Equilibrium, Acid Base Chemistry

Equilibrium, Acid Base Chemistry & Aqueous Equilibria Dynamic Nature of Chemical Equilibria • Chemical equilibria occurs when the forward rate of the reaction (kf) is equal to the reverse rate of the reaction (kr). • Substances are being consumed & made at the same rate. • Processes in both the forward and reverse direction are said to be reversible. 3H2(g) + N2(g) 2 NH3(g) Equilibrium is established • The numerical value of Kc, the equilibrium constant, is determined from the equilibrium expression (see below). • It is a mathematical ratio between products & reactants. Kc = [NH3] 2 [H2]3 [N2] products reactants • Its numerical value is stoichiometically dependent. • It is a unit less value. • It is influenced temperature. Reference: imgarcade.com • Only equilibrium concentrations (M) or pressures (atm) values are to be entered into an equilibrium expression to calculate Kc values. The Relationship Between Equilibrium Constants & Chemical Reactions • Given the following chemical reaction and its equilibrium expression: CO(g) + H2O(g) CO2(g) + H2(g) [CO2] [H2] Kc forward = [CO][H2O] Kc forward = 0.64 • The reverse of the above chemical reaction: CO2(g) + H2(g) Kc reverse = CO(g) + H2O(g) Kc reverse = 1.56 [CO] [H2O] [CO2][H2] • When the chemical reaction is reversed, the equilibrium constant is inverted. 1 1 Kc reverse = Kc reverse = = 1.56 Kc forward 0.64 Equilibrium & Equilibrium Constants • The numerical value of Kc indicates whether the chemical reaction is product favored or reactant favored. • Kc>1: For reactions that reach equilibrium AFTER MOST of the REACTANT molecules have been CONSUMED. • This indicates that the position of EQUILIBRIUM FAVORS the formation of PRODUCTS (Kc>1). • If the numerical value of Kc >> 1, there will be many more product molecules present than reactant molecules. • Kc< 1: For reactions that reach equilibrium when only a SMALL percentage of the REACTANT molecules are CONSUMED. • This indicates that the position of EQUILIBRIUM FAVORS the REACTANTS (Kc<1). • When the value of Kc << 1, there will be many more reactant molecules present than product molecules. • Kc = 1: For reactions that reach equilibrium when the concentrations of the reactants and product remain CONSTANT (K=1). • This DOES NOT MEAN the concentrations of the reactants and products are of equal amounts (e.g., concentrations or pressures). Homogeneous Equilibria vs. Heterogeneous Equilibria • Homogeneous equilibrium: • Chemical reactions occurring between solutes in liquid solutions and/or gaseous state. • Reactants and products are in the same phase (gas or aqueous). • Example: CO(g) + H2O(g) CO2(g) + H2(g) [CO2] [H2] • Kc = [CO][H2O] • Heterogeneous equilibrium: • Chemical reactions where reactants and products are in different phases. • The only phases ALLOWED in equilibrium mass action expressions are aqueous (aq) and/or gaseous (g). • Concentrations/amounts of pure solids (s) & pure liquids (l) in solution do not change during a chemical reaction. • Because their concentrations do not change, solids and liquids are NOT included in the equilibrium expression. • Example: PbCl2 (s) Pb2+ (aq) + 2Cl- (aq) • Kc = [Pb2+] [Cl−]2 Problem: Write an equilibrium mass action expression for the following chemical reactions. 1. AgCl (s) Ag+ (aq) + Cl- (aq) 2. CaO (s) + CO2 (g) 3. 2H2O(l) CaCO3(s) H3O+ (aq) + OH- (aq) 4. C(s) + 2S (g) CS2(g) Determining a Kc value for Unknown Chemical Reaction Equilibrium constant numerical values can be determined two ways. 1. Using the mass action expression and the equilibrium concentrations of reactants and products of the chemical reaction. 