 Ka3 37 38  Calculate the concentration of each species at equilibrium 0.01 M HNO3 and 0.01 M CH3COOH (Ka =1.8x10-5)  % of ionization = conc. of ionized molecules at eq. x 100 initial conc. of molecules  The concentration of water in dilute aqueous solutions is enormous when compared with hydrogen and hydroxide. Hence, its concentration can be taken as constant.  Remember that the numerical value of equilibrium constant is an indication of how far to the right a reaction proceeds.  The greater the magnitude of the equilibrium constant, the more complete the reaction is. Generally, for dissociation of polyprotic acids the numerical value of Ka1 is larger than that of Ka2, which is larger than Ka3. This is reasonable when one considers that successive protons are removed from species with progressively higher negative charges. 39 pH and dissociation of water  Self ionization/auto ionization H2O + H2O  H3O+ + OH- Keq = ([H3O+] [OH-])/[H2O]2 Keq [H2O]2 = [H3O+] [OH-] KW = [H3O+] [OH-]  Kw, like all equilibrium constants, is temperature dependent. At 25oC, Kw = 1.0 X 10-14  Since the only source of [H3O+] and [OH-] is water, [H3O+] = [OH-] = 1.0 X 10-7  pH + pOH =14  Example: Calculate the concentration of H+ and OH- and pH for: a/ 0.005 M HCl b/ 0.005 M HCN (Ka = 6.2x10–10 ) a/ HCl is a strong acid and completely dissociates HCl → H+ + Cl0.005 M 0.005M 0.005M pH = -log [H3O+] = -log 0.005 M = 2.3 pH + pOH =14 pOH =14 – 2.3= 11.7 40 b/ 0.005 M HCN( Ka = 6.2x10–10 ) HCN ↔ H+ + CN- Initial conc. 0.005 M 0 0 Equil.conc. 0.005 M - x x x Ka = [H3O+] [CN-] = 6.2x10–10 [HCN] 6.2x10–10 = X2 suppose 0.o05-x  0.005 0.005M-X x2 = (6.2x10–10 ) (0.005M) = 0.031x10-10 x = 0.176x10-5 M = [H3O+] = [CN-] pH = -log [H3O+] = -log 0.176x10-5 M = 5.75 [OH-] = Kw/[H3O+] = 5.6x10-10 M pOH = 8.25 41 Calculate the pH of a 0.20 M aqueous NH3 solution. Given that Kb for NH3 = 1.75 x 10-5. The equation for the ionization of aqueous ammonia is NH3 + H2O ↔ NH4+ + OH0.2 0 0 0.2 –x x x Substitution into the ionization constant, Kb, expression gives Kb = [NH4+][OH-] [NH3] 1.75x10-5 = (x)(x) 0.20-x x2 = (1.75x10-5) (0.20) = 3.5x10-6 M x = 1.87x10-3 M Then [OH-] = x = 1.87x10-3 M, pOH = 2.73, and pH = 11.27. 42 2.3.6 Common ion effect  Equilibria can be markedly affected by adding one or more of the species present, as predicted from Le Chatelier’s principle.  It is the effects of decreasing the dissociation of a weak electrolyte by the addition of an ion common to one ion of the electrolyte. This phenomenon is called common ion effect.  Example: HNO2  H+ + NO2 If we add HCl to the solution, it will provide H+ to the ionic equilibrium, and shift the reaction to the left.  If we add NaNO2, it will increase [NO2-] and shift the reaction to the left. Effects  The equilibrium will shift to the left direction.  The concentration of the uncommon ion decreases.  Degree of dissociation of the weak acid decreases. 43 common ion effect on weak bases Example: NH3 + H2O ↔ NH4+ + OH When we add NaOH, it will provide OH- NaOH → Na+ + OH When we add NH4NO3, it will provide NH4+ NH4NO3 ↔ NH4+ + NO3-  Calculate the pH and degree of dissociation of 0.1M HAc, 0.1M HAc + 0.1M HCl, Ka = 1.75 x 10-5  A solution is prepared to be 0.1M HAc and 0.2M sodium acetate, what is the pH of this solution? (Ka = 1.75 x 10-5)  What is the pH of a solution that is 0.1M NH3 and 0.2M NH4Cl? Kb = 1.8 x 10-5 44 a/ Calculate the pH of solution that is 0.01 M in HNO2 and b/the above solution that contains 0.05 M in NaNO2 (Ka = 4.6x10-4) Solution a/The pH of the solution in the absence of common ion effect initial conc. at equilibrium HNO2 + H2O  H3O+ + NO20.01 M 0 0M 0.01 – x x x Ka = [H3O+] [NO2–] = 4.6x10–4 [HNO2] 4.6x10–4 = (x) (x) assume that 0.01–x 0.01 0.01-x 4.6x10–4 = x2 0.01 x2 = 4.6 x10-6 M [H3O+] = [NO2–] = 2.1x10-3 pH = -log 2.1x10-3 = 2.7 45 B. Henderson–Hasselbalch equation are pKa = -log Ka = -log 4.6 X 10-4 = 3.35 NaNO2 → Na+ + NO20.05 0.05 0.05 HNO2 + H2O  H3O+ + NO2initial conc. 0.01 M 0M 0.05 M at equilibrium 0.01 – x x 0.05+x Ka = [H3O+] [NO2–] = 4.6x10–4 [HNO2] 4.6x10–4 = (x) (x+0.05) assume that x+0.05  0.05 and 0.01–x 0.01 0.01-x 4.6x10–4 = 0.05x 0.01 x =0.92 x10-4 M = [H3O+] [NO2–] = x+0.05M = 0.92 x10-4 + 0.05  0.05M [HNO2] = 0.01-x = 0.01- 0.92 x10-4  0.01 pH = pKa + log [NO2-] = 3.35 + log (0.05)/(0.01) = 4.05 [HNO2] in the presence of common ion, NO2– , the pH increases from 2.7 to 4.05 46 2.3.7 Buffer Solutions  A buffer solution is a solution that resists slight changes in pH when H3O+ (a strong acid) or OH- ions (a strong base) are added.  It is a solution of acid-base pair/conjugate A-B. 1. 2. 3. Weak acid and its conjugate base Weak base and its conjugate acid Double salts, e.g., ammonium acetate i/ A buffer solution containing a weak acid and its salt Ex.CH3COOH/CH3COO-, HCN/CN-, HNO2/NO2-, HF/F- 47 Example: Addition of NaCN on HCN HCN  H+ + CNNaCN  Na + + CNThe common ion = CNCN- + H+  HCN a/ addition of a strong acid CN- combines with H3O+(added acid) to form non-ionized HCN and H2O, so long as we do not exceed the amount of buffer reserve. CN- + H3O+  HCN + H2O Because the added H3O+ is consumed by CN- ion; there is very little changes in the pH. b/ addition of a strong base When a strong base is added to the buffer, the OH- is neutralized by HCN, OH- + HCN  CN- + H2O Once again, there is a little change in the pH. 48 HA + H2O ↔ H3O+ + AKa = [H3O+] [A-] [HA] [H3O+] = Ka[HA] [A-] acid  + log [H3O ] = log Ka + log base Multiplying both sides of the equation by -1 gives salt  pH = pKa + log acid  or anion  pH = pKa + log acid  This equation is known as the Henderson–Hasselbalch equation. anion  pH is determined by the value of log acid  because pKa is constant. For example, HF + H2O ↔ H3O+ + FpH = pKa + log [F-] [HF] 49 ii) A buffer solution containing a weak base and its salt B + H2O  BH+ + OHKb = [BH+] [OH–] [B] [OH–] = [B]Kb [BH+] BH   pOH = pKb + log B  Kw = [B]Kb [H+] [BH+] [H+] = [BH+]Kw [B]Kb BH   pH = 14 - pKb - log B or pH = 14 - pKb - log salt  base 50  What is the pH of a solution that contains 0.4M formic acid and 1M sodium formate? Ka = 1.8 x 10-4 pH = pKa + salt  log acid  pH = 4.14 Two important characteristics of buffer 1. 2. pH maintenance Buffer capacity – the number of moles of strong acid/base which is required to change the pH of a solution by one unit. pH ± 1  Example: calculate the buffer capacity of solution which contains 0.1M in HAc and 0.1M in NaAc. 51 2.3.8 Hydrolysis of Salts  Salt hydrolysis is defined as the process in which a salt reacts      with water to give back the acid and the base. Salt + Water → Acid + Base All salts are strong electrolytes and thus ionize completely in the aqueous solution. If the acid (HA) produced is strong and the base (BOH) produced is weak, we can write the equation as: B+ + A- + H2O → H+ + A- + BOH B+ + H2O → H+ + BOH In this case, the cation reacts with water to give an acidic solution – cationic hydrolysis. If the acid produced is weak and the base produced is strong, we can write: B+ + A- + H2O → HA + B+ + OHA- + H2O → OH- + HA Here the anion reacts with water to give basic solution – anionic hydrolysis. 52 Hydrolysis of Salts cont’d…  Therefore, salt hydrolysis may be defined as the reaction of the cation or the anion of the salt with water to produce acidic or basic solution.  Depending upon the relative strengths of the acid and the base produced, the resulting solution may be acidic, basic or neutral. Hence salts are divided into four categories. 1. Salts of strong acids and strong bases  No hydrolysis  The solution is neutral (neither ion of such a salt reacts to disturb the H3O+/ OH- balance in water).  pH = 7 Ex. NaCl, K2SO4 53 2. Salts of weak acids and strong bases The anion hydrolyzed The solution is basic pH is greater than 7 Ex. NaF, NaAc, Na2CO3, Na3PO4    Calculate the pH and percent hydrolysis of 0.10 M solution of sodium cyanide, NaCN. CN- + H2O  HCN + OH0.1 0 0 0.1-x x x Ka of HCN is 6.2x10–10, from the relationship Ka x Kb = Kw, Kb = [HCN] [OH-] [CN-] 1.6x10-5 = (x)(x) 0.1-x x = 1.27x10-3 M = [OH-] pOH = 2.89 and pH = 11.11 % hydrolysis = [HCN] x100 = 1.27x10-3 x100 = 1.27 % which is less than 5%. [CN-] 0.1 Ex.2. What is the pH of 0.025 M NaF 54 3.Salts of strong acids and weak bases  The cation hydrolyzed  The solution is acidic  pH < 7 Ex. NH4Cl, CuSO4, AlCl3, CaCl2 Calculate the pH of 0.03 M NH4Cl. (Kb = 1.75x10-5) Solution NH4+ + H2O  NH3 + H3O+ 0.03-x x x Ka = [NH3][H3O+] = 5.71x10-10 [NH4+] 5.71x10-10 = (x)(x) 0.03-x x2 = (5.71x10-10 ) (0.03) x = 4.13x10-6 M = [H3O+] pH = - log[H3O+] = -log (4.13x10-6 M) = 5.38 55 4. Salts of weak acids and weak bases  Both the cation and anion hydrolyzed  The solution can be basic or acidic  Ex: NH4Ac, (NH4)2CO3, AlPO4 1. If Ka = Kb, neutral Ex. CH3COONH4 2. If Ka> Kb, acidic Ex. NH4F (Ka= 7.2 X 10-4,Kb= 1.8 X 10-5 ) 3. If Ka < Kb, Basic Ex NH4CN 56 Hydrolysis constant  The general equation for the hydrolysis of a salt (BA) may be written as: BA + H2O  HA + BOH Applying the law of chemical equilibrium, we get K = [HA] [BOH] [BA] [H2O] K [H2O]= Kh = [HA] [BOH] [BA] Kh = [free acid] [free base] [unhydrolyzed salt] Degree of hydrolysis is defined as the fraction of the total salt which is hydrolyzed. h = [No of moles of the salt hydrolyzed] [Total No of moles of the salt taken] 57 Hydrolysis constant cont’d…  58 Hydrolysis constant cont’d…  59 2.4 Solubility product principle 4.4.1 Solubility, solubility equilibria and solubility product Heterogeneous state  The equilibrium is between the liquid and solid state. Liquid  solid  A saturated solution is in a dynamic equilibrium with solid electrolyte. MX(slightly soluble)  M+(aq) + X-(aq)  According to solubility product principle, for a saturated solution of slightly soluble electrolytes, the ionic product of two opposing ions is called the solubility product constant, Ksp, i.e, MX(slightly soluble)  M+(aq) + X–(aq) Ksp = [M+] [X–]  The forward reaction is dissociation while the backward reaction refers to precipitation. . 60 Cont’d… Solubility  Solubility is the amount of the solute that is dissolved in a specified volume of solution.  It is usually expressed as either g/L or g/100 mL  The factors that affect solubility includes: temperature, pressure (inversely) and nature of solute. General expressions for slightly soluble substances  For a reaction, MmXn(s)  mM+n(aq) + nXm-(aq) Ksp = [M+n]m[Xm-]n  Solution can be: unsaturated, saturated or supersaturated.  Example: calculate the Ksp for AgCl, if a saturated solution of AgCl contains a. 1 x 10–5 M AgCl b. 1.86 x 10–3 gram of the salt in 1 liter 61  The solubility of slightly soluble electrolyte can be determined in terms of mole/lit in Ksp expression.  Example 1: calculate the solubility of Mg(OH)2. Given Ksp = 6 x10–10 in terms of: a. M b. g/100 mL Example 2. Calculate the solubility of 0.02 M Cu2S (Ksp = 3x10-49) 62 2.4.2 common ion effect and salt effect on solubility  The solubility of a slightly soluble electrolyte decreases when a strong electrolyte with a common ion is added to its solution.  The increase in solubility of a slightly soluble electrolyte by the presence of another electrolyte without any common ion is known as the salt effect. 63 2.4.3 Factors affecting solubility i. Temperature- increase in temperature generally increases the solubility of a substance ii. Nature of solvent Ex Iodine + water → less soluble Iodine + methanol → more soluble NaCl + Water → NaCl + ethanol → iii. Common ion effect – common ion decreases the solubility of a salt Ex. Addition of sodium fluoride on hydrofluoric acid iv. Salt effect(diversion effect)- increases the solubility of salt. Ex,. Addition of sodium sulphate on acetic acid v. Effect of acidity - the solubility of the salt of weak acid depends up on pH of the solution. As a result, the solubility of the salt will increase. Ex. Solubility of AgOOCCH3 on acidic solution (pH greater than 7) vi. Formation of complexes Ex. AgCl  Ag+ + ClAg+ + NH3  Ag(NH3)+ Solubility of AgCl increases 64 Precipitation       Liquid  Solid MX (s)  M+ + XIn a saturated solution: [M+] [X- ] = Ksp In a supersaturated solution: [M+] [X- ] > Ksp In an unsaturated solution: [M+] [X- ] < Ksp Precipitation of solid takes place in a supersaturated solution. The minimum concentrations of the ions that must be present in a solution, for the precipitation of a slightly soluble electrolyte, is the concentrations at which the ion product equals the Ksp. In a qualitative analysis, precipitation of a given ion is regarded as complete when its concentration is not less than 1 x 10–6 M to 1 x 10–5 M. 65 Factors affecting formation of precipitation i. Solubility of the precipitate  Solubility and precipitation are inversely related to one another.  Low solubility and Ksp form a precipitate first. ii. Amount of precipitating agent  Usually a slight excess of precipitating agent is used for complete precipitation.  For appreciably soluble precipitates, more excess of precipitating ions must be added to increase precipitation through common ion effect. However, with large excess of the precipitating ion, the salt effect may set in, and decreasing the extent of the precipitation rather than increasing it. iii. pH  Suitable media is again one determining factor for the formation of precipitate.  