Ionic Equilibrium

INTRODUCTION  According to Conductivity Substances are of 2 types 1. Non-Conductor : Those substances which do not show the flow of current or electricity. Ex. Non - metals, plastic rubber, wood etc. Exception – Graphite is a non-metal but show conductivity due to motion of free electrons. 2. Conductors – Those substances which show conductivity or flow of current are called conductors and these are of 2 types : (a) Metallic Conductor Those conductor which show conductivity due to motion of free electrons. Eg. All metals, Graphite (b) Ionic Condutors : Those conductor which show conductivity due to movement of free ions. Ions are in free state in the solutions of ionic compounds. On passing electric current through the solution, ions move towards oppositely charged electrodes, i.e., the cation moves towards cathode (negative electrode) and the anion moves towards anode (positive electrode). Due to this reason, they are called cations and anions respectively. The current flows through the solution due to the movement of the ions. Cathode + – Anode Ag+ NO3– Movement of ions through the solution of electrolyte (AgNO3) towards oppositely charged electrodes DEGREE OF DISSOCIATION  When an electrolyte is dissolved in a solvent (H2O), it spontaneously dissociates into ions.  It may dissociate partially ( <<< 1) or sometimes completely (  1) Ex. NaCl + aq  Na+ (aq) + Cl– (aq) – + CH3COOH ‡ˆ ˆˆ ˆ† ˆˆ CH3COO (aq) + H (aq)   The degree of dissociation () of an electrolyte is the fraction of one mole of the electrolyte that has dissociated under the given conditions. No.of moles dissociated = No. of moles taken initially According to Strength Electrolyte are of 2 types : 1. Strong Electrolyte : Those ionic conductors which are completely ionized in aqueous solution are called as strong electrolyte. Ex. Na+Cl–, K+Cl–, etc. For strong electrolyte ionisation is 100%. i.e. = 1 Ex. (a) Strong acid H2SO4, HCl, HNO3 HClO4, H2SO5, HBr, HI (b) Strong base KOH, NaOH, Ba(OH)2 CsOH, RbOH (c) All water soluble salts NaCl, KCl, CuSO4.......... IONIC EQUILIBRIUM 2. Weak Electrolytes : Those electrolytes which are partially ionized in aqueous solution are called as weak electrolytes. For weak electrolytes the value of is less than one. Ex. (a) Weak acid  HCN, CH3COOH, HCOOH, H2CO3, H3PO3, H3PO2, B(OH)3,  H3BO3 (boric acid) (b) Weak base NH4OH, Cu(OH)2, Zn(OH)2, Fe(OH)3, Al(OH)3 (c) All sparingly soluble salts like AgCl, AgBr, PbSO4, etc. FACTORSAFFECTING THE VALUE OF DEGREE OF DISSOCIATION (1) Temperature  On increasing temperature ionization increases so, increases (2) Nature of electrolyte (i) Strong electrolyte (ii) Weak elecrolyte = 100 % < 100 % (3) Nature of solvent If Dielectric constant of solventincreases, then the value of increases. H2O = 81 D2O = 79 C6H6 = 2.5 CCl4 = 0 Ex. Sol. Ex. Sol. 5. Which one is greater 1 or 2 for the following equation : (i) NH4OH + H2O 1 (ii) NH4OH + D2O 2 Dielectric constant of H2O is more than that of D2O , so1> 2 Which one is greater 1 or 2 for the following equations : (i) HCN + CCl4 1 (ii) HCN + C6H6 2 (A)1 > 2 (B)2 > 1  (CCl4) = 0 and (C6H6) = 2.5 So, 2 > 1 Mixing of Ions (C)1= 2 Common Ion Effect Common Ion Effect + Ans. (B) Odd Ion Effect Odd Ion Effect NH4 OH฀ NH4 + OH + – NH4 OH ฀ NH4 + OH On mixing NH4 Cl On mixing HCl + (D)None – – NH4 Cl  NH4 + Cl + – HCl  H + Cl Due to mixing of common ion concentration Due to mixing of odd ions concentration – of ammonium ion will increase therefore of OH will decrease  Equilibrium will equilibrium will shift in backward direction shift in forward direction i.e. rate of i.e. rate of backward reaction increases forward reaction increases, means a means  decreases. increases Common Ion Effect : The decrease in degree of inonisation of a weak electrolyte in presence of a strong electrolyte having a common ion is called common ion effect. For example when HCl is added in a solution containing CH3COOH, the following equilibria exists : CH3COOH(aq.) ƒ CH3COO– (aq.) + H+ (aq.) ...