2. Combining two or more known chemical reactions and their equilibrium constants to obtain the unknown chemical reaction’s constant. NOTE: K constants follow the operational rules of multiplication and division. Problem: Calculate the equilibrium constant numerical value and determine whether the reaction is at equilibrium, product favored, or reactant favored. Given the following: 2 SO3(g) 2 SO2(g) + O2(g) at 25oC Equilibrium concentration: [SO3] = 1.1M, [SO2] = 0.90M & [O2] = 0.65M Problem: Given the following chemical reactions with their equilibrium constants, predict the numerical value of the equilibrium constant (Kc3)for the reaction: 2NO2(g) + Cl2(g) Given: 1. ClNO2 (g) + Cl(g) 2. 2NO2 (g) + 2Cl(g) 2ClNO2(g) Kc3 = ? NO2 (g) + Cl2(g) 2ClNO2 (g) Kc1 = 0.33 Kc2 = 4.24 Strategy: Given the following chemical reactions with their equilibrium constants, predict the numerical value of the equilibrium constant (Kc3)for the reaction: 2NO2(g) + Cl2(g) 2ClNO2(g) Kc3 = ? Given: 1. ClNO2 (g) + Cl(g) 2. 2NO2 (g) + 2Cl(g) NO2 (g) + Cl2(g) 2ClNO2 (g) Kc1 = 0.33 Kc2 = 4.24 1. Equation #1 needs to be reversed so that the reactants and products are in the correct order and Kc1 must reflect this change. Kc1 value is now 1/Kc1 2. Equation #2 reactants and products are written in the proper order, but their amounts are twice that is needed to obtain the desired chemical reaction Kc3. To reflect this change Kc2 value will be the its square root. Kc2 value is now √Kc2 3. To obtain Kc3, multiple (1/Kc1) x (√Kc2) Q, The Quotient Constant vs. K, The Equilibrium Constant • Kc is the equilibrium constant that is associated with a reaction that is at equilibrium. • Its numerical value is computed from the equilibrium concentrations of the reaction’s products and reactants and the mass action expression. aA + bB cC + dD [C]ceq [D]deq Kc = a [A] eq][B]beq • Qc is the reaction quotient value. • Its numerical value is calculate using a Q expression and the concentrations of the reaction’s products and reactants. [C]c [D]d aA + bB cC + dD Qc = = a b [A] ][B] Q, The Quotient Constant vs. K, The Equilibrium Constant • To determine if a reaction concentrations are equilibrium concentration a Q constant is calculated. • The calculate reaction quotient value is Q to compare with Kc numerical value to determine if the reaction is at equilibrium. • If Q = K, then reaction is at equilibrium. • If Q < K, then the reaction is not at equilibrium and will need to shift to the product side (right). • If Q > K, then the reaction is not at equilibrium and will need to shift to the reactant side (left). Problem: Given the following information, determine whether the reaction is at equilibrium and if not, what direction must the reaction shift to achieve equilibrium. CO(g) + H2O(g) CO2(g) + H2(g) Kc = 0.64 Reactants/ Products Experiment #1 (M) Experiment #2 (M) CO 0.0203 0.011 H2O 0.0203 0.011 CO2 0.0040 0.037 H2 0.0040 0.046 Kp: The Equilibrium Constants for Reactions Involving Gases • The concentration of a gas in a mixture can be given as a concentration (M) or as a partial pressure (atm). • If it is given in atm, then a Kp rather than a Kc constant is used. • For the reaction: aA(g) + bB(g) cC(g) + dD(g) then Kp = 𝐶 𝐴 𝑐 𝑎 𝐷 𝐵 𝑑 𝑏 • The mathematical relationship between Kp & Kc is: Kp = Kc x (RT)Dn • Where Dn is defined as the (S moles of products – S moles of reactants) • The values of Kp and Kc will usually have different numerical values but will follow the same trend. • Meaning if Kc is less than 1 so will Kp value. • The values of Kp and Kc will have identical numerical values (i.e., Kp = Kc) when Dn = 0 Problem: The Kc value for CS2(g) + 4H2(g) CH4(g) + 2H2S(g) at 900 oC is 0.28. 1. Determine the Kp value for this reaction under the same conditions. 