pH affects not only extent of precipitation but also the kind of precipitate.  If the media is not suitable, the reactant forms complex, instead of forming the required precipitate 66 2.4.4 Applications of the solubility product principle and precipitation reaction  There are three general methods of qualitative analysis that are based on solubility equilibria and precipitation. i. Fractional precipitation ii. Precipitate dissolution iii. Precipitate inter-conversion Please read their details from reference books!! 67 2.5 Complexation Equilibria 2.5.1 complex ion and ligands Introduction  Complex compound is a distinct chemical species with properties different from its constituents. M + L  ML  Where M is a metal ion-electron pair acceptor and L is a ligandelectron pair donor.  A metal ion is an electron acceptor and occupies the central position.  It is a transition metal ion having a vacant d-orbital. Example: Cu2+, Fe3+, Pt4+  A ligand is an electron donor. It may be anion or neutral molecule that contains lone pair of electrons. Ex: Cl-, H2O,etc  Since the lone pair of electron is donated by one anion or neutral molecule, the bond is called coordinate covalent bond 68 2.5.2 Complex formation equilibria with unidentate and multidentate ligands  Depending on the way how they attached to the Mn+ (metal ion), ligands can be classified as: monodentate and multidentate ligands. 1. Monodentate ligands  A ligand that is attached to the central metal ion through one of its atom. Example: [PtCl6]2-, Pt4+ + 6Cl-  [PtCl6]2 Number of ligands attached at the center is 6. Hence its coordination number is 6.  [Al (H2O)6]3+ ; Al3+ + 6H2O  [Al (H2O)6]3+  Mn+ + xL  [MLX]n-x  Donor atom is the atom in the ligand which forms a bond directly to the central Mn+ Example: oxygen in H2O; nitrogen in NH3 69 Cont’d… 2. Multidentate ligands  A ligand which is attached to the central metal ion through more than one of its donor atom a. Bidentate ligand: two donor atoms, e.g., ethylene diamine b. Tridentate ligand: Diethylene triamine c. Hexadentate ligand: is the most widely used ligand and it has six donor atoms. Example EDTA, 2 nitrogen and 4 oxygen atoms - coordination number 6. Coordination number  The number of donor atoms that are attached with the central metal ion.  Example platinum has a coordination number of 6.  It give us the following information  Structure of the complex  We have Pt = 6 ligands; if L=6 the structure is called octahedral  If L=4, tetrahedral (no lone pair of electrons) or square planer (lone pair)  If we know coordination number, we can write the chemical formula M + nL  MLn 70 2.5.3 Factors affecting stability of complexes  Nature of the metal ion  Charge of the metal ion: the greater the charge on the central metal ion, greater will be the interaction between the central metal ion and the ligands which surround it. As a result, the stability of the complex should increase.  Size of the metal ion: as the size of the central metal ion decreases, the stability of its complex increases. For cations of same charge, then the stability of their complexes (irrespective of the ligand) increases with decrease in the ionic radius of the central metal ion. The greater the charge and the smaller the size of a central metal ion, the greater is the stability of its complexes. Example: Mn2+ < Fe2+ < Co2+ < Ni2+ < Cu2+ (ionic radii decreases)  Class of the metal ion: the stability of a complex also depends upon the class to which the metal belongs. Metal ions may be divided into two types, class “a” and class “b”, depending on the strength of their complexes with certain ligands. Class “a” metals form the most stable complexes with ligands containing N and O donors (ammonia, amines, water, ketones, alcohols) and also with F- and Cl-. Class “b” metals form their most stable complexes with ligands having donor atoms P, S, I , Br, SCN and CN. 71 Cont’d…  Nature of the ligand: the greater the base strength of a ligand (ease of donation of a complex), the greater will be its tendency to form a stable complex. Example Ni2+ form complexes with NH3 and H2O. But NH3 being a stronger base than H2O, forms more stable complex than water. Similarly, CNand CO form stable complexes. For negative ligands the higher the charge and smaller the size, the more stable is the complex formed. Example: stability increases as follows for halides F- > Cl- > Br- > I-  Presence of ring: a chelate is a cyclic complex formed when a cation is bonded by two or more donor groups contained in a single ligand. Example: in copper complex of glycine, copper bonds to both the oxygen of the carboxyl group and the nitrogen of the amine group. Generally, the stability of complex is greatly increased by chelation, i.e., formation of ring structure. Complexes formed by chelating ligands like (en) or EDTA are far more stable than those formed by monodentate ligands like water or ammonia. In this case, the number of chelate rings and the size of the rings further affect the stability of a complex. The larger the number of chelate rings in a complex, the greater is the stability of a complex. Moreover, more chelate rings means higher stability of a complex. 72 Cont’d… 2.5.4 Effect of complexation on solubility Please refer this part from books!! 73 2.6 Oxidation–Reduction (Redox) reactions  It is a chemical process which involves transfer of electrons from one species to another. 4.6.1 Redox Reactions, Reducing and Oxidizing agents Oxidation is an increase in oxidation state  corresponds to the loss, or apparent loss, of electrons. Reduction is a decrease in oxidation state  corresponds to a gain, or apparent gain, of electrons.  Because oxidation and reduction occur simultaneously, they are referred to as oxidation–reduction or redox reactions. 97 Cont’d…  Oxidizing agents (also called an oxidant) are species that  oxidize other substances  contain atoms that are reduced  gain (or appear to gain) electrons  Reducing agents (a reductant) are species that  reduce other substances  contain atoms that are oxidized  lose (or appear to lose) electrons  For example consider the following reaction:  ). 75 Ex. 2[Fe(CN)6]4− + Cl2 → 2[Fe(CN)6]3− + 2Cl−  Reducing agent [Fe(CN)6]4− it donates an electron to chlorine to form chloride  Oxidizing agent chlorine because it converts Fe(II) to Fe(III) 76  2Fe3+(aq) + H2C2O4(aq) + 2H2O(l)  2Fe2+(aq) +2CO2(g) + 2H3O+(aq)  Iron is reduced since its oxidation state changes from +3 to +2. Oxalic acid, on the other hand, is oxidized since the oxidation state for carbon increases from +3 in H2C2O4 to +4 in CO2.  The separate half-reactions that individually describe the oxidation and the reduction processes. H2C2O4(aq) + 2H2O(l) → 2CO2(g) + 2H3O+(aq) + 2eFe3+(aq) + e- → Fe2+(aq) 77     Features of Redox reaction The number of electrons donated by the reducing agent must be equal to the number of electrons accepted by the oxidizing agent Any redox equation can be splitted into two half reaction Any redox reaction involves the formation of a new reducing and oxidizing agent from the oxidant and reductant taken for the reaction It may serve as a source of electric current if it takes place in a galvanic cell. 78 2.6.2 Redox Reactions in electrochemical cells and electrode potential Electrochemistry involves  Electron transfer between physically separated oxidizing and reducing agents  Interconversions between chemical energy and electric energy.  All electrochemical reactions are oxidation–reduction reactions.  An electrochemical cell is composed of two compartments or half-cells, each composed of an electrode dipped in a solution of electrolyte. There are two types of electrochemical cells. 1. Electrolytic cells  are those in which electrical energy from an external source causes non-spontaneous chemical reactions to occur. Electrical energy chemical energy 79  2. Voltaic (Galvanic) cells  are those in which spontaneous chemical reactions produce electricity and supply it to an external circuit.  Chemical rxn → electrical energy (electricity)  Spontaneous reaction occurs  galvanic cells are commonly used as batteries. Electrodes  Electrodes are conducting materials in which electron/s transfer into or from the surrounding solution.  electrodes can be classified as : 80  Cathode is the electrode at which reduction occurs as electrons are gained by some species.  Anode is the electrode at which oxidation occurs as electrons are lost by some species.  Electrode Electrochemical cell Galvanic cell Electrolytic cell Anode negative Positive Cathode positive negative 81  The oxidized and the reduced agents are separated from one another by salt bridge.  A salt bridge containing an inert electrolyte, such as KCl, connects the two half-cells.  The ends of the salt bridge are fixed with porous frits, allowing ions to move freely between the half-cells and the salt bridge. The use of Salt bridge  which prevents the spontaneous mixing of the aqueous solutions in each compartment  allows the migration of ions in both directions to maintain electrical neutrality. This movement of ions in the salt bridge completes the electric circuit. 82 Anode: Zn(s)  Zn2+(aq) + 2e– oxidation the zinc electrode will slowly dissolve to produce zinc ions (Zn+2), which enter into the solution containing Zn+2 (aq) and SO4-2 (aq) ions. The electrode on the right is the cathode, where reduction occurs cathode: Cu2+ (aq) + 2e-  Cu (s) reduction copper atoms accumulate on the surface of the solid copper electrode. The electrochemical cell electrons flow from the anode to the cathode, thus the over all reaction. Zn(s) + Cu2+ (aq)  Cu(s) + Zn2+(aq) 83 Shorthand representation  Anode/solution/cathode  Zn(s) | ZnSO4 (C1) || Cu SO4 (C2) | Cu(s) Anode cathode  (||) indicates the salt bridge, the contents of which are normally not indicated.  A vertical slash (|) indicates a phase boundary where a potential develops.  C1 and C2 represent the concentration of the ionic species at the anode and cathode respectively. 84 The standard hydrogen electrode  Every oxidation must be accompanied by a reduction.  Therefore, it is impossible to determine experimentally the     potential of any single electrode. The conventional reference electrode is the standard hydrogen electrode (SHE). This electrode contains a piece of metal coated with a grainy black surface of inert platinum metal, immersed in a 1 M H+ solution. Hydrogen, H2, is bubbled at 1 atmospheric pressure through a glass envelope over the platinized electrode. By international convention, the standard hydrogen electrode is arbitrarily assigned a potential of exactly 0.0 volt. H2 → 2H+ + 2e- exactly 0.0V (SHE as anode) 2H+ + 2e- → H2 exactly 0.0V (SHE as cathode) 85  By coupling SHE with another electrode it is possible to determine the electrode potential.  Because the defined electrode potential of the SHE contributes exactly 0.0 volt to the sum, the voltage of the overall cell then determines the standard electrode potential of the other half-cell.  This is its potential with respect to the standard hydrogen electrode, measured at 25°C when the concentration of each ion in the solution is 1 M and the pressure of any gas involved is 1 atm.  By agreement, the standard cell potential is for a halfcell as a reduction process. 86 Standard electrode potentials  Electrode potential is a measure of the tendency of the ions to give off or take on electron. Many electrodes involve metals or non-metals in contact with their ions.  negative electrode potential indicates that the electrode has a greater tendency to undergoes oxidation than SHE  positive electrode potential indicates that the electrode has a greater tendency to undergo reduction (the more positive the value, the better the oxidizing agent) than SHE. 87  E°reac = E°red – E°ox = E°cathode – E°anode  where both E°red and E°ox are standard-state reduction potentials. Electrodes with Positive Eoreduction  Reduction occurs more readily than the reduction of 2H+ to H2.  Electrode acts as a cathode versus the SHE.  Electrodes with Negative Eoreduction  Reduction is more difficult than the reduction of 2H+ to H2.  Electrode acts as an anode versus the SHE. 88  The more positive the standard-state reduction potential, the more favorable the reduction reaction  Ex. under standard-state conditions, the reduction of Cu2+ to Cu (E° = +0.34V) is more favorable than the reduction of Zn2+ to Zn (E° = –0.76V).  If E°cell > 0, then the process is spontaneous (galvanic cell)  If E°cell < 0, then the process is non-spontaneous (electrolytic cell). 89 90  Examples:1. Calculate the cell potential for the reaction     Zn(s) /Zn2+(aq) //Cu2+(aq)/ Cu (s) The reaction: Cu2+(aq) + Zn(s)  Cu (s) + Zn2+(aq) The standard reduction potential for each half reaction is Zn 2+(aq) + 2e-  Zn(s) E º = -0.76 V Cu2+(aq) + 2e- Cu (s) E º = 0.34 V From the above electrochemical reaction, zinc - oxidized and copper reduced, therefore E°reac = E°red – E°ox E°red = 0.34 V and E°ox = -0.76 V E°reac =(0.34 V) –(-0.76 V) = 1.1 V Since the potential is positive, the reaction is spontaneous and carried out in galvanic cell. 91 2. Can Zn2+(aq) oxidize Sn under standard state conditions?  