(1) C - C C (C1 + C) HCl(aq.)  H+(aq.) + Cl–(aq.) C1 C1 C1 ...(2) when HCl is added the concentration of H will increase in solution and therefore equilibrium (1) will shift backward as a result of which the degree of ionisation of CH COOH will decrease + Ka = α[C1 + cα] 1– α 3 Ex. Calculate concentration of each species at equlibrium containing – (i) 0.1 M NH4OH only (ii) 5.35 g/L NH4Cl along with 0.1 M NH4OH (kb = 1.8 × 10–5) Sol. (i) NH4OH(aq) 0.1 – 0.1 ƒ NH4+(aq) + OH– (aq) 0.10.1 0.1 2   = kb  (1  ) 1.8 10 5 = 1.8 104 = 1.34 × 10–2. 0.1 kb = C   = 0.0134 [NH4+] = [OH–] = c = × 1.34 × 10–2 = 1.34 × 10–3 (ii) ƒ NH4OH 0.1 – 0.1  NH4+ + OH– (0.1  + 0.1) 0.1 NH4Cl 0.1 M NH4 + Cl 0.1 0.1 + – 5.35 g/L 53.5 g kb = 0.1(1  )  0.1 0.1(1  ) kb = 0.1(1  ).1 (1  ) a<<1 kb = 0.1 ×  = 1.8 × 10–5 = 1.8 × 10–4 = 0.00018  1 +  1 and 1 – 1 IONIC EQUILIBRIUM Ex. The solubility product of Mg(OH)2 is 1.2 x 10¯11. What minimum OH¯ concentration must be attained (for example, by adding NaOH) to decrease the Mg2+ concentration in a solution of Mg(NO3)2 to less than 1.1 x 10–10 M? Sol. Ksp expression: Ksp = [Mg2+] [OH¯]2 We set [Mg2+] = 1.1 x 10–10 and [OH¯] = x. Substituting into the Ksp expression: 1.2 x 10–11 = (1.1 x 10–10) (x)2 x = 0.33 M Any sodium hydroxide solution greater than 0.33 M will reduce the [Mg2+] to less than 1.1 x 10–10 M. Ex. Calculate the number of moles of Ag 2CrO 4 that will dissolve in 1.00 L of 0.010 M K 2CrO 4 solution. Ksp for Ag2CrO4 = 9.0 x 10-12. Sol. 1) Concentration of dichromate ion from potassium chromate : 0.010 M 2) Calculate solubility of Ag+: Ksp = [Ag+]2 [CrO42–] 9.0 x 10-12 = (x)2 (0.010) x = 3.0 x 10-5 M Since there is a 2:1 ratio between the moles of aqueous silver ion and the moles of silver chromate that dissolved, 1.5 x 10-5 M is the molar solubility of Ag2CrO4 in 0.010 M K2CrO4 solution. Since we were asked for the moles of silver chromate that would disolve in 1.00 L, the final answer is: 1.5 x 10-5 mol Weak Electrolyte : A weak electrolyte is a substance which forms ions in an aqueous solution but does not dissociate completely. When dissolved, a weak electrolyte does not disperse completely into ions. The solution instead contains both ions and molecules. Some examples of weak electrolytes are carbonic acid, acetic acid and ammonia OSTWALD'S DILUTION LAW (for weak electrolyte's)  For a weak electrolyte A+B– dissolved in water, if is the degree of dissociation then AB initial conc. conc-at eq. ‡ˆ ˆˆ ˆ† ˆˆ C A+ + B– 0 C(1 – ) 0 CC Then according to law of mass action, Ka = Cα 2 Cα.Cα [A + ][B – ] = = = dissociation constant of the weak electrolyte (1– α) C(1– α) [AB] If is negligible in comparison to unity 1 –  ; 1. so keq = 2C =  1  volume concentration k eq C mass    as concentration  volume    as concentration increases decreases at infinite dilution reaches its maximum value, unity i.e. A weak electrolyte will as a strong electrolyte at infinite dilution.   