2. Is the reaction product favored or reactant favored. Justify answer 3. Determine the Kp value for 2CH4(g) + 4H2S(g) 2CS2(g) + 8H2(g) at 900oC. Le Châtelier’s Principle: Shifting Equilibria • Le Châtelier’s principle examines and predicts how a system, at equilibrium, responses to a change in experimental conditions. • The chemical reaction will either shift (to product side or to the reactant side) to alleviate (i.e. move away from the stress). • A system at equilibrium can be altered by changes in its: • Temperature • Effects Kc • Concentration (M) • Pressure – for gases in atmospheres (atm) • The outcome on the system’s equilibria to these external changes (i.e., stresses) is governed by LE CHÂTELIER’S PRINCIPLE. “...If a system at equilibrium is disturbed, the system tends to shift its equilibrium position to counter the effect of the disturbance.” Le Châtelier’s Principle: Temperature Changes • Changes in temperature not only affect the chemical reaction’s equilibria but causes the numerical value of Kc to change. • Consider the amount of fizz (e.g., the escape of carbon dioxide) in a can of soda. • An increase in temperature puts a stress on the reactant side. CO2 (aq) + HEAT CO2(g) + H2O(l) Kp = P(CO2) / [CO2] • The system will move away from the stress, shifting right towards the product side of the reaction. • P(CO2) increases as [CO2] decreases due to the escape of carbon dioxide from the solution. • The equilibrium ratio changes so Kp value changes. Le Châtelier’s Principle: Temperature • Temperature: • Endothermic reaction: heat is treated as a pseudo reactant, since energy is entering the system. • An increase in temperature puts a stress on the reactant side, so the system will move away from the stress and shift right to the product side. • A decrease in temperature puts a stress of the reactant side by depleting the heat so the reaction will shift to the reactant side (left) to replenish. • Exothermic reaction: heat is treated as a pseudo product for energy is leaving the system. • An increase in temperature puts a stress on the product side, therefore the system will move away from the stress and shifts left to the reactant side. • A decrease in temperature puts a stress of the product side by depleting the heat so the reaction will shift to the product side (right) to replenish. Le Châtelier’s Principle: Concentration, Pressure & Catalysts • Concentration: • Changes in reactants and/or products concentrations do not change the equilibrium constant value (Kc ) for the ratio is maintained. • However, decreasing or increasing the concentration of the reactants and/or products does change the equilibrium position of the chemical reaction. • The reaction will establish a new equilibrium position by shifting to alleviate the stress. • Pressure: • Changes in reactants and/or products pressures do not change the equilibrium constant value (Kp) for the ratio is maintained. • The equilibrium position of the reaction does change. • An increase in pressure or a decrease in volume will cause the system to shift to the side with the FEWEST gas moles. • A decrease in pressure or an increase in volume will cause the system to shift to the side with the MOST gas moles. • Catalysts: • Do NOT affect a system’s equilibrium position or the equilibrium constant numerical value. • No shift will occur. • A catalyst only affects the RATE of approach to equilibrium. Table: Summary of Changes Stress/Disturbance Observed changes for a new equilibrium position to be obtained Direction of shift to obtain new equilibrium position. Effect on the equilibrium constant numerical value (Kp or Kc) Reactant added. Added reactant is partially consumed. Towards products (right). No effect. Product added. Added product is partially consumed. Towards reactants (left). No effect. Increase in pressure (i.e., decrease in volume) of gas. Pressure decreases. Shifts towards the side of the reaction with the fewest moles of gas particles. No effect. Decrease in pressure (i.e., increase in volume) of gas. Pressure increases. Shifts towards the side of the product with the greatest number of moles of gas particles. No effect. Temperature increase Heat is absorbed. Endothermic: towards the product side (right). Exothermic: towards the reactant side (left). Changes. Temperature Decrease Heat is given off. Endothermic: towards the reactant side (left). Exothermic: towards the