The reaction : Zn2+(aq) + Sn(s)  Zn (s) + Sn2+(aq)  The cell can be represented as Sn(s)/Sn2+ // Zn2+/ Zn  The standard potential Zn2+(aq) + 2e-  Zn(s) Eº = -0.76 V Sn2+(aq) + 2e-  Sn(s) Eº = -0.14 V  Since the Sn2+/Sn potential is less negative than Zn2+/Zn, Zn is a better reducing agent than Sn2+.  i.e, Zn2+ will not oxidize Sn cell rather it is in electrolytic cell. This is further confirmed by calculating the cell potential, E°reac = E°red – E°ox = (-0.76 V)-(-0.14 V) = -0.62 V  Since the sign is negative, the reaction is not spontaneous in the direction written.  This reaction will take place in electrolytic cell 92 3. What is the standard EMF of a galvanic cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3 solution? Solution  Standard reduction potential Cr3+(aq) + 3e-  Cr(s) Eº = -0.74 V Cd2+(aq) + 2e-  Cd(s) Eº = -0.40 V  From the given standard potential, the one with more negative (-0.74V) has a higher tendency to oxidize than other in galvanic cell. Therefore, the reaction of the galvanic cell will be 2Cr(s) + 3Cd2+(aq)  2Cr3+(aq) + 3Cd(s) Cr(s)/ Cr3+ // Cd2+ /Cd(s) E°reac = E°red – E°ox = (-0.40 V) – (-0.74 V) = + 0.34 V 93 2.6.3 Dependence of electrode potential on concentration   the general equilibrium redox reaction equation involving A, B, C, and D, with stoichiometric coefficients a, b, c, and d is given as follows: aA + bB  cC + dD  For this reaction, the reaction quotient is Q = [C]c[D]d [A]a[B]b  The relationship between electrochemical potential and the concentrations of reactants and products can be determined by –nFE = –nFE° + RT ln Q where E° is the electrochemical potential under standard-state conditions. Dividing the equation by –nF leads to the well-known Nernst equation.  E = E° - RT ln Q nF = E° - RT ln [C]c[D]d nF [A]a[B]b 94 Cont’d…  Substituting appropriate values for R = 8.314 J mol-1 k-1 and F = 96,485 C mol–1 assuming a temperature of 25 °C (298 K), and switching from ln to log form gives the potential in volts as E = Eo – 0.0592 log Q n  In general, half-reactions for standard reduction potentials are written xOx + ne_  yRed  The Nernst equation for any cathode half-cell (reduction halfreaction) is  E = Eo – 0.0592 log [Red]y n [Ox]x 95 Cont’d… Example: Zn2+ + 2e-  Zn E0 = -0.763 V  the corresponding Nernst equation is E = Eo – 0.0592 log 1 (for reduction half reaction) 2 [Zn2+]  We substitute the E0 value into the equation to obtain  E = -0.763 V - 0.0592 log 1 2 [Zn2+] 96 Cont’d… Example: A zinc electrode is submerged in an acidic 0.80 M Zn2+ solution which is connected by a salt bridge to a 1.30 M Ag+ solution containing a silver electrode. Determine the cell potential for the electrochemical cell Zn + 2Ag+(1.3 M)  Zn2+(0.80 M) + 2Ag at 298K. Solution: The standard reduction potential for the half reactions is: Zn2+ + 2e- → Zn E0red = -0.76 V Ag+ + e- → Ag E0red = +0.80 V Ecell = E0cell - log Q where, Q = [Zn2+]/[Ag+]2 E0cell = E0red - E0ox = 0.80 V – (-0.76 V) = 1.56 V Ecell = 1.56 V – 0.059 log ([Zn2+]/[Ag+]2 ) 2 Ecell = 1.57 V . The reaction proceeds spontaneously (takes place in galvanic cell). 97 2.6.4. Calculating equilibrium constant from electrode potential  The standard-state electrochemical cell potential, E° provides an alternative way of expressing the equilibrium constant for a redox reaction. -RT ln K = -nFEo Eo = RT ln K = 0.059 log K nF n log K = n Ecell = n ( Eored- Eoox) 0.059 0.059 where K is an equilibrium constant that defines the reaction’s equilibrium position. 98 Cont’d…  Example: 1. Cu2+(aq) + Zn(s)  Cu (s) + Zn2+(aq) E Zn2+/Zn = EoZn2+/Zn - 0.059 log 1 ----------------------------(1) 2 [Zn2+] E Cu2+/Cu = Eo Cu2+/Cu - 0.059 log 1 -------------------------(2) 2 [Cu2+] At equilibrium equation (1) and (2) are equal, i.e., EoZn2+/Zn - 0.059 log 1 = Eo Cu2+/Cu - 0.059 log 1 2 [Zn2+] 2 [Cu2+]  Eo Cu2+/Cu - EoZn2+/Zn = 0.059 log [Zn2+] 2 [Cu2+] [Zn2+] = K [Cu2+] log [Zn2+] = 2 (EoCu2+ - EoZn2+ ) [Cu2+] 0.0592  log K = 2 (EoCu2+ - EoZn2+ ) = 2(1.1) = 37.16 0.0592 0.0592 K= 1.4x1037 99 Cont’d… 2. Ni + Cr3+  Ni2+ +Cr Cr3+ +3e-  Cr Eo = -0.49 V Ni2+ + 2e-  Ni Eo = 0.25 V log K = 2x3(EoCr3+ - EoNi2+ ) = 6(-0.49-0.25) = -75 0.0592 0.0592 K = 1075 Example3. Calculate K for the reaction 2Fe3+ + 3I-  2Fe2+ + I3From the standard reduction potential, we find that 2Fe3+ + 2e  2Fe2+ 0.771 V I3- + 2e  3I- 0.536 V   log K = 2(EoFe3+ - EoI3-) = 2 (0.771 – 0.536 ) = 7.939 0.0592 0.059 K = 8.7x107 100 Cont’d… In general for a reaction Aox + ae-  Ared Box +be-  Bred (1) (2) To balance the reaction multiply (1) by b and (2) by a bAred + aBox  bAox + aBred logK = ab(EoB - EoA) 0.0592 101 UNIT 3. STATISTICAL EVALUATION OF ANALYTICAL DATA 3.1. Mean, Median, Standard deviation and Variance  Two common ways of estimating central tendency are the mean and the median. Mean  Mean is the numerical average obtained by dividing the sum of the individual measurements by the number of measurements  in1 X i (3.1) X n where Xi is the ith measurement, and n is the number of replicate measurements 102 Standard deviation and Variance  Two common measures of spread are standard deviation and variance.  Standard deviation (s), describes the spread of individual measurements about the mean and is given as s  n i 1 Xi  X  2 n 1 Relative standard deviation, sr, : s sr  x 103  Variance is another common measure of spread, which is the square of the standard deviation.  The standard deviation, rather than the variance is usually used because of the unit for standard deviation is the same as that for the mean value.  Variance = s2 104  Example  Consider the following results that are obtained for seven replicate measurements for the mass of iron in a sample.  Measurement 1 Mass (g) 2 3 3.080 3.094 3.107 4 5 6 7 3.056 3.112 3.174 3.198 Calculate the a. mean b. standard deviation c. % relative standard deviation d. variance. Solution a) To calculate the mean, we add the results for all measurements together as 3.080 + 3.094 + 3.107 + 3.056 + 3.112 + 3.174 + 3.198 = 21.821 and divide by the number of measurements = 21.821/7 = 3.117 g 105  b) To calculate the standard deviation, we obtain the difference between the mean value and each measurement, square the resulting differences, and add them to determine the sum of the squares. (3.080 – 3.117)2 = (–0.037)2 = 0.00137 (3.094 – 3.117)2 = (–0.023)2 = 0.00053 (3.107 – 3.117)2 = (–0.010)2 = 0.00010 (3.056 – 3.117)2 = (–0.061)2 = 0.00372 (3.112 – 3.117)2 = (–0.005)2 = 0.00003 (3.174 – 3.117)2 = (+0.057)2 = 0.00325 (3.198 – 3.117)2 = (+0.081)2 = 0.00656 0.01556 106  The standard deviation is calculated by dividing the sum of the squares by n – 1, where n is the number of measurements, and taking the square root. s = 0.051 c) The percent relative standard deviation is sr = 0 051/3.117 = 0 016 sr (%) = 0.016 × 100% = 1.6% d) The variance is just the square of the absolute standard deviation. Variance = s2 = (0.051)2 = 0.0026 107 3.2. Accuracy and Precision of Measurements Accuracy Accuracy is a measure of how close a measured value (X )with the true, or expected value, µ. Accuracy is usually expressed as either an absolute error E=X-µ or a percent relative error, Er, as Er = X   x 100  Precision  Is a measure of the degree of agreement among measured values  Precision is a measure of the spread of data about a central value  may be expressed as the range, the standard deviation, or the variance. 108 Precision  Precision is commonly divided into two categories: repeatability and reproducibility. a/ Repeatability is the precision obtained when all measurements are made by the same analyst during a single period of laboratory work, using the same solutions and equipment. b/Reproducibility is the precision obtained under any other set of conditions, including that between analysts, or between laboratory sessions for a single analyst.  Since reproducibility includes additional sources of variability, the reproducibility of an analysis can be no better than its repeatability. 109 3.3. Errors in Analytical Results  Measurement errors are classified in to two as a) Systematic error (determinate error)  Errors affecting the accuracy of an analysis are called determinate and are characterized by a systematic deviation from the true value;  that is, all the individual measurements are either too large or too small.  A positive determinate error results in a central value that is larger than the true value,  a negative determinate error leads to a central value that is smaller than the true value.  Determinate errors may be divided into four categories: sampling errors, measurement errors, method errors, and personal errors. 110  1. Sampling errors is an error which occurs when our sampling strategy fails to provide a representative sample.  This is especially important when sampling heterogeneous materials.  Determinate errors associated with selecting a sample can be minimized with a proper sampling strategy. 2. Instrumental error (Measurement error) - is an error due to limitations in the equipment and instruments used to make measurements.  Analytical instruments and equipments, such as glassware and balances, are usually supplied by the manufacturer together with a statement of the instruments and equipments maximum measurement error, or tolerance. 111  This type of error can be minimized by calibration.  Many instruments, in particular, drift out of calibration over time. This complication can be minimized by frequent recalibration. 3. personal error is an error which can include the ability to see a change in the color of an indicator used to signal the end point of a titration; biases, such as consistently overestimating or underestimating the value on an instrument’s readout scale; failing to calibrate glassware and instrumentation; and misinterpreting procedural directions. Personal errors can be minimized with proper care 112 4. Methodic error is an error due to limitations in the analytical method used to analyze a sample, and occur when the assumption about the relationship between the sample response and the amount of analyte present in the sample are invalid. 113 b) Random error (indeterminate error)  Errors affecting the distribution of measurements around a central value are called indeterminate and are characterized by a random variation in both magnitude and direction.  Indeterminate errors need not affect the accuracy of an analysis.  Since indeterminate errors are randomly scattered around a central value, positive and negative errors tend to cancel, provided that enough measurements are made. In such situations the mean or median is largely unaffected by the precision of the analysis. 114  Although it is impossible to eliminate indeterminate error, its effect can be minimized if the sources and relative magnitudes of the indeterminate error are known.  Indeterminate errors may be estimated by an appropriate measure of spread. Typically, a standard deviation is used, although in some cases estimated values are used. The contribution from analytical instruments and equipment are easily measured or estimated. 115 3.4. Confidence Limit Confidence Limit for Populations Confidence limit, confidence interval, is the range of results around a mean value that could be explained by random error. In general, an equation which used to determine the confidence limit for a single member of a population Xi = µ ± zσ  where the factor z accounts for the desired level of confidence and σ is the population standard deviation. Values reported in this fashion are called confidence limit or confidence intervals. Confidence intervals can be quoted for any desired probability level, several examples of which are shown in Table 3.1. 116 Table 3.1. Confidence Intervals for different probability levels z 0.50 1.00 1.50 1.96 2.00 2.50 3.00 3.50 probability level (%) 38.30 68.26 86.64 95.00 95.44 98.76 99.73 99.95 117 Confidence Intervals for Samples What is confidence interval? Earlier we introduced the confidence interval as a way to report the most probable value for a population’s mean, µ, when the population standard deviation, σ, is known. It is also possible to construct confidence intervals for samples by replacing σ in equation given above with sample standard deviation, s, and the term z in equation by the variable t, and with some modification as 118  Values for t at the 95% confidence level are shown in Table 3.2. Note that t becomes smaller as the number of the samples (or degrees of freedom) increase, approaching z as n approaches infinity X ts n 119 = Table 3.2. Values of t for the 95% probability level Degrees Freedom Values of t of 1 2.71 2 3 4 5 6 7 8 9 10 11 4.30 3.18 2.78 2.57 2.45 2.36 2.31 2.26 2.23 2.18 Example: What is the 95% confidence interval for the masses of Seven United States Pennies in circulation? Penny 1 2 3 4 5 6 7 Mass (g) 3.080 3.094 3.107 3.056 3.112 3.174 3.198 ts X n  2.450.051 3.117  7 = 3.117±0.047 g 120 3.5. Testing for Significance (t-test and F-test)  A significance test is designed to determine whether the difference between two or more values is too large to be explained by indeterminate error  1. t-testComparing sample mean to population mean One approach for validating a new analytical method is to analyze a standard sample containing a known amount of analyte, µ. The method’s accuracy is judged by determining the average amount of analyte in several samples and using a significance test to compare it with µ. 121  The equation for the test (experimental) statistic, texp, is derived from the confidence interval for µ X ts n Rearranging this equation texp = X x n s The value of texp is then compared with a critical value, tcrit, which is obtained from tables for a given degree of probability and number of degrees of freedom n-1. a/ If texp is greater than tcrit the difference between mean(X ) and µ is significant and cannot be explained by indeterminate sources of error. There is evidence, therefore, that the results are affected by a determinate source of error. 