ACIDS BASES AND SALTS Arrhenius Concept Arrhenius Acid : Substance which gives H+ ion on dissolving in water (H+ donor) Ex. HNO3, HClO4, HCl, HI, HBr, H2SO4, H3PO4 etc. H3BO3 is not Arrhenius acid. (it is a lewis base) + + H+ ion in water is extremely hydrated (in form of H3O+, H5O2 , H7O3 , general form H+ (H2O)n The structure of solid HClO4 is studied by X-ray, It is found to be consisting of H3O+ and ClO4–    – + HClO4 + H2O ‡ˆ ˆˆ ˆ† ˆˆ H3O + ClO4 (better representation) Arrhenius Base : Any substance which releases OH– (hydroxyl) ion in water (OH– ion donor) – – – OH – ion is present also in hydrated form of H3O2 , H7O4 , H5O3 .general form OH– (H2O)n First group elements (except Li.) form strong bases Limitation of Ostwald Dilution Law (1) It is not applicable for strong electrolyte (2) It is not applicable for saturated solution.   Modified Arrhenius Concept It rectifies most of the above limitations (i) Water is weak electrolyte and ionises to a very weak extent. H 2O ‡ˆ ˆˆ ˆ† ˆˆ H+ + OH– H+ + H2O ‡ˆ ˆˆ ˆ† ˆˆ H3 O+ H2O + H2O ‡ˆ ˆˆ ˆ† ˆˆ H3 O+ + OH– Above reaction is called Autoionisation or selfionisation of water. (ii) Water is neutral in nature i.e. [H3O+] = [OH–] (iii) The substances which increase the H3O+ ion concentration act as acids and while those which increase OH– ion concentration act as bases. Ex. (a) SO2 + H2O  H2SO3 ‡ˆ ˆˆHˆˆ2 Oˆˆ†ˆ H3O+ + HSO3– Acid (b) NH3 + H2O  NH4OH ‡ˆ ˆˆHˆˆ2 Oˆˆ†ˆ NH4+ + OH– Base IONIC EQUILIBRIUM Basicity or Protocity of an Acid Types of bases Mono acidic base Gives single OH– per molecule e.g CsOH, RbOH NH4OH Di acidic base Gives two OH– per molecule e.g Ba(OH)2, Ca(OH)2 Tri acidic basic Gives three OH– per molecule e.g Fe(OH)3 It is number of H+ ions furnished by a molecule of an acid. An acid may be classified according to its basicity. Thus we may have, (i) Mono basic or Mono protic acids like HCl, HNO3, CH3COOH, HCN etc. (ii) Dibasic or Diprotic acids like, H2SO4, H2CO3, H2SO3, H2S etc. (iii) Tribasic or Triprotic acids like H3PO4,H3AsO4 etc. Acidity or Hydroxity of a Base Mono basic acid Gives single H+ per molecule e.g HA (HNO3, HClO4, HCl) Types of acids Di basic acid Gives two H+ per molecule e.g H2A (H2SO4, H3PO3) Tri basic acid Gives three H+ per molecule e.g H3A (H3PO4) It may be defined as the number of OH– ions furnished by a molecule of a base. A base can be, (i) Mono acidic or Monohydroxic like NaOH, NH4OH, AgOH etc. (ii) Diacidic or dihydroxic like Ba(OH)2, Mg(OH)2, Ca(OH)2, Sr(OH)2 etc. (iii) Triacidic or trihydroxic like Fe(OH)3, Al(OH)3 etc. Strength of Acid or Base (i) Strength of Acid or Base depends on the extent of its ionisation. Hence equilibrium constant Ka or Kb respectively of the following equilibria give a quantitative measurement of the strength of the acid or base. + – (ii) HA + H2O ‡ˆ ˆˆ ˆ† ˆˆ H3O + A ; Ka = [H 3O+ ][A – ] [HA] (iii) Similarly B + H2O ‡ˆ ˆˆ ˆ† ˆˆ BH+ + OH– ; Kb = [BH + ][OH – ] here H2O is szolvent. [B] K EE YYPP OOI INN TT SS (i) The other ways to represent above equilibrium is : H O 2 + – (a) HA ‡ˆ ˆˆ ˆˆ ˆ† ˆˆ H + A ; Ka = [H + ][A – ] [HA] H2O [B+ ][OH – ] + – (b) BOH ‡ˆ ˆˆ ˆˆ ˆ† ˆˆ B + OH ; Kb = [BOH] (ii) The larger the value of Ka or Kb, the more complete the ionisation, the higher the concentration of H3O+ or OH– and stronger is the acid or base. EXAMPLE BASED ON : ARRHENIUS CONCEPT Ex. The characteristics of an acid is : (A) turns blue litmus to red. (B) turns phenolphthalein pink from colourless. (C) decompose carbonates (D) oxy compounds of non-metals Sol. (A), Statement (A) indicates characteristic of acid. Ex. Arrhenius theory