product side (right). Changes. Problem: Given the following chemical reaction at equilibrium and room temperature, 25 oC: 2H2(g) + CO(g) ⇌ CH3OH(g) ΔH = −90.2 kJ a. Write the equilibrium expression for the reaction. b. Predict the direction the reaction will shift if more H2 is added. c. Predict the direction the reaction will shift if CO is removed. d. Predict the direction the reaction will shift if CH3OH is removed. e. Predict the direction the reaction will shift if the volume is decreased. f. Predict the direction the reaction will shift if the temperature is increased. g. Predict the direction the reaction will shift if Xenon is added to the system. h. Predict the direction the reaction will shift if a Pt catalyst is added to the system. Equilibrium Calculations Equilibrium calculations can be grouped into three different types. • 1st type: Solve for the Kc or Kp numerical value. • The equilibrium concentrations (M) or pressures (atm) for reactants & products for reaction are given. • 2nd type: Solve for the equilibrium concentrations (M) or pressures (atm). • The Kc or Kp numerical value for a reaction is given. • 3rd type: Determining whether the reaction is at equilibrium. • The concentrations (M) or pressures (atm) for reactant & products and corresponding Kc or Kp numerical value are given. • Calculate reaction quotient value Q to compare with Kc numerical value to determine if the reaction is at equilibrium. • If Q = K then reaction is at equilibrium. • If Q < K: then the reaction is not at equilibrium and will need to shift to the product side (right). • If Q > K: then the reaction is not at equilibrium and will need to shift to the reactant side (left). Initial, Change & Equilibrium: ICE Strategy • ICE is a problem-solving strategy that can be used whether you are solving for equilibrium concentrations of products and reactants when given the Kc value or when determining whether a chemical reaction is at equilibrium (Q problems). • It is a way to organize data in a logical format. • The format that is used is referred to as ICE tables. Given: A(aq) ⇌ 2B(aq) where Initial [A] = 3.0 M Initial Change Equilibrium [A] [B] 3.0 0 -x +2x 3.0 - x +2x The symbol x represents changes in the chemical species from their initial conditions in order to reach their equilibrium concentrations. Problem 1: Determine the equilibrium constant value (Kc) given the following information: I2 (aq) + I- (aq) I3- (aq) at 25 oC [I-]o = 0.001M [I2]o = 0.001M [I3-]o = 0 M [I2]eq = 6.61 x 10-4 M Problem 2: Determine the equilibrium concentration of NO given the following information: N2 (g) + O2 (g) 2 NO(g) at 25 oC Kc = 4.1 x 10-4 [N2]eq = 0.036M [O2]eq = 0.0089M [NO]eq = ?M Problem 3: Given the information below, determine: a. Whether or not the reaction is at equilibrium? b. If it is not at equilibrium what direction must it shift. c. Set up an equilibrium expression need to determine the equilibrium concentrations of all chemical species. Given: Conditions: 2 H2S (g) 2 H2 (g) + S2 (g) 0.105 atm = (H2) 0.075 atm = (S2) 0.375 atm = (H2S) Kp = 1.85 x 10-4 at 750 oC Approximation Equilibrium Calculations • Using the quadric equation is not a simple task when calculating equilibrium constants. • Approximations: • When the equilibrium constant numerical value is less than one (K < 1), the position of equilibrium favors the reactants. • For relatively large initial concentrations of reactants, the reactant concentration will not change significantly when it reaches equilibrium. • [A]equilibrium = ([A]initial − x)  [A]initial • Assuming that the reaction will be proceeding forward, an approximation can be used to simplify the math needed to solve the problem. approximated value of x ( ) x 100 < 5% initial concentration • If the approximate value of x is less than 5% of the initial concentration, the approximation is valid. Problem: The equilibrium concentrations of chemical species for a 0.15 M HCN (aq) solution. HCN(aq) ⇌ H+(aq) + CN−(aq) Kc = 4.9 × 10−10 Everyday Acid & Bases Acids • Citric acid • Formic acid • Sea slug venom (sulfuric) Bases • Caffeine • Codeine • Theobromine • (chocolate) Reference: bewellbuzz.com; www.math-science-resources.com; www.owlcation.com; Chemistry.stackexchange.com; www.rxlist.com;chemistry.about.com;www.isowhey.com.au Acid Strengths: Strong versus Weak Strong Acids or Bases Weak Acids or Bases • The term strong implies COMPLETE (~100%) dissociation in of the molecule in water. • Strong electrolyte • The term weak implies PARITIAL (< 5%) dissociation in water. • Weak electrolyte • Example: Strong acid ionization: HNO3(aq) + H2O(l) → H3O+ + NO3- • Example: Weak acid ionization: • There are SIX strong acids: HCl, HBr, HI, HNO3, H2SO4 & HClO4 Example: Strong base ionization: Ca(OH)2 (aq) + → 2 OH- + Ca2+ • Metal hydroxide compounds are strong bases. CH3COOH(aq) + H2O(l) H3O+ + CH3COO- • Example: Weak base ionization NH3 (aq) + H2O (l) NH4+ + OH- Brönsted-Lowry Acid & Base Chemistry • A Brönsted-Lowry acid donates a proton (H+ ion) in water. • ACIDS are species DONATE H+ IONS. • A Brönsted-Lowry base will accept a proton (H+ ion). • BASES are species that ACCEPT H+ IONS. • In acid–base reactions: • Base species that pick up a hydrogen ion form a CONJUGATE acid. • Acid species that lose a hydrogen ion form a CONJUGATE base. Reference: chemistry.com.pk Problem: Determine the conjugate acid or conjugate base for each of the following, if any: conjugate acid conjugate base a. H2O b. HCO3c. C2H5NH d. NH3 e. HSf. HSO3g. H3PO4 Answer: species conjugate acid conjugate base a. H2O H3O+ OH- b. HCO3- H2CO3 CO32- c. C2H5NH C2H5NH2+ C2H5N- d. NH3 NH4+ NH2- e. HS- H2S S2- f. HSO3- H2SO3 SO32- g. H3PO4 does not have a conjugate acid HPO4- Water: Autoionization & Amphoterism Autoionization: when a substance, such as water, reacts with itself. 2 H2O (l) H3O+(aq) + OH-(aq) Reference: Chemistry.libretexts.org • Kw is the ion product the describes the autoionization of water. • Kw = [H3O+][OH-] = 1 x 10-14 at 25 oC • Process is endothermic; increasing temperature results in an increase in ion concentration and the Kw value (i.e., extent of ionization increases). • Water is an amphoteric substance, because it can either donate a proton (i.e., H+ ion) or accept a proton Solvent Leveling Effect • Water is a very weak electrolyte (Kw = 1 x 10-14 at room temperature) • Only about two out of every 109 molecules in a sample of pure water are ionized (i.e., autoionize) at 25 °C. auto ionization... 2 H2O (l) H3O+(aq) + OH-(aq) • The strongest acid species present in water (i.e., solvent) is the hydronium ion, H3O+ (i.e. ,hydrogen ion, H+). • The strongest base species present in water (i.e., solvent) is the hydroxide ion, OH• This is referred to the leveling effect of the solvent. Polyprotic Donors: Acids with more than one ionizable hydrogen atoms. • Polyprotic acids are acids that have more than one hydronium ion (proton or H+) that dissociates in water. • Each H+ that dissociates has its own Ka (pKa) value. • First Ka1 >>>> Ka2 • Ka1 is the result of the first H+ ion dissociating from a neutral compound. • Ka2 and Ka3 are the next H+ ions to dissociate. The second hydrogen ion is dissociating from an anion (negatively charged species) and requires more energy, and the third hydrogen ion dissociates from the anion of the second ionization also requiring more energy than the previous dissociations. • The ionization of these subsequent hydrogens results in succeeding smaller Ka values. • Examples: H2CO3 H3PO4 H2SO4 H2SO3 H3PO3 carbonic acid phosphoric acid sulfuric acid sulfurous acid phosphorous acid weak acid weak acid strong acid weak acid weak acid Poly Bases: Bases with the ability to accept more than one hydrogen atom. • Poly bases are base that have the ability to accept more than one hydrogen atom. • Each H+ that the base species accepts has its own Kb (pKb) value. • First Kb1 >>>> Kb2 • Kb1 is the result of the first H+ ion being accepted from a neutral compound. • Kb2 is the next base ionization values associated with the acceptance of a H+ ion. • The second hydrogen ion accepted is from the cation (positively charged species) The ionization of these subsequent hydrogens results in larger Kb values. • Example: Na2CO3 (aq) CO3 2- (aq) (weak base) + H2O (l) HCO3- (aq) (conj. acid) + OH-(aq) Kb1 = 2.3 x 10-8 HCO3- (aq) (weak base) + H2O (l) H2CO3 (aq) (conj. acid) + OH-(aq) Kb2 = 2.1 x 10-4 What is pH? What is pOH? • It is a measurement of the acidity (pH) or basicity (pOH) of a solution based on the concentration (e.g., amount) of H3O+ ion (acidity) present or OH- ion (basicity) present. • pH: The lower the pH number value, the more acidic the solution; the higher the pH value, the more basic the solution. • 1 pH unit corresponds to a factor of 10 difference in acidity. • pOH: The lower the pOH number value, the more basic the solution; the higher the pOH value, the more acidic the solution. • 1 pOH unit corresponds to a factor of 10 difference in basicity. Reference: acidsandbases8.weebly.com Formulas for pH & pOH calculations pH pOH • pH = −log[H3O+] • pOH = −log[OH−] • [H3O+] = 10−pH • [OH−] = 10−pOH • Kw = [H3O+] [OH-] • 14 = pH + pOH where Kw = 1 x10-14 Problem: A 0.0020 mol of Be(OH)2 is added to of pure water to make a 500.00 mL solution. Calculate [H3O+], [OH-], pH and pOH of the solution. Acid & Bases: Strength • The relative strength of acids and bases is indicated by their ionization constant; Ka & Kb respectively. • The larger the Ka & Kb value the stronger the acid or base. • Acids: HA (aq) + H2O (l) • Bases: B (aq) + H2O (l) A-(aq) + H3O+(aq) A− [H3O+] Ka = [HA] BH+(aq) + OH-(aq) BH+ [OH−] Kb = [B] • To solve for the pH and equilibrium concentrations of weak acids or weak bases you must use the ICE strategy. Determining pH of an Aqueous Solution • Experimentally, the pH of a solution can be determined using pH paper, litmus paper and/or pH electrode-meter set-up. • pH values of strong acids and strong bases DO NOT HAVE to be determined using Ka & the ICE strategy or Kb & the ICE strategy. • Reason: strong acids and strong bases completely ionizes (~100%) in water. Therefore, the concentration of the acid or base is also the concentration of the H3O+ and OH- ions, respectively. • The theoretical pH of an aqueous solution of weak acids or weak bases can be determined using Ka & the ICE strategy. • To solve for their pH values, you need only use the appropriate pH or pOH formula. • pH = -log [H+] or pOH = - log [OH-] Problem #1: Determine the hydronium ion and pH of formic acid. Given: formic acid (HCOOH) initial concentration is 0.534 M HCOOH (aq) + H2O (l) Ka = 1.8 x 10-4 H3O+(aq) + HCOO-(aq) Problem #2: Determine the pH of trimethylamine. Given: Trimethylamine ((CH3)3N) initial concentration is 0.250 M (CH3)3N (aq) + H2O (l) Kb = 6.3 x 10-5 (CH3)3NH+(aq) + OH-(aq) Hydrolysis of Salts & pH: Neutral & Basic Salt Solutions • A salt’s composition dictates the pH of the solution formed from a neutralization reaction. • The salt solution’s is determined by the origins of the salt’s cation& anion components. • Remember: • Strong acids and strong bases produce relatively WEAK conjugates. • Weak acids and weak bases produce relatively STRONG conjugates. • Neutral Salt solution • The cation is the conjugate acid of a strong base. • The anion is the conjugate base of a strong acid. Examples: KCl • Basic Salt solution • The cation is the conjugate acid of strong base. • The anion is the conjugate base of a weak acid Examples: BeF2 • Acidic Salt solution (case 1) • The cation is the conjugate acid of weak base. • The anion is the conjugate base of a strong acid. Examples: NH4Cl • Acidic Salt solution (case 2) • The is a highly charged metal ion. • The anion is the conjugate base of a strong acid. Example: Fe(NO3)3 Mg(NO3)2 LiBr Na(C2H3O2) Ca3(PO3)2 Problem: Predict the pH range for the following salt solutions. a. KBr b. NaHCO3 c. NH4Cl d. K2HPO4 e. FeBr3 f. LiCN Hydrolysis of Salt Solution: Quantitative pH • When a strong acid and strong base react, the salt solution that is formed will have a pH of 7. • Reason: the cation and anion of the salt formed is neutral. The pH of the solution is due to the autoionization of water. • To quantitatively determine the pH of a non neutral salt (i.e. acidic or basic) the strategy ICE must be used. • The ionization equation used for ICE will be the salt’s interaction with water. • The ionization constant Ka or Kb used depends on the salt. • If the salt is basic then Kb is used • If the salt is acidic then Ka is used. • To go from Ka to Kb or Kb to Ka the relationship Ka x Kb = Kw is used. Problem: Aniline (C6H5NH2) is an amine that is used to manufacture dyes. It is isolated as aniline hydrochloride, (C6H5NH3+)Cl, as a salt that is prepared by reacting aniline with hydrochloric acid. Determine the pH of a 0.233 M aniline hydrochloride solution. Given: Kb = 4.6 x 10-10 Buffer Solutions • Buffer implies a mitigation device or action which lessen the effect between two opposing forces. Acidic Acid Buffer: weak acid – salt of its conjugate base • A chemical buffer is a solution whose composition (i.e., weak acid and its conjugate base salt) aids in maintaining the solution’s pH when extra acid or base is added. • Chemical buffers play an important role in biological systems (i.e., regulating the acidity of blood and kidney system) and ecosystems (i.e., maintain seawater and freshwater acidity). • The conjugate acid-base pairs’ interaction with the added acid (H+) or base (OH-) is why a buffer maintains its pH when additional acid or base is added to the solution. Reference: collective.chem.cmu.edu • How efficient a chemical buffer can resist changes in its pH is dependent on the concentration (molarity (M)) of each of the buffer solution’s unique components: • weak acid & salt of its conjugate base (for an acidic buffer) • weak base and the salt of its conjugate acid. • It is the ratio of this conjugate acid-base pair that is the key behind the workings of a buffer. Note: HAc is acetic acid & Ac- represents the acetate ion Reference: Chem libretexts.org Chemistry behind the workings of a Buffer • Adding acid or base to the buffer puts a “stress” on the system and according to Le Chatelier’s’ Principle, a system under stress moves away from the stress obtaining a new equilibrium position. • When additional base (OH- ion) is added, it is the weak acid (HAc) component of the buffer that counteracts the effect whereas if acid (H+ ion) is added to the buffer, it is the conjugate base (Ac-) ion which neutralizes the effect. • OH- + (added) • H+ + (added) HAc → (wa) Ac- → (cb) H2O (l) + Ac(cb) HAc (wa) • The system establishing a new equilibrium position, maintaining its pH. • This “attack/counterattack” mechanism that is required to maintain the solution’s pH will occur if sufficient quantities of each of the buffer components (i.e., weak acid and its conjugate base) exist in solution. • Once one of the buffer components concentration drops to insufficient levels, the buffering action ceases. Henderson-Hasselbalch Relationship: [𝐁𝐮𝐟𝐟𝐞𝐫 𝐁𝐚𝐬𝐞] pH = pKa + log [𝐁𝐮𝐟𝐟𝐞𝐫 𝐀𝐜𝐢𝐝] • The Henderson-Hasselbalch equation is a mathematical relationship that can be used to calculate the molarities of each of the buffer components required to make a buffer solution at a given pH. • The Henderson-Hasselbalch equation is ONLY to be used for determining the pH of a buffer solution or at the ½ neutralization point of a titration curve. • The Henderson-Hasselbalch Equation is not to be used when the problem is asking for you to determine the pH for a non buffer solution. • The Henderson–Hasselbalch equation works when the ‘x’ is small” approximation is applicable. • Criteria that needs to be met: • The initial concentrations of acid and salt are not very dilute. • The Ka is small. NOTE: The ICE strategy ALWAYS WORKS. Problems: Determine the pH given the following information: a solution of 0.075M HNO2 & 0.030M NaNO2. Ka = 4.5×10−5 Problem: You want to buffer a solution at pH = 9.30. Given the following possibilities, select which combination would meet this criteria. Possible conjugate acid-base pairs available Ka KHSO4 / K2SO4 CH3COOH / NaCH3COO HCN / LiCN 1.2 x 10-2 1.8 x 10-5 4.0 x 10-10 Double Displacement & Ksp Connection • Double Displacement reaction: • Example: AB + CD AD + CB • Al(NO3)3 (aq) + 3 KOH (aq) Al(OH)3(s) + KNO3 (aq) • Net ionic: Al3+ (aq) + 3OH- (aq) Al(OH)3(s) • Ksp is the equilibrium constant for the dissolution of a slightly soluble ionic solid into its ions. • Ksp chemical reaction: Al(OH)3(s) Al3+ (aq) + 3OH• Equilibrium expression: Ksp = [Al3+][OH-]3 Ksp & Solubility • Solubility of a compound is the amount of solute that will completely dissolve in a given volume of solvent (water) at a specified temperature. • Molar solubility of a compound is defined as the number of moles (amount of solute) that will dissolve in ONE liter of water (solvent). • In other words, it is the molarity (M) of a saturated salt solution. • The numerical value for the solubility product constant Ksp is about the maximum ion concentrations (M) for a saturated solution to occur. • Calculated numerical values(Qsp) GREATER than the actual Ksp value indicate formation of a solid (precipitate). • Generally, the larger the Ksp value (i.e., negative value, 10-x) the more insoluble the salt. • Ksp values can be compared but ONLY if the stoichiometric dissolution is the same. • Example: PbI2 vs. CuCl2 YES Cu3(SO4)2 vs. PbBr2 NO Ksp Calculations Types • There are basically four types of Ksp calculation problems. • 1st: Determining the numerical value for Ksp when given molar solubility values (Molarity) of the product(s). • 2nd: Determining the molarity of each of the products when given Ksp. • 3rd: Determining whether a solution is saturated, unsaturated or supersaturated when given the molar solubility values of the products and the Ksp value for the reaction. • This is a Qsp problem, similar to the Q problems worked in the equilibrium chapter. • 4th: Common ion effect on solubility when given Ksp value. Problems: Determining Ksp or Calculating molar solubility values. 1. CaF2 is the mineral fluorite. In a saturated solution, the equilibrium concentration of Ca2+ is 2.15 x 10-4 and for F- ion it is 4.30 x 10-4 what is the solubility product value for CaF2(s) . 2. Determine the solubility product value for Mg(OH)2 (s) given that the equilibrium concentration of Mg2+ is 1.31 x 10-4. 3. The Ksp for Ca(OH)2(s) is 1.3 x 10-6. Determine the molar solubility for Ca(OH)2 at equilibrium. Precipitation • Precipitation is when the concentrations of the ions exceed the solubility of the ionic compound and a solid is formed. • Predicting precipitation: • Comparing Ksp to the reaction quotient, Qsp. • The Qsp value is determined from the molar solubility (e.g. concentrations) of the ions and the calculated Qsp value is compared to the Ksp value of the reaction. • Qsp calculations are similar calculation when we compared Q value to Kc in the equilibrium chapter. • Qsp vs. Ksp • Qsp = Ksp: the solution is saturated; no precipitation • Qsp < Ksp: the solution is unsaturated; no precipitation • Qsp > Ksp: the solution is above saturation; the salt above saturation will precipitate. • NOTE: Some solutions with Qsp > Ksp will not precipitate unless disturbed. These are called supersaturated solutions discussed in the Solutions chapter. Problems: The first step in the preparation of magnesium metal is the precipitation of Mg(OH)2 from sea water by the addition of lime, Ca(OH)2. The concentration of Mg2+(aq) in sea water is 0.0537 M. Will Mg(OH)2 precipitate if enough Ca(OH)2 is added to give a [OH–] is 0.0010M. Mg(OH)2(s) Ksp = 8.9×10−12