122 b/ If texp is less than or equal to tcrit  then the confidence interval for the data could be attributed to indeterminate error.  There is no significant difference b/n mean and µ Example 1. Before determining the amount of Na2CO3 in an unknown sample, a student decides to check her procedure by analyzing a sample known to contain 98.76% w/w Na2CO3. Five replicate determinations of the %w/w Na2CO3 in the standard were made with the following results 98.71% 98.59% 98.62% 98.44% 98.58%. Is the mean for these five trials significantly different from the accepted value at the 95% confidence level? 123  texp =   X x n s = 98.76  98.59 0.0973 x 5 = 3.91 The critical value for tcrit for 95% confidence level and 4 degrees of freedom is 2.78. Since texp is greater than tcrit at the 95% confidence level the difference between the measured mean and µ is significant and cannot be explained by indeterminate sources of error. There is evidence, therefore, that the results are affected by a determinate source of error. 124 2. F-test -Comparing Two Sample Variances  The F-test is used for the comparison of variances for two samples, A and B, by the equation; Fexp  SA SB 2 2 where A and B are defined such that s2A is greater than or equal to s2B . If the Fexp< Fcrit. A critical value, Fcrit gives the largest value of F that can be explained by indeterminate error. There is no significant difference 125 Example Tow different methods A and B were applied for the analysis of sulfur in a drug sample. Method A provides results with s = 0.051 for seven analysis and method B provides results with s = 0.037 for five analysis. Determine whether there is a difference in the precisions of the two methods at 95% probability level. Solution We find that the variances are s2A = 0.00259 and s2B = 0.00138. And the test statistic is Fexp  SA SB 2 2 = 0.0026 0.00137 = 1.89 126  The critical value for, Fcrit, is 9.197. Since Fexp is less than Fcrit  There is no significance difference level to suggest that the difference in precisions is significant. 127 3.6. Rejection test (Q-test)  On occasion, a data set appears to be skewed by the presence of one or more data points that are not consistent with the remaining data points.  Such values are called outliers.  The most commonly used significance test for identifying outliers is Q-test.  The Q-test compares the difference between the suspected outlier and its nearest numerical neighbor to the range of the entire data set.  Data are ranked from smallest to largest so that the suspected outlier is either the first or the last data point. The test statistic, Qexp, is calculated using following equation if the suspected outlier is the smallest value (X1); 128  Qexp = X2 - X1 Xn – X1 or if the suspected outlier is the largest value (Xn) using the equation; Qexp = Xn – Xn-1 Xn – X1 If Qexp is greater than Qcrit, then the outlier may be rejected. When Qexp is less than or equal to Qcrit the suspected outlier must be retained. Qcrit is obtained from tables for a given degree of probability and number of measurements n. 129  Example. The following masses, in grams, were recorded in an experiment to determine the average mass of a U.S. penny. 3.067 3.049 3.039 2.514 3.048 3.079 3.094 3.109 3.102 Determine if the value of 2.514 g is an outlier at = 0.05 (95 %). Solution To begin with, place the masses in order from smallest to largest 2.514 3.039 3.048 3.049 3.067 3.079 3.094 3.102 3.109 and calculate Qexp Qexp = X 2  X 1 3.039  2.514   0.882 = 3.109  2.514 X 9  X1 The critical value for Q(0.05, 9) is 0.493. Since Qexp > Q(0.05, 9) the value is assumed to be an outlier, and can be rejected 130 The critical value for Q(0.05, 9) is 0.493. Since Qexp > Q(0.05, 9) the value is assumed to be an outlier, and can be rejected. 3.039 3.048 3.049 3.067 3.079 3.094 3.102 3.109 Is 3.039 rejected? Qexp = X2 - X1 Xn – X1 = 3.048 – 3.039 = 0.1286 3.109 – 3.039 Qexp < Qcrit, 0.1286 < 0.526 3.039 couldn’t rejected Is 3.109 retained? Qexp > Qcrit 0.1 < 0.526 it should not be rejected 131 132 133 UNIT 4. SOLUTIONS AND THEIR CONCENTRATIONS Solution  Is a homogeneous mixture.  consists of solute and solvent.  Solute + solvent solution Solute the one which is dissolved Solvent the one which dissolves the solute solid-liquid solution Solvent is the major component of a solution solute the minor component of the solution. liquid-liquid solution Ex. water (50 mL) + ethanol(60ml) solution solute solvent water (60 mL) + ethanol(50ml) solution solvent solute 134 4.1 Types of solutions 1. Gaseous solution- includes gases or vapors dissolved in one another. Ex. Air. 2. Liquid solution- contains a liquid solvent in which gas, liquid, or solid is dissolved. Ex. Salt solution. 3. Solid solution- is a mixture of solids spread equally throughout one another. Ex alloys Depending on the amount of solute in relative to the amount of solvent in the solution, a solution can be further classified as: Dilute solution: a solution that contains a small proportion of solute relative to solvent, or Concentrated solution: is a solution that contains a large proportion of solute relative to solvent. 135 Dilution is the process of converting concentrated solution in to diluted solution During dilution Volume of solvent increases Amount of solute added remains constant Moles constant Concentration decreases Mcon x Vcon = Mdil x Vdil 136 1. A chemist starts with 50.0 mL of a 0.40 M NaCl solution and dilutes it to 1000 mL. What is the concentration of NaCl in the new solution? (Mdil) (Vdil) = (Mcon) (Vcon) (Mdil) (1000 mL) = (0.40 M) (50.0 mL) Mdil = (0.40 M) (50.0 mL) = 0.02 M (1000 mL) 2. A chemist wants to make 500 mL of 0.050 M HCl by diluting a 6.0 M HCl solution. How much of that solution should be used? Vcon = (Mdil) (Vdil) Mcon Vcon = 0.050 M x 500 mL = 4.2 mL 6M 137 4.2 Different ways of expressing concentrations  relative and absolute way of expressing concentration Concentration = amount of solute amount of solution The most common units of concentrations are as follows: 1. Molarity (M) The molarity of a solution is defined as the number of moles of solute per one liter of solution. Molarity = moles of solute Volume of solution in liters Moles of solute = actual mass of solute molecular weight solute actual mass of solute (g) Molarity = molecular weight of solute (g/mol) Volume of solution in liters 138  Example: 1. A 2 molar solution of CdBr2 would contain 2 moles of CdBr2 in each liter of solution. 2. If 24.5 grams of H2SO4 are dissolved in water to give 2000 ml of solution, calculate its molarity. actual mass of solute (g) Molarity = molecular weight solute (g/mol) Volume of solution in liters M = 24.5 g / 98 g/mol = 0.125 M 2L 3. How many ml of a 1.5 M NaBr solution would be required to give 0.5 moles of NaBr? M = no.of moles/ volume Volume = moles / M = 0.5 moles / 1.5 moles= 0.333 L = 333.3 ml 139 2. Normality, N Normality is the number of equivalents of solute per liter of solution (N). A "Normal" solution (1 N) =1 gram equivalent/liter of solution. N = no. of equivalents/Volume of sol.in liter no. of equivalents = actual wt/ Equivalent weight Equivalent weight (EW)=molecular weight of solute(g) n N = Weight of solute (g) equi. weight of solute(g) × Volume of soln(in L) 140 Determining the equivalent weight of a solute i/ neutralization reaction Example: HCl  H+ + Cln = 1 EW = 36.5/1 = 36.5 H2SO4  2H+ + SO42- n = 2 EW = 98/2 = 49 H3PO4  3H+ + PO43n = 3 EW = 98/3 = 32.67 NaOH  Na+ + OHn = 1 EW = 40/1 = 40 Ca(OH)2  Ca2+ + 2OHn = 2 EW = 74/ 2 = 37 ii/ Redox reactions The equivalent weight of a substance undergoing oxidation or reduction is equal to its molecular weight divided by the total number of electrons gained or lost per molecule in that particular reaction. Equivalent weight (EW) = molecular weight of solute(g) No. of e- gained or lost 1. Fe3+ + e-  Fe2+ 141 Equivalent weight of Fe = 58.85 = 58.85 g/eq  2. Sn4+ + 2eSn2+ Equivalent weight Sn = 118.69 = 59.34 g/eq 2 3. Potassium Dichromate (K2Cr2O7) Cr2O72-+ 14H+ + 6e-  2Cr3+ + 7H2O Equivalent weight K2Cr2O7 = 294.19 = 49.03 g/eq 6 iii/ dissolution reactions The equivalent weight of a substance undergoing dissolution reactions is equal to its molecular weight divided by the total number of moles of ions furnished by the substance. AlCl3  Al3+ + 3ClEW = 132.9 = 44.3 g/eq 3 2+ Pb(NO3)2  Pb + 2NO3 EW = 331 = 165.5 g/eq 2 142 1. Calculate the normality of a sodium chloride solution prepared by dissolving 2.92 grams of NaCl in 500.0 ml of water solution. Solution: Given: Weight of solute (NaCl) =2.92 g Equivalent weight = 58.5 = 58.5 g/eq 1 Volume of solution = 500 mL = 0.5 L Required: N = ? N = Weight of solute (g) equivalent weight of solute(g) × Volume of soln(in L) N = 2.92 g/ 58.5 g/eq = 0.099 N 0.5 L 143 2. How many grams of sodium sulphate (Na2SO4) are necessary to prepare 200 mL of 0.5 N solutions for precipitation reaction? Solution: Given: Normality = 0.5 N Molecular wt = 142 Equivalent weight = 142 /2 = 71 g/eq Volume of solution = 200 mL = 0.2 L Required: Weight of solute (NaCl) =? N = Weight of solute (g) equivalent weight of solute(g) × Volume of solution(in L) Weight of solute (g) = N x equivalent weight of solute (g) × Volume of solution (in L) = 0.5 eq/L x 71 g/eq x 0.2 L = 7.1 g 3. How many mL of a 0.825 N HNO3 solution are needed to supply 15.7 grams of HNO3? Solution: Given: Normality = 0.825 N, Weight of solute (HNO3) = 15.7 g Molecular wt = 63 g, Equivalent weight = 63 /1 = 63 g/eq Required: Volume of solution =? ans 0.816 mL 144 3. Molality (m) Molality (m) is the number of moles of solute in one kilogram of solvent. Molality = no. of moles Mass of solvent (Kg) No. of moles = mass of solute/ molecular weight actual mass of solute (g) Molality = molecular weight solute (g/mol) Mass of solvent (Kg) 145 Example.1. what is the molality of a solution made by dissolving 23.2 g of sodium nitrate in 250.0 g of water? Given: Weight of solute (NaNO3) = 23.2 g Molecular wt = 85 g/ mol Mass of solvent (water) = 250.0 g = 0.25 Kg Required; m=? ; moles = 23.2 g = 0.273 mol m = no. of moles Mass of solvent (Kg) 85 g/ mol m = 0.273 mol = 1.09 m 0.25 Kg Ex.2. How many grams of silver bromide must be dissolved in 725 grams of water to make a 0.220 m(molal) solution? Answer: 29.99 g 146 4. Percent expression a/ Weight percent (w/w) - is the grams of solute per 100 grams of solution (% w/w). % (w/w) = wt of solute x 100 Wt of solution = wt of solute + wt of solvent Wt of solution Example: 1. 20 g of sodium chloride in 100 g of solution is a 20% by mass solution. 2. Calculate the % by mass solution of ethyl alcohol in water contains 10 g of ethyl alcohol and 80 g of water. Solution: wt of solute (ethyl alcohol) = 10 g wt of solvent (water) = 80 g % (w/w) = wt of solute x 100 = 10 g x 100 = 11.11 % Wt of solution 90 147 3. How many grams of a 10 % AgNO3 solution must be weighed out to have 20 grams of AgNO3? Ans: 200 g solution AgNO3 b/ weight-to-volume percent (w/v). Grams of solute per 100 mL of solution (% w/v). % (w/v) = mass of solute x 100 Volume of solvent Example. What is the percent of 10 g of ethyl alcohol in 500 mL of solution? % (w/v) = mass of solute x 100 = 10 g x 100 = 2 g/ mL = 2000 g/L Volume of solvent 500 mL Trace concentrations are usually given in smaller units, such as parts per million (ppm) and parts per billion (ppb). Parts per million (ppm): is the number of grams of solute in a million (106) grams of solution. Parts per billion (ppb): is the number of grams of solute in a billion (109) grams of solution. 1 ppm = wt of solute x 106 = µg/g = mg/kg Wt of solution 1 ppm = 1mg/L = 1µg /mL 1 ppb = wt of solute x 109 = ng/g = µg /kg wt of solution 148 Example: 1. A 2.6 g sample of plant tissue was analyzed and found to contain 3.6 µg Zn. What is the concentration of Zn in the plant in ppm? ppb? Solution: ppm =wt of solute x 10 = 3.6 x10-6g x 106 = 1.4 µg/g = 1.4 ppm wt of solution 2.6 g ppb = wt of solute x 10 = 3.6 x10-6g x 109 = 1400 ng/g = 1400 ppb wt of solution 2.6 g 2. Two grams of MgBr2 are dissolved in water to give 500 ml of solution. Calculate the ppm MgBr2 of this solution. ppm = wt of solute x 106 = 2 g x 106 = 4000 ppm volume of solution 500 mL 149 4. Mole fraction Mole fraction of solute = moles of solute total moles of solution = moles of solute moles of solute + moles of solvent Mole fraction of solvent = moles of solvent total moles of solution = moles of solvent moles of solute + moles of solvent If a solution is composed of two substances A and B, then the mole fraction expression will be; XA = nA nA + nB XB = nB nA + nB (XA + XB = 1) 150 What is the mole fraction of each component in a solution in which 3.57 g of sodium chloride, NaCl is dissolved in 25.0 g of water? Solution Moles of NaCl = 3.57 g = 0.0611 mole 58.44 g/mol Example: Moles of water = 25.0 g = 1.39 mole 18 g/mol XNaCl = 0.0611 mole = 0.0421 (0.0611 + 1.39) mol XH2O = 1.39 mole = 0.958 (0.0611 + 1.39) mol XNaCl + Xwater = 1. The sum of the mole fractions for a solution will equal 1. 