of acid-base is not applicable in : (A) aqueous solution (B) in presence of water (C) non-aqueous solutions (D) none of the above Sol. (C), since Arrhenius theory is only applicable to aqueous medium. Ex. Select the suitable reason (s) for higher strength of an acid or base : (A) higher value of Ka or Kb (B) higher extent of ionisation (C) (A) and (B) both (D) Larger number of replaceable H atoms. Sol. (C), Ka or Kb and degree of ionisation are the measure of strength of an acid or base. Ex. The basicity of phosphorous acid is : (A) 1 (B) 2 (C) 3 (D) 4 + Sol. (B), Phosphorous acid has two replaceable H ions. • BRONSTED - LOWERY CONCEPT : (CONJUGATE ACID - BASE CONCEPT) (PROTONIC CONCEPT) Acid : substances which donate H+ are Bronsted Lowery acids (H+ donor) Base : substances which accept H+ are Bronsted Lowery bases (H+ acceptor) • Conjugate Acid - Base Pairs These are the species which differ by one H+ ion. Conjugate acid are the species fromed after adding one H+ to a base or removal of one OH– Ex. NH3 + H+  NH4+ base Ex. NaOH  base Conjugate acid Na+ + OH– Conjugate acid Conjugate base are the species fromed after removal of one H+ ion from an acid Ex. HCl  H+ acid + Cl– Conjugate base IONIC EQUILIBRIUM Ex. H2SO4  H+ + HSO4– acid Conjugate base In a typical acid base reaction – + HX + B ‡ˆ ˆˆ ˆ† ˆˆ X + HB HX acid + HB+ + Conjugate acid B base X– Conjugate Base Conjugate pair   Forward reaction - Here HX being a proton donor is an acid B being a proton acceptor is a base. Backward reaction - Here HB+ being a proton donor is an acid X– being a proton acceptor is a base. Acid Base H 2O + NH3 Conjugate Base ‡ˆ ˆˆ ˆ† ˆˆ ˆ‡ ˆˆ ˆ† ˆˆ H 3O + + Cl– + ‡ˆ ˆˆ ˆ† ˆˆ NH4 H 3O + SO4  HCl   HSO4    Conjugate acid - base pair differ by only one proton. Strong acid will have weak conjugate base and vise versa. Reaction will always proceed from strong acid to weak acid or from strong base to weak base. – + Conjugate Acid [Fe(H2O)6]3+ + Acid HCl H2SO4 – HSO4 H2O Ex : H 2O Conjugate base Cl– – HSO4 2– SO4 OH– + Base NH3 H2O RNH2 + –2 [Fe(H2O)5(OH)]2+ Conjugate acid + NH4 H3O+ + RNH3 Amphoteric : Substances which can act as acid as will as base are known as amphoteric + – HCl + H O ‡ˆ ˆˆ ˆ† ˆˆ H O + Cl 2 base 3 + – NH3 + H2O ‡ˆ ˆˆ ˆ† ˆˆ NH4 + OH acid Amphiprotic : An amphiprotic molecule (or ion) can either donate or accept a proton, thus acting either as an acid or a base. Water, amino acids, hydrogen carbonate ions and hydrogen sulfate ions are common examples of amphiprotic species. Since they can donate a proton, all amphiprotic substances contain a hydrogen atom. Ex.  – – H2O + S2– ‡ˆ ˆˆ ˆ† ˆˆ OH + HS + 2– H2O + HSO4– ‡ˆ ˆˆ ˆ† ˆˆ H3O (aq) + SO4 (aq) According to this concept, neutralisation is a process of transfer of a proton from an acid to a base. (a) + – CH3COOH + NH3 ‡ˆ ˆˆ ˆ† ˆˆ NH4 + CH3COO (b) – NH4+ + S2– ‡ˆ ˆˆ ˆ† ˆˆ HS + NH3 (c) + 2+ [Fe(H2O)6]3+ + H2O ‡ˆ ˆˆ ˆ† ˆˆ H3O + [Fe(H2O)5(OH)]  An acid–base reaction always proceeds in the direction of formation of the weak acid and the weak base. In the equilibrium, HA + ‡ˆ ˆˆ ˆ† ˆˆ H2O Strong acid Strong base H 3O + + Weak acid A– Weak base In general "The conjugate base of a strong acid is always a weak base and the conjugate base of a weak acid is always a strong base."  A number of organic compounds containing oxygen, can accept protons and thus act as bases. Ex. (a)    – C2H5 OH + H2SO4 ‡ˆ ˆˆ ˆ† ˆˆ C2H5 O H2 + HSO4  Ethanol (b)  (Oxonium ion)    – (C2H5)2 O + HCl ‡ˆ ˆˆ ˆ† ˆˆ (C2H5)2 O  H + Cl   Ethylether Oxonium