151 4.3 preparation of solutions  Preparation of solution from the stock solution  Required information from the label includes: assay, density/sp.gr/wt per ml and M.M. %w/w = (m.solute/m.solution)*100 M = (%w/w*ρ*10)/M.M  Preparation of solutions from solid reagents M= m.solute/(Vsol(L)*M.M) 152 4.4 Electrolytes, weak electrolytes and non-electrolytes  An electrolyte may be defined as substance which either in molten state or in aqueous solution can conduct electricity. Acids ,bases and salts are mostly electrolytes. There are two types of electrolytes. • STRONG ELECTROLYTES: the electrolytes which are almost completely ionized in aqueous solution. Eg.H2SO4 ,HCl • WEAK ELECTROLYTES: the electrolytes which are ionized to a small extent in aqueous solution. Eg.CH3COOH,NH4OH 153 4.5 Activity and activity coefficients  the presence of diverse salts will cause an increase in dissociation of weak electrolyte or in the solubility of a precipitate.  The cations of analyte attract anions of the diverse electrolyte and the anions of the analyte attract the cations of the diverse electrolyte.  The attractions of the ions of the equilibrium reaction by the dissolved electrolyte effectively shields them and decreasing their effective concentration and shifting the equilibrium. 154  As the charge on either the diverse salt or the ions of the equilibrium reaction is increased, the diverse salt effect generally increases.  This “effective concentration” of an ion in the presence of an electrolyte is called the activity of the ion.  Ions activities can be calculated by Debye and Hückel  First step is calculation of the so called ionic strength, using following formula: µ= 1 CiZi 2 2  where Ci is a molar concentration of ith ion present in the solution and Zi is its charge. Summation is done for all charged molecules present in the solution. 155  Example: Calculate the strength of a/ 0.2 M solution of KNO3 b/ 0.5 M Al2(SO4)3. Solution a/ For 0.2 molar of KNO3 KNO3  K+ + NO3- 0.2 M 0.2 M 0. 2 M µ = 2 1 2 2 CiZi ½ [0.2 2 x 1 + 0.2 x 1 ] = 0.2 = b/ 0.5 M Al2(SO4)3 Al2(SO4)3  2Al3+ + 3SO420.5 M 1 M 1.5 M µ = ½ (CAl3+Z2Al3+ + CSO43- Z2 SO42-) = ½(1 x 32 + 1.5 x 22) = 7.5 The ionic strength of a solution of strong electrolyte containing singly charged ions is the same as with the molar concentration of the electrolyte. However, it is greater than the molar concentration of the electrolyte for solution containing ions of multiple charges. 156 UNIT 5 TITRIMETRIC METHODS OF ANALYSIS 5.1. Fundamentals of Titrimetry Titration is the process by which the standard solution is gradually added to the solution of analyte until quantitatively (completely) react. In a titration, the analyte reacts with a reagent added as a solution of known concentration. The volume of the titrant (reagent) required to completely react with the analyte is measured. Since the concentration of the reagent as well as the reaction b/n the analyte and reagent is known, the amount of analyte can be calculated. Requirements for a titration: 1. The reaction must be stoichiometric. Thus, there must be a well defined and known reaction between the analyte and the titrant. 2. The reaction should be rapid. 3. There should be no side reaction and the reaction should be specific. 4. There should be a marked change in some property of the solution when the reaction is complete. 5. The reaction should be quantitative 157 5.1.1. Definition of terms  Direct titration: is a known volume of analyte is titrated with a standard solution of a reagent or a known volume of a reagent solution can be titrated with the analyte solution.  Back titration: is a titration in which a reagent is added to a solution containing the analyte, and the excess reagent remaining after its reaction with the analyte is determined by a titration.  Titrant: is the reagent added to a solution containing the analyte.  Equivalent point: is a point of titration where the reagent added is chemically equivalent to the analyte (i.e. the reaction is at the stoichiometric ratio).  End point: is a point at which a physical change (color) associated with the equivalent point appears. Ideally, the end point should coincide with the equivalence point. 158 Standard solution  Solutions of accurately known concentrations are called standard solutions.  Standardization is the process by which one determines the concentration of a solution by measuring accurately the volume of the solution required to react with an exactly known amount of a primary standard.  For a given compound to be considered as a primary standard, it should fulfill the following criteria. 1. It must not react with or absorb the components of the atmosphere, such as water vapor, oxygen, and carbon dioxide. 2. It must react according to one invariable reaction. 3. It must have high percentage purity. 4. It should have a high formula weight to minimize the effect of error in weighing. 5. It must be soluble in the solvent of interest. 6. It should be nontoxic. 7. It should be readily available (inexpensive). 8. It should be environmentally friendly. 159 The common examples of primary standards:  a/ potassium hydrogen phthalate (KHP): used to standardize a base using phenolphthalein as an indicator.  b/ potassium hydrogen iodate (KH(IO3)): it is a strong acid used to standardize bases.  c/ sodium carbonate (Na2CO3) which is used to standardize strong acids to an end point of pH 4-5 .  Benzoic acid, sodium oxalate; borax and mercuric oxide are other primary standards. Secondary standard: is a solution whose concentration has been accurately determined by titration of primary standard and which is used to standardize other solutions.  Its purity is established by chemical analysis.  There are generally four classes of volumetric analysis a. acid-base titration b. Precipitation titration c. Complexometric titration and d. Redox titration 160 Neutralization titrations can be conducted in non-aqueous media under the following two conditions.  When the solubility of the analyte in water is too low: Example HCOOH and CH3COOH are soluble in water, but not C17H35COOH  When the strength of the analyte as an acid or a base is insufficiently low: Example: CH3COOH is weak acid in water; however, it is strong acid in liquid ammonia  Indicators are added to the analyte solutions to signal the end-points of titration. The end-points are characterized by observable physical changes such as: appearance or disappearance of color, change of color or appearance or disappearance of turbidity.  The behavior of: acid-type indicator – HIn + H2O  H3O+ + InBase-type indicator – In + H2O  InH+ + OH The hydronium ion concentration determines the ratio between the acid and conjugate base forms of the indicator. 161 Acid base indicators  Indicators are pH dependent.  Example: an aqueous solution of litmus is:  Red color at pH < 5  Blue color at pH > 8  Water color at pH between 5 and 8  Transition range: the pH at which the indicator will have intermediate      color. It is useful guide for the selection of an indicator for a particular type of neutralization reaction. Titration curve: is a plot of the reagent’s volume versus analyte concentration or some other quality related to concentration. There are two types of titration curves. Sigmoidal curve: p-function versus reagent volume Linear segment curve: instrument reading versus reagent volume Concentration changes during titration. One important characteristics of the equivalent point is the major changes in concentration. The relative concentrations of the reagent and analyte change abruptly when the equivalence point is approached. 162 Acid base indicators cont’d…  Indicators for acid-base reaction are substances that change color over a definite pH range.  Most neutralization indicators are weak organic acids or weak organic bases for which the molecules and ions have different colors.  One form may be colorless and the other colored or both may be colored.  EX. for methyl orange the acid color is red and the basic color is yellow. Thus, when both forms are present, the mixture of yellow and red appear orange. Assume the indicator is a weak acid, HIn; and the unionized form, HIn, is red while the ionized form, In- , is blue: HIn  H+ + In(red) (blue)  The pH of the indicator can be written in terms of HendersonHasselbalch equation pH = pKIn + log [In-] [HIn] 163 i/ If the color of the unionized form to be observed, [In-] = 1 [HIn] 10 pH = pKIn + log [In-] [HIn] pH = pKIn + log 1 10 pH = pKIn - 1 ii/ If only the color of the ionized form is observed, [In-] = 10 [HIn] 1 pH = pKIn + log [In-] [HIn] pH = pKIn + log 10 1 pH = pKIn + 1 164  In going from one color to the other, the pH has changed from pH = pKIn - 1 to pH = pKIn +1.  That is, pH = pKIn ± 1. This is a pH change of 2 and most indicators require a transition range of about two pH units. pH = (pKIn + 1) - (pKIn - 1) = 1+1 = 2  For example, methyl red (pKIn = 5) is effective indicator when a pH range 4.2 - 6.2 and shows red in acidic solution (pH nearly equals to 4) and yellow in basic solution (pH near to 6). In acidic solution, pH = pKIn – 1 = 5 – 1 = 4, and In a basic solution, pH = pKIn + 1 =5+1=6 165 166 5.2. Acid Base Titration 5.2.1. Acid Base Titration Curves A. Titrating Strong Acids with Strong Bases The titrant is a strong base; the analyte is a strong acid. Both the titrant and the analyte are completely ionized. H3O+(aq) + OH–(aq)  2H2O(l) Mostly carried out in aquous media Before addition of NaOH (before titration) before any base is added, the total concentration of hydronium ion is the sum of the concentration contributed from the strong acid and from water. [H3O+]total = [H3O+]acid + [H3O+] water . Since the acid is strong, the concentration of [H3O+]acid = concentration of strong acid The contribution to hydronium ion from the weak electrolyte (water) is negligible. Therefore, [H3O+] = conc. of the strong acid pH = –log [H3O+] = –log [s.acid] 167 Before equivalent point Before the equivalence point,  the acid is present in excess  the pH is determined by the concentration of excess acid.  The concentration of hydronium ion can be determined by the concentration of the acid left un-reacted. [H3O+] = initial moles of acid - moles of base Total volume = Ma Va –Mb Vb Va +Vb where the subscript ‘a’ indicates the acid and the subscript ‘b’ indicates the base 168 At the equivalence point At the equivalence point,  the moles of acid and the moles of the base are equal.  Since neither the acid nor the base is in excess,  [H3O+] = [OH-], the pH is determined by the dissociation of water. Kw = [H3O+] x [OH-] = 1x10-14, Since, [H3O+] = [OH-] [H3O+]2 = 1x10-14 [H3O+] = 10-7 = [OH-] pH = -log [H3O+] = -log 10-7 = 7 Thus, the pH at the equivalence point is 7.00 169 After equivalent point  The base is in excess  the pH is determined by the concentration of excess OH–. [OH-]excess = total no. of moles of base - no. of moles of acid Total volume = Mb Vb – Ma Va Va + Vb  The [H3O+] in this solution is not contributed from the acid because it is completely reacted  only water is the source of hydronium ion.  Hydronium ion concentration can be calculated from the ionization of water; [H3O+] = Kw [OH-] 170 Example.  Titration of 50.0 mL of 0.100 M HCl with 0.200 M NaOH. HCl + NaOH  Na+ + Cl- + 2H2O  calculate the volume of NaOH needed to reach the equivalence point. At the equivalence point, Moles HCl = Moles NaOH MaVa = MbVb The volume of NaOH needed to reach the equivalence point, therefore, is MaxVa = 0.1Mx50mL = 25 mL Vb  Mb 0.2M 171 Before equivalent point i/ after addition of 5 mL of NaOH Given: milimoles of acid = Ma x Va = 0.1 M x 50 mL = 5 milimole milimoles of base = Mb x Vb = 0.2 M x 5 mL = 1 milimole Total volume = Vacid + Vbase = 50 + 5 = 55 mL Required: [H3O+]= ? [H3O+] = Ma Va- MbVb = 5-1 = 4 = 0.073 M Va +Vb 55 55 Or, pH= - log[H3O+] = -log 0.073 = 1.14 H3O+ + OH-  H2O initial mole 5 Equilibrium 4 1 _ [H3O+] = 4 = 0.073 M 55 172 ii/ if the volume of NaOH = 10 mL mmole of acid= 5 mmole of base = 0.2 M x 10 mL = 2 Vtotal = (50 +10) mL = 60 mL [H3O+] = Ma Va- MbVb = 5 - 2 = 3 = 0.05 M Va + Vb 60 60 pH = - log [H3O+]=-log 0.05= 1.3 173  iii/ if the volume of NaOH =20 ml mmole of acid= 5 mmole of base= 0.2 M x 20 mL = 4 Vtotal = (50 mL+20 mL) = 70 mL [H3O+] = Ma Va - MbVb = 5 – 4 = 1 = 0.0143 M Va + Vb 70 70 pH = - log [H3O+] = -log 0.0143 M = 1.85 At the equivalent point At the equivalent point neither HCl nor NaOH is in excess and so the concentrations of hydronium and hydroxide ions must be equal. [H3O+] = Kw = 10-7 = [OH-] pH= - log [H3O+] = - log 10-7 = 7 174 After equivalent point  i. addition of 26 mL of NaOH mmoles of acid = 5 mmoles of base added = 0.2 M x 26 mL = 5.2 mmoles of base unreacted = 5.2 - 5 = 0.2 [OH-] = MbVb - MaVa = 5.2 mmol - 5 mmol = 2.6x10-3 M Va + Vb 76 ml pOH = -log [OH-] = -log 2.6x10-3 M = 2.59 pH + pOH =14 pH = 14- pOH = 14 - 2.59 = 11.41 or, [OH-] x [H+] = 1x10-14 [H+] = 1x10-14 = 3.84 x10-12 M, pH = -log 3.84x10-12 M = 11.41 2.6x10-3 175 Figure 5.1 Acid–base titration curve for 50.0 mL of 0.100 M HCl with 0.200 M NaOH. 176 B/ The titration of a strong base with a strong acid Titration curves for strong bases are derived in an analogous way to those for strong acids.  before equivalent point  the solution is basic  the hydroxide ion concentration is the same as the concentration of the base un-reacted.  at the equivalent point  The solution is neutral  beyond the equivalent point  The solution becomes acidic in the region Example. Calculate the pH during the titration of 20 mL of 0.2 M NaOH with 0.1 M HCl after the addition of the following volume of reagent. A/ 0 mL B/ 10 mL C/ 30 mL D/ 40 mL E/ 50 mL 177 A/ Before addition of HCl (V HCl = 0 mL) NaOH  Na+ + OHpOH = –log [OH-] = –log [NaOH] = –log (0. 200) = 0.70 0.2 M 0.2 M pH = 14 - 0.7 = 13.3 B/ after addition of 10 mL of HCl (it is before equivalent point) Before the equivalence point, NaOH is in excess and the pOH is determined by the concentration of excess NaOH. [OH-] = initial moles of base - moles of acid Total volume = MbVb - MaVa Va + Vb milimoles of base = Mb x Vb = 0.2 M x 20 mL = 4 milimoles of acid = Ma x Va = 0.1 M x 10 mL = 1 Total volume= Vacid + Vbase= 20 mL + 10 mL = 30 mL Required: [OH-]= ? [OH-] = MbVb - Ma Va = 4 - 1 = 3 = 0.1 M Va +Vb 30 30 pOH = - log[OH-] =-log 0.1 M =1.0 M pH = 14 - 1 = 13 178 C/ after addition of 30 mL of HCl (it is also before equivalent point) milimoles of base = Mb x Vb = 0.2 M x 20 mL = 4 milimoles of acid = Ma x Va = 0.1 M x 30 mL = 3 [OH-] = MbVb - Ma Va = 4 - 3 = 1 = 0.02 M Va +Vb 50 50 pOH= - log[OH-] = -log 0.02 M = 1.70 M pH = 14 - 1.70 = 12.3 D/ after addition of 40 mL of HCl (at the equivalent point) milimoles of base = Mb x Vb = 0.2 M x 20 mL = 4 milimoles of acid = Ma x Va = 0.1 M x 40 mL = 4 Milimoles of HCl = milimoles of NaOH [H3O+] = [OH-]. Kw = [H3O+] x [OH-] = 1x10-14 Since, [H3O+] = [OH-] [H3O+]2 = 1x10-14 [H3O+] = 10-7 = [OH-] pH = -log [H3O+] = -log 10-7 = 7 179 E/ addition of 50 mL HCl (after equivalent point) After equivalent point the solution will contain excess acid. [H3O+]= MaVa - MbVb = 5 mmol - 4 mmol = 0.0143 M Va + Vb 70 ml pH = - log[H+] = -log 0.0143 M = 1.84 180 C/ Titrtaion of Weak Acid with a Strong Base  Example: the titration of 50.0 mL of 0.100 M acetic acid, CH3COOH, with 0.100 M NaOH. The volume of NaOH needed to reach the equivalence point; moles CH3COOH = moles NaOH MaVa = MbVb Vb = MaVa/Mb = (0.1 M) (50 mL)/ 0.1 M = 50 mL Before adding any NaOH The pH is that for a solution of 0.100 M acetic acid can be determined by ionization constant method:  H3O+(aq) + CH3COO–(aq) CH3COOH(aq) + H2O(l) Ka = [CH3COO-] x [H3O+] Ka for acetic acid = 1.75 x10-5 [ CH3COOH] CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO–(aq) Initial conc. 0.1 M 0 0 At equilibrium 0.1 M- x x x 181 Ka = [CH3COO-] [H3O+] [CH3COOH] 1.75 x10-5 = (x)(x) 0.1 x2 = 1.75x10-6 x = [H3O+] = 1.32 x10–3 M pH = - log [1.32x10–3 M] = 2.88 Before equivalent point Adding NaOH converts a portion of the acetic acid to its conjugate base, CH3COO–. CH3COOH(aq) + OH–(aq)  H2O(l) + CH3COO–(aq) The pH of a buffer solution can be calculated using the Henderson– Hasselbalch equation . 182 pH = pKa + log [salt] = pKa + log [CH3COO–] [acid] [CH3COOH] The concentration of acetic acid is; [CH3COOH] = moles of un-reacted CH3COOH = MaVa - MbVb Total volume Va + Vb The concentration of acetate ion is; [CH3COO-] = moles of NaOH added = MbVb Total volume Va + Vb 183 i. after addition of 10 mL of NaOH [CH3COOH] = (0.1 M)(50 mL) - (0.1 M)(10 mL) = 0.0667 M 50 mL + 10 mL [CH3COO-] = (0.1 M)(10 mL) = 0.0167 M 10 mL + 50 mL The pH of the solution: pH = pKa + log [CH3COO-] = 4.76 + log 0.0167 = 4.16 [CH3COOH] 0.0667 ( pKa = -log Ka) 184 ii. after addition of 20 mL of NaOH [CH3COOH] = (0.1 M)(50 mL) - (0.1 M)(20 mL) = 0.043 M 50 mL + 20 mL [CH3COO-] = (0.1 M)(20 mL) = 0.0286 M 20 mL + 50 mL The pH of the solution: pH = pKa + log [CH3COO-] = 4.76 + log 0.0286 M = 4.58 [CH3COOH] 0.043 M 185 At equivalent point (50 mL of NaOH) At the equivalence point, moles of acetic acid = moles of sodium hydroxide The concentration of CH3COO–; [CH3COO-] = MbVb = 0.1 M x 50 mL = 0.05 M Va + Vb 100 mL The pH of the solution can be determined by the amount of the conjugate base, acetate ion hydrolyzed shown in the reaction: CH3COO–(aq) + H2O(l)  OH–(aq) + CH3COOH(aq) Before calculating the pH we need to determine the concentration of OH– . Ka x Kb = Kw Kb = Kw ; Ka CH3COOH = 1.75 x 10-5, then Kb for CH3COO– Ka Kb = 1x10-14 = 5.71x10–10 1.75 x10-5 186 CH3COO–(aq) + H2O(l)  OH–(aq) + CH3COOH(aq) 0.05 M – x x x Kb = [CH3COOH] [OH-] [CH3COO-] 5.71x10–10 = (x) (x) 0.05 – x x2 = (5.71x10–10)(0.05) x = [OH–] = 5.34x10–6 M The concentration of H3O+, therefore, [H3O+] [OH–] = 1x10-14 [H3O+] = 1x10-14 = 1x10-14 = 1.87x10–9 [OH–] 5.34x10–6 pH = 8.73 187 Beyond equivalent point After the equivalence point NaOH is present in excess [OH-] = MbVb - MaVa Va + Vb pOH = -log [OH-] i/ addition of 60 mL of NaOH mmole of acid = 5 mmole of base = 60 mL x 0.1 M = 6 mmol CH3COOH(aq) + OH–(aq) H2O(l) + CH3COO–(aq) Initial mmol 5 6 0 Equil. mmol 0 1 5 [OH-] = MbVb - MaVa = 6 – 5 = 1 = 0.0091 Va + Vb (60 + 50) mL 110 mL pOH = - log[OH-]= 3.04, pH = 14 - 3.04 = 11.96 188 ii. after addition of 70.00 ml of NaOH CH3COOH(aq) + OH–(aq)  H2O(l) + CH3COO–(aq) Initial mmol 5 7 Equil. mmol 0 2 5 Th [OH-] = MbVb - MaVa = 7 mmol - 5mmol = 0.0167 M Va + Vb 70 + 50 ml pOH = - log[OH-]= -log 0.0167 M = 1.7 PH = 14-1.77= 12.22 189 Figure. Titration curve for 50.0 mL of 0.100 M acetic acid (pKa = 4.76) with 0.100 M NaOH 190 Note that: The pH at the equivalent point for the titration of i. Strong acid with strong base = 7 (neutral) ii. Strong acid with weak base < 7 (acidic) iii. Weak acid with strong base > 7 (basic) iv. Weak acid with weak base = 7, < 7 or > 7 depending on the extent of ionization of the acid or base. i.e. If Ka = Kb, pH = 7; if Ka > Kb, pH < 7 and if Ka < Kb, pH > 7. 191 5.2.6 Titration of polyprotic acids  Acids that contain more than one acidic (ionizable) hydrogen (proton) are called polyprotic or polybasic acids.  The dissociation of polyprotic acids occurs in a stepwise fashion, one proton lost at a time. For example, the generic triprotic acid will dissociate as shown in Reactions (1) through (3).  The equilibrium constants for these reactions are symbolized by Ka. The trailing subscript "a" indicates that the equilibrium constant describes an acid dissociation reaction. The trailing subscript "n" is written as a number and indicates which proton is being dissociated. 192 Titration of polyprotic acids cont’d… H3A + H20 <-----> H3O+ + H2A- Ka1 = [H3O+] [H2A-] / [H3A] (1) H2A- + H20 <-----> H3O+ + HA2- Ka2 = [H3O+] [HA2-]/ [H2A-] (2) HA2- + H20 <-----> H3O+ + A3- Ka3 = [H3O+] [A3-] / [HA2-] (3)  Reactions such as those shown in Reactions (4) and (5) are sometimes used as the basis of chemical calculations, but these reactions are generally not considered to be representative of what is actually occurring in solution.  These reactions can be useful at times even though they are not rigorously correct chemically.  The equilibrium constant for Reaction (4) is the product of Ka1 and Ka2 while the equilibrium constant for Reaction (5) is the product of Ka1, Ka2, and Ka3. Prove this to yourself. H3A + 2H20 <-----> 2H3O+ + HA2- (4) H3A + 3H20 <-----> 3H3O+ + A3- (5) 193 Titration of polyprotic acids cont’d…  Remember that the numerical value of equilibrium constant is an indication of how far to the right a reaction proceeds. The greater the magnitude of the equilibrium constant, the more completes the reaction. Generally the numerical value of Ka1 is larger than that of Ka2, which is larger than Ka3. This is reasonable when one considers that successive protons are removed from species with progressively higher negative charges.  Polyprotic acids are called polybasic because they have more than one conjugate base. A conjugate base is a species that is produced when an acid loses a proton. For example, the species H2A-, HA2-, and A3- are all conjugate bases of H3A. The completely deprotonated acid, A3-, is a weak base. The other two conjugate bases are also weak bases, but in addition they have one or more acidic protons. As a result these species are amphiprotic or amphoteric, which means that they can act as either an acid or a base. Molecules that are partially deprotonated polyprotic acids are called acid salts. A common acid salt is sodium bicarbonate, NaHCO3, which is found in baking soda. 194 Titration of polyprotic acids cont’d…  When a strong base is added to a solution of a polyprotic acid, the protons of the acid are neutralized in a stepwise fashion. That is, Reaction (6) will occur first until all of the H3A is used up. Then Reaction (7) will begin and will continue until all of the H2A- is gone. Then Reaction (8) will occur until all of the HA2- is gone. This will only be true when the successive dissociation constants are different by a large enough factor and when all of the acidic species are strong enough. For example, phosphoric acid has Ka values that are different by a large enough factor to allow it to react with a strong base in a stepwise fashion, however, the value of Ka3 for phosphoric acid is so small that the last proton of phosphoric acid is extremely difficult to remove, and the reaction of phosphoric acid that is analogous to Reaction (8) essentially will not occur in aqueous solution.  H3A + OH- <-----> H2A- + H20  H2A- + OH- <-----> HA2- + H20  HA2- + OH- <-----> A3- + H20 (6) (7) (8) 195 Titration of polyprotic acids cont’d…  The titration curve for the titration of a polyprotic acid such as phosphoric acid will have the form shown in Figure 1. This curve is a plot of pH (log[H30+]) versus the volume of base added. The first point on the curve corresponds to a solution of phosphoric acid only. The pH at this point is due solely to the phosphoric acid in the solution. As soon as some base is added, some H2PO4- is produced. The solution now contains both phosphoric acid (a weak acid) and its conjugate base dihydrogen phosphate, and thus it is a buffer. A buffer is a solution that is resistant to change in pH upon dilution and addition of acid and base. This will be the situation from the initial point to the first equivalence point, and therefore this region of the curve is called the first buffer zone. The pH of the point midway between the first and second equivalence points is equal to the negative logarithm of Ka1, pKa1. At the first equivalence point, all of the H3PO4 has been neutralized and only H2P04- is present in the solution. When more base is added, some of the H2P04- is neutralized and some HP042- is produced. 196 Titration of polyprotic acids cont’d…  This solution contains both dihydrogen phosphate (a weak acid) and its conjugate base hydrogen phosphate, and thus is a buffer. At all points between the first and second equivalence points, the solution will contain these two species, and therefore this portion of the curve is called the second buffer zone. The pH of the solution at the point halfway between the second and third equivalence points is equal to pKa2. At the second equivalence point the only species in the solution is HPO42-. The region between the second and third equivalence points is called the third buffer zone because the solution contains both the weak acid HPO42- and its conjugate base PO43-. Halfway between the second and third equivalence points the pH of the solution is equal to pKa3. At the third equivalence point, only the weak base PO43- is in the solution. After the third equivalence point, the solution contains phosphate ion and excess unreacted OH-. 197 Titration of polyprotic acids cont’d… Titration curve for titration of phosphoric acid with sodium hydroxide. The point marked "a" is the initial point. The pH of the solution equals pKa1, pKa2, and pKa3 at points b, d, and f, respectively. Points c, e, and g are the first, second, and third equivalence points, respectively. The points between a and c, between c and e, and between e and g are the first, second, and third buffer zones, respectively. 198 5.3. Precipitation titration  Is a type of reaction in which the analyte and titrant form an insoluble precipitate.  Precipitation reaction should occur rapidly and quantitatively. Unfortunately, the rate of most precipitation reactions are slow.  the most important precipitating agent is silver nitrate, which is used for the determination of the halides, like anions SCN-, CN-, CNO- and several divalent inorganic anions.  Titrimetric methods based up on silver nitrate are called argentometric method 5.3.1. Titration Curves The titration curve for a precipitation titration follows the change in either the analyte’s or titrant’s concentration as a function of the volume of titrant. For example, in an analysis for Cl– using Ag+ as a titrant Ag+(aq) + Cl–(aq)  AgCl(s)  Precipitation titration curve is influenced by the concentration of Ag+ and Cl- and by KSP value.  The titration curve may be a plot of pAg or pCl as a function of the titrant’s volume. 199  Example: calculate the titration curve for the titration of 50.0 mL of 0.0500 M Cl– with 0.100 M Ag+. The reaction in this case is: Ag+(aq) + Cl–(aq) AgCl(s) Ksp = 1.8x10–10 = [Ag+][Cl–]  calculate the volume of Ag+ needed to reach the equivalence point. From the stoichiometry: moles Ag+ = moles Cl– at the equivalence point Or MAg VAg = MCl VCl VAg = MClVCL = 0.05M 50mL  25mL MAg 0.1M Before equivalent point Before the equivalence point Cl– is in excess. i. addition of 10 mL of Ag+ The concentration of un-reacted Cl– after adding 10.0 mL of Ag+ is [Cl–] = moles of Cl- excess = MCl- VCl- - MAg+VAg+ Total volume VCl + VAg [Cl–] = (0.05 M) (50 mL) - (0.1 M) (10 mL) = 2.5x10-2 50 mL + 10 mL pCl = -log[Cl–] = -log (2.5x10-2) = 1.60 200 The change in concentration for Ag+ Ksp = [Ag+] [Cl–] = 1.8x10-10 [Ag+] = Ksp = 1.8x10-10 = 7.2x10-9 [Cl–] 2.5x10-2 pAg = - log[Ag+] = -log7.2x10-9 = 8.14 ii. After addition of 20 mL of [Ag+] [Cl–] = moles of Cl- excess = MCl- VCl- - MAg+VAg+ Total volume VCl + VAg [Cl–] = (0.05 M) (50 mL) - (0.1 M) (20 mL) = 7.14x10-3 50 mL + 20 mL pCl = log[Cl–] = -log (7.14x10-3) = 2.15 The change in concentration for Ag+ Ksp = [Ag+] [Cl–] = 1.8x10-10 [Ag+] = Ksp = 1.8x10-10 = 2.52x10-8 M [Cl–] 7.14x10-3 pAg = - log[Ag+] = -log(2.52x10-8 M) = 7.60 201 Equivalent point At the equivalent point, the concentration of Ag+ and Cl- are equal. The concentration of Ag+ and Cl- can be calculated using solubility product of AgCl. Ksp = [Ag+] [Cl–] = 1.8x10-10 [Ag+] =[Cl–] = 1.34x10-5 pCl = -log[Cl–] = -log 1.34x10-5 = 4.87 pAg =- log[Ag+] = -log 1.34x10-5 = 4.87 At the equivalent point, therefore, pCl and pAg are equal. Beyond equivalent point After the equivalence point, the titration mixture contains excess Ag+. i. After adding 35.0 mL of titrant (Ag+) [Ag+] = moles excess Ag+ total volume 202 [Ag+] = (0.1 M)(35.0 mL) – (0.05 M)(50.0 mL) 35 mL + 50 mL [Ag+] = 1.18x10-2 M pAg = -log1.18 x10-2 M = 1.93. The concentration of Cl– is; [Cl–] = Ksp = 1.8x10-10 = 1.5x10-8 M [Ag+] 1.18 x10-2 M pCl = -log[Cl–] = -log1.5x10-8 M = 7.82 pCl + pAg is always equal to pKsp (= -log Ksp). Precipitation titration curve for 50.0 mL of 0.0500 M Cl– with 0.100 M Ag+ (a) pCl versus volume of titrant; (b) pAg versus volume of titrant 203 End point detection methods of precipitation titration There are three distinct methods for the titration of chloride with silver based up on the different types of indicators. These are: 1. Reagents which can react with the excess titrant used that yield a colored solid (precipitate), Mohr Method. 2. The indicator reacts with excess titrant to form a colored soluble complex. The reaction product is a complex of titrant and the indicator, Volhard Method. 3. Indicator adsorbed on the surface of the precipitate formed during titration. The adsorbed indicator has a different color than the dissolved indicator. The adsorptions of an indicator does not occur until an excess of titrant is added, Fajans Method 204 The Mohr Method  In Mohr method silver nitrate is used to titrate the chloride solution to which chromate salt has been added to act as an indicator.  This indicator produces a yellow solution.  When the precipitate of the chloride is complete, the excess silver ion reacts with the indicator at the equivalent point to precipitate silver ion to form red silver chromate. CrO42- + 2Ag+  Ag2CrO4 (yellow) (brick-red) The concentration of Ag+ at equivalent point is 1.3x10-5 The concentration of chromate ion required to start the formation of Ag2CrO4 Ag2CrO4 → 2Ag+ + CrO4-2 Ksp = [Ag+]2[CrO42-] [CrO42-] = Ksp= 1.2 x 10-12 [Ag+]2 (1.35x10-5)2 [CrO42-] = 6x10-3M 205  The practical problem associated with this concentration of chromate at this high final concentration is that the brick-red color cannot be detected. This problem can be overcome by using a more dilute solution of the indicator, and applying blank titration.  The Mohr titration must be neutral or slightly basic (pH = 7-9). Why is the Mohr method restricted within this pH range? i. If the solution is too acidic,  the chromate reacts with H+ to form hydrogen chromate ions, which reduces the concentration of chromate ion or it needs more Ag+ to form Ag2CrO4 precipitate. It will lengthen the equivalent point. CrO42- + H+  HCrO41- ii. If the solution is basic (at pH > 10), there are enough hydroxide ions present to cause the formation of silver hydroxide. 2Ag+ + 2OH-  2AgOH  Ag2O + H2O The presence of an excess of either sodium hydrogen carbonate or borax in the solution tends to maintain the hydrogen ion conc. within suitable limits. 206 2. The Volhard method  The Volhard method is an indirect or back-titration method in which an excess of a standard solution of silver nitrate is added to a chloride solution. The excess silver nitrate is then determined by titration with a standard solution of ammonium or potassium thiocyanate with ferric ion as an indicator. Cl- + Ag+  AgCl + excess Ag+ excess Ag+ + SCN-  AgSCN(s) The end point is detected by adding iron (III) as a ferric alum (ferric ammonium sulfate – FeNH4(SO4)2.12H2O, which forms a soluble red complex with first excess of titrant. The first excess of thiocyanate reacts with the ferric ion to form the intensely colored FeSCN complex, Fe3+ + SCN-  [FeSCN]2+ The standard silver solution is added to the acidified chloride solution to precipitate silver chloride. Since AgSCN (Ksp =1x10-12) is less soluble than AgCl (Ksp = 1.8x10-10), it can react with and dissolve the precipitated AgCl, AgCl(s) + SCN-  AgSCN(s) + ClFor this reason, the precipitated chloride must be removed from contact 207 with solution before titration with thiocyanate. 3/ The Fajans Method (adsorption indicator Method)  Some organic compounds (fluoresceine and eosin) are adsorbed      strongly on certain precipitates, and undergo a change in colour up on adsorption so that they can be used as indicator for selected titrtaions. In the titration of chloride by silver, the precipitated silver chloride is formed in a solution that contains excess Cl- and consequently adsorbs Cl- in the primary layer to become negatively charged. These charged particles then attract cations at the second layer which can be symbolized as AgCl: Cl-. The particles will remain negatively charged untill the exact equivalence point at which they become neutral and tend to coagulate. The first excess silver, however, the particles adsorb silver in strongly and becomes postively charged. The postive particles attract anions in to the second layer. If the chloride is added to a silver solution, the particles are postively charged before equivalent point and become negatively just beyond the equivalent point. 208  Fluoresceine is a weak acid (HC20H11O5 or HFI), which is       dissolved and exists in solution as fluoresceinate, FI-. HFI  H+ + FIBefore the end point, the precipitate of AgCl has a negative surface charge due to the adsorption of excess Cl–. The anionic indicator (FI-) is repelled by the precipitate and remains in solution where it has a greenish yellow color. After the end point, the precipitate has a positive surface charge due to the adsorption of excess Ag+. The anionic indicator (FI-) is now adsorbed to the precipitate’s surface where its color is pink. The color change can be explained as being due to the precipitation of AgFI up on the surface. Since the molecular or unionized form of Fluoresceine is not attracted to the particles and will not be adsorbed, the indicator must be in the form to be effective. 209  ClCl- ClAgCl(s) Cl- ClBefore equivalent point Ag + Ag+ Ag+ AgCl(s) Ag+ Ag+ After equivalent point  This method is a rapid, accurate, and reliable method of end-point detection; however, its use is limited to few reactions which form colloidal precipitates rapidly. 210 5.5 Redox Titration  This titration involves electron transfer reaction between two redox species  a titration in which the reaction between the analyte and titrant is an oxidation/reduction reaction;  for example, the reaction between cerium(IV) and iron(II) to form cerium(III) and iron(III). 5.5.1. Redox Titration Curves Consider, for example, a titration in which the analyte in a reduced state, Ared, is titrated with a titrant in an oxidized state, Tox. The titration reaction is Ared + Tox  Aox + Tred 211  The electrochemical potential for the reaction is the difference between the reduction potentials for the reduction and oxidation half-reactions Erxn = Ereduced – Eoxidized Erxn = ETox/Tred – EAox/Ared Before the equivalence point It consists of a mixture of appreciable quantities of both the oxidized and reduced forms of the analyte, but very little unreacted titrant. The potential, therefore, is best calculated using the Nernst equation for the analyte’s half-reaction. E = Eo – 0.059 log [Tred] n [Tox] Eo = standard reduction potential n = number of electrons lost or gained 212 At the equivalent point The reaction’s electrochemical potential, Erxn, is zero, and Ereduced = Eoxidized ETox/Tred = EAox /Ared Example: Calculate the potential for the titration of 50.0 mL of 0.100 M Fe2+ with 0.100 M Ce4+ in a matrix of 1 M HClO4. The reaction is Fe2+(aq) + Ce4+(aq)  Ce3+(aq) + Fe3+(aq) 213  First we need to calculate the volume of Ce4+ needed to reach the equivalence point. From the reaction, Moles Fe2+ = moles Ce4+ MFe2+ x VFe2+ = MCe4+ x VCe4+ VCe4+ = MFe2+ x VFe2+ = (0.1 M) (50 mL) = 50.0 mL MCe4+ (0.1 M) Before the equivalence point i/ after adding 5 mL of Ce4+ mmole of Fe2+ = 0.1 M x 50 mL= 5 mmole of Ce4+ = 0.1 M x 5 mL= 0.5 Fe2+(aq) + Ce4+(aq) Initial mmole 5 Equil. mmole 4.5 0.5 0  Ce3+(aq) + Fe3+(aq) 0.5 0.5 214 The concentration of species at equilibrium, = 4.5 mmole = 0.0818 M [Fe2+] = mmole total volume 5 mL+50 mL [Fe3+] = mmole = 0.5 mmole = 0.0091M total volume 5 mL+50 mL The potential can be calculated using Nernestian equation, Eo Fe3+/ Fe2+ = 0.77 V Ecell = Eo – 0. 059 log[reduced] n [oxidized] = Eo – 0. 059 log [Fe2+] 1 [Fe3+] E Fe3+/Fe2+ = 0.77 V - 0. 059 log [0.0818 M] = 0.71 V 1 [0.0091M] 215  ii/ after adding 10 mL of Ce4+ mmole of Fe2+ = 0.1 M x 50 mL = 5 mmole of Ce4+ = 0.1 M x 10 mL = 1 Fe2+(aq) + Ce4+(aq) Initial mmole 5 1 Equilibrum mmole 4 0  Ce3+(aq) + Fe3+(aq) 1 1 The concentration of species at equilibrium, = 4 mmole = 0.067 M [Fe2+] = mmole total volume 10 mL+50 mL [Fe3+] = mmole = 1 mmole = 0.0167 M total volume 10 mL+50 mL Ecell = Eo – 0. 059 log [reduced] n [oxidized] E Fe3+/Fe2+ = 0.77 V - 0. 059 log [0.067 M] 1 [0.0167 M] = 0.73 V 216  At the equivalence point The moles of Fe2+ and the moles of Ce4+ added are equal. As a result, the potentials for the two half-reactions are the same. Eeq = Eo Fe3+/Fe2+ - 0. 059 log [Fe2+] ………….(1) 1 [Fe3+] Eeq = Eo Ce4+/Ce3+ - 0. 059 log [Ce3+]…………(2) 1 [Ce4+] Add up these two equations: 2Eeq = Eo Fe3+/Fe2++ Eo Ce4+/Ce3+ - 0. 059 log [Ce3+][Fe2+] 1 [Ce4+][Fe3+] At the equivalence point, [Fe2+] = [Ce4+] and [Fe3+] = [Ce3+], thus the above equation becomes 2Eeq = Eo Fe3+/Fe2++ Eo Ce4+/Ce3+ - 0. 059 log 1 (log 1 = 0) 1 1 Eo Ce4+/Ce3+ = 1.7 V 217  Eeq = Eo Fe3+/Fe2+ + Eo Ce4+/Ce3+ = 0. 77 V + 1.7 V = 1.23 V 2 2 In general, E at the equivalence point is given by E = nred Eored + nox Eoox 2 Where Eored = standard potential for reduced substance Eoox = standard potential for oxidized substance nred = no. of electrons involved in reduction reaction nox = no. of electrons involved in oxidation reaction. 218  After the equivalence point After equivalent point, the potential determined by the concentration of the excess Ce4+ and Ce3+ produced. i/ after adding 60.0 mL of titrant (Ce4+) The potential can be calculated using the Nernst equation for the titrant’s halfreaction. mmole of Fe2+ = 0.1 M x 50 mL= 5 mmole of Ce4+ = 0.1 M x 60 mL= 6 Ecell = Eo – 0. 059 log [reduced] n [oxidized] Fe2+(aq) + Ce4+(aq)  Ce3+(aq) + Fe3+(aq) Initial mmole 5 6 0 0 Equil. mmole 0 1 5 5 The concentration of species at equilibrium, [Ce3+] = mmole = 5 mmole = 0.0091 M total volume 60 mL + 50 mL 219 [Ce4+] = mmole total volume = 1 mmole = 0.045 M 60 mL + 50 mL E Ce4+/Ce3+ = Eo – 0. 059 log [Ce3+] 1 [Ce4+] E Ce4+/Ce3+ = 1.7 V – 0. 059 log 0.0091 M 1 0.045 M E Ce4+/Ce3+ = 1.66 V ii/ after addition of 70 mL of Ce4+ mmole of Fe2+ = 0.1 M x 50 mL= 5 mmole of Ce4+ = 0.1 M x 70 mL= 7 Ecell = Eo – 0. 059 log [reduced] n [oxidized] Initial mmole Equil. mmole Fe2+(aq) + Ce4+(aq) 5 7 0 2  Ce3+(aq) + Fe3+(aq) 0 0 5 5 220  The concentration of species at equilibrium, [Ce3+] = mmole total volume = 5 mmole = 0.042 M 70 mL+50 mL [Ce4+] = mmole total volume = 2 mmole = 0.017 M 70 mL+50 mL E Ce4+/Ce3+ = Eo – 0. 059 log [Ce3+] 1 [Ce4+] E Ce4+/Ce3+ = 1.7 V – 0. 059 log 0.042 M 1 0.017 M E Ce4+/Ce3+ = 1.68 V 221 E(V) Volume of Ce4+ (mL) Figure. Redox titration curve for 50.0 mL of 0.100 M Fe2+ with 0.100 M Ce4+ in 1 M HClO4.  222 Oxidation – Reduction Indicators  General redox indicators do not participate in the redox titration, but whose oxidized and reduced forms differ in color.  When added to a solution containing the analyte, the indicator imparts color that depends on the solution’s electrochemical potential .  The relationship between a redox indicator’s change in color and the solution’s electrochemical potential is easily derived by considering the half-reaction for the indicator. 223 Inox + ne–  Inred Where, Inox = the indicator’s oxidized form, and Inred = is the reduced form The Nernst equation expresses this reaction as E = Eo Inox/Inred – 0.0592 log [Inred] (1) n [Inox] If we assume that the indicator’s color in solution changes from that of Inox to that of Inred when the ratio . E = Eo Inox/Inred – 0.059 log 10 n E = Eo Inox/Inred – 0.059 n If we assume that the indicator’s color in solution changes from Inred to that of Inox when the ratio 224 If we assume that the indicator’s color in solution changes from Inred to that of Inox when the ratio E = Eo Inox/Inred – 0.059 log 1/10 n = Eo Inox/Inred + 0.059 n E = Eo Inox/Inred + 0.0592 n The potential change required to produce the full color change of a typical general indicator can be found by: E = Eo Inox/Inred ± 0.0592 n 225  A typical general indicator exhibits a detectable color change when a titrant causes the system potential to shift from E = Eo Inox/Inred + 0.0592 to E = Eo Inox/Inred - 0.0592 n n  One can select the appropriate indicator for a given redox titration : The Eo Inox/Inred must be near the equivalence point. For instance, if the potential of a titration at the equivalent point were 0.5 V, then the most suitable indictor would be methylene blue. 226 Selected general redox indicators Indicator Oxidized color Reduced colour Eo (V) indigo tetrasulfonate Blue colorless 0.36 methylene blue Blue colorless 0.53 diphenylamine Violet colorless 0.75 diphenylamine sulfonic acid red-violet colorless 0.85 tris(2,2’-bipyridine)iron pale blue red 1.120 227 UNIT 6. GRAVIMETRIC METHODS OF ANALYSIS 6.1. Principles and Types of Gravimetric Analysis Gravimetric analysis is the quantitative isolation of a substance by precipitation and the weighing of the precipitate. It is based on the measurement of mass. There are three gravimetric methods of analysis. 1. Precipitation gravimetry- is a gravimetric method in which the signal is the mass of a precipitate. The precipitate is washed free of impurities, and converted to a product of known composition (usually by heating to obtain a more stable form). 2. Electrogravimetry- the signal is the mass of an electrodeposit on the cathode or anode in an electrochemical cell. The analyte is deposited as a solid film on one electrode in an electrochemical cell. 3. Volatilization gravimetry- it is a gravimetric method in which the loss of a volatile species gives rise to the signal. The analyte or its decomposition products are volatilized and the volatile product is collected and weighed. For example, the gravimetric determination of water content in barium chloride dihydrate: BaCl2.2H2O + heat  BaCl2 + 2 H2O (g)  For the most part, gravimetric analyses are highly accurate and precise because they use the analytical balance as measurement tool. 228 6.2. Properties of precipitates and precipitating agents The precipitate must:  be of low solubility so that no significant loss of the analyte occurs during filtration and washing.  be unreactive with constituents of the atmosphere.  be of high purity  have known composition after it is dried or if necessary ignited.  be in a form that is easy to separate from the reaction mixture (readily filtered and washed free of contaminants). 229 Particle size and filterability of precipitates  The precipitates made up of large particles are generally desirable in gravimetric analysis because large particles are easy to filter and wash free of impurities and are usually purer than are finely divided precipitates. The size of the precipitate’s particles determines the ease and success of filtration. Smaller, colloidal particles are difficult to filter because they may readily pass through the pores of the filtering device, while large, crystalline particles are easily filtered. Mechanism for Precipitation  Precipitation formed by two distinct events: a/ nucleation: is the process by which minimum number of atoms, ions, or molecules join together to form a stable solid. Sometimes these nuclei form on the surface of suspended solid contaminants such as dust particles. 230 b/ Particle growth The particle size of a precipitate is influenced by different variables, *like solubility of the precipitate, *temperature, *reactants’ concentrations and *rate at which reactants are mixed. The net effects of these variables are related to the particle size by relative supersaturation (RSS).  Relative supersaturation is a measure of the extent to which a solution, or a localized region of solution, contains more dissolved solute than that expected at equilibrium (RSS). 231 RSS = QS S Where Q = the concentration of the solute in solution at the instant the reagents are mixed S = the concentration of the solute in a saturated solution or the solute’s expected concentration at equilibrium, and Q – S = is the concentration of the solute in excess of that found in a saturated solution.  As the particle size of the precipitate increases, the impurities decrease and the ease of filtration increases.  The particle size of the precipitate is inversely proportional to the relative supersaturation.  At large RSS, the rate of nucleation is much greater than the rate of growth and large number of small particles is formed. 232  At low RSS,  the rate of particles growth tends to predominate and deposition of solid on existing particles occurs to the exclusion of further nucleation, a crystalline suspensions results. Techniques to minimize (to increase the particle size of precipitate) RSS  Precipitation from a very dilute solution (Q is kept small)  Slow addition of the precipitating reagent (Q is kept small)  Precipitation from a hot solution or increasing the precipitate’s solubility  In general, a precipitate’s solubility usually increases at higher temperatures, and adjusting pH may affect a precipitate’s solubility if it contains an acidic or basic anion.  Temperature and pH, therefore, are useful ways to increase the value of S. 233  Colloidal suspensions are often stable for indefinite periods and are not directly usable for gravimetric analysis because their particles are too small to be readily filtered.  These particles can be converted in to filterable solids by the process of coagulations or agglomeration  Coagulation can be induced in two ways:  by increasing the concentration of the ions responsible for the secondary adsorption layer  by heating the solution. 234 Steps in Gravimetric Analysis  The steps required in gravimetric analysis, after the sample has been dissolved can be summarized as follows: 1. Preparation of the solution  it is necessary to eliminate interfering materials.  the solution must be adjusted to maintain low solubility of the precipitate and to obtain it in a form suitable for filtration.  the pH of the solution must be adjusted because it influences the solubility of the precipitate and the possibility of the interferences from other substances. Example: CaC2O4 is insoluble in basic medium, but soluble at low pH because oxalate ion combines with H+ to form oxalic acid. 235 2. Conditions for analytical precipitation  The precipitate should first be sufficiently insoluble that the amount lost due to solubility will be negligible.  It should consist of large crystals so that they can be easily filtered.  Larger crystals have less chance of adsorbing impurities.  When a precipitate is performed, a slightly excess of precipitating agent is added to decrease the solubility of the substance (common ion effect) and to assure complete precipitation. 3. Digestion of the precipitate  Digestion is the process in which a precipitate is heated for an hour or more in the solution from which it was formed (the Mother liquor-the liquid remaining after a substance has crystallized out).  When a precipitating agent is allowed to stand in the presence of the mother liquor, the large crystals grow at the expense of the small ones. The small particles tend to dissolve and reprecipitate on the surfaces of the larger crystals. 236  When coagulated particles are filtered, they retain the adsorbed primary and secondary ion layers along with the solvent. If washed in water, the extent of solvent (water) molecules between the layers is increased, causing a secondary layer to be loosely bound and the particles revert to the colloidal state. This process is called peptization. How can one remove impurities from the solution during precipitation?  Soluble compounds which cause contaminations of the precipitate are removed out of the solution by the process called co-precipitaion. 237  There are different ways of co-precipiation: a/ Surface adsorption Soluble compounds are carried out of the solution on the surface of coagulated colloidal. b/ Occulation A material that is not part of the crystal structure is trapped with a crystal. Example. Water may be trapped in pockets when AgNO3 crystals are formed and this can be driven off by melting. c/ Post precipitation It is the process where by a precipitate is allowed to stand in contact with the mother liquor, a second substance will slowly form a precipitate with the precipitating agent. Example CuS will precipitate with acid solution in the presence of Zn2+, but eventually ZnS will precipitate. 238 4. Washing and Filtering  The precipitate is often washed to remove impurities adsorbed onto the surface of the particles.  Washing may be done with a solution of the precipitating agent, to avoid re-dissolving a slightly soluble salt.  With many precipitates, peptization occurs during washing. Here part of the precipitate reverts to the colloidal form [e.g. AgCl(colloidal) AgCl(s)]. This results in the loss of part of the precipitate because the colloidal form may pass through on filtration.  Peptization can be reduced with careful technique and washing with a solution of appropriate pH and ionic strength.  Co-precipitated impurities can be removed by washing the precipitate after filtering. When a precipitate is washed, a test should be made to determine when the washing is complete. 239 5. Drying or igniting the precipitate  The collected precipitate that is to be in a form suitable for washing must be heated to remove water and the adsorbed electrolyte from the wash liquid. 6. Weighing  After the precipitate is allowed to cool (preferably in a desiccator to keep it from absorbing moisture), it is weighed (in the crucible). The mass of the crucible is subtracted from the combined mass, giving the mass of the precipitated analyte.  Since the composition of the precipitate is known, it is simple to calculate the mass of analyte in the original sample. 240 7. Gravimetric Calculations The results of a gravimetric analysis are generally computed from two experimental measurements: The mass of sample; and the mass of a product of known composition When the product is the analyte X, the concentration is given by GF =  g  massofanalyte    mol  x amoleofanal yte   g  bmoleofprec ipitate  massofprecipitate    mol  = g of sample g of precipitate Where GF= gravimetric factor which states the relationship between chemically equivalent amounts of two substances. a and b are the number of moles in the numerator and denominator, respectively. In Gravimetric analysis, the amount of a substance is generally expressed in terms of percent composition by weights of the analyte in the sample. % substance sought = wt of substance sought (g) wt of sample (g) 241 wt of substance sought (g) = wt of precipitate(g) x f.wt of substance sought (g/mol) f.wt of precipitate(g/mol) = wt of precipitate (g) x GF (g sought/ g precipitate) % sought = wt of precipitate (g) x GF (g sought/ g precipitate) x 100 wt of sample (g) Example. Calculate the grams of analyte per gram of precipitate for the following conversion Analyte P K2HPO4 Bi2S3 precipitate Ag3PO4 Ag3PO4 BaSO4 242  Solution a/ P (g) = At. Wt of P (g/mol) x 1 (mol P) Ag3PO4 (g) F.wt Ag3PO4(g/mol) 1 (mol Ag3PO4) = 31 g/mol x 1 = 0.0435 g P/g Ag3PO4 = GF 711 g/mol 1 That is, every gram (1 g) of Ag3PO4 contains 0.0435 g of P. b/ g K2HPO4 = F.wt of K2HPO4(g/mol) x 1 (mol K2HPO4) F.wt = formula weight g Ag3PO4 F.wt of Ag3PO4(g/mol) 1 (mol Ag3PO4) = 136 g/mol x 1 = 0.19135 g K2HPO4 711 g/mol 1 g Ag3PO4 c/ g Bi2S3 = F.wt of Bi2S3 (g/mol) x 1 = 514.15 x 1 = 0.73429 g Bi2S3 = GF g BaSO4 F.wt of BaSO4 (g/mol) 3 233.40 3 g BaSO4 243 Example: A 0.4960 g sample of a CaCO3 is dissolved in an acidic solution. The calcium is recipitated as CaC2O4.H2O and the dry precipitate is found to weigh 0.6186 g. What is the percentage of CaO in the sample? Solution wt of CaCO3 sample = 0.4960 g, wt of CaC2O4.H2O = 0.6186 g Unknown: % CaO in the 0.4960 g sample of CaCO3 Wt of CaO = (wt of CaC2O4.H2) x 1 mole CaO x 1 mole Ca FWCaO 1 mole Ca 1 moe CaC2O4.H2O FW precipitate = (0.6186) 1 mole CaO 56.08 (g/mol) 1 moe CaC2O4.H2O 146.12 (g/mol) Wt of CaO = 0.2374 g % CaO = wt of CaO x 100 = 0.2374 g x 100 = 47.87 % wt of sample 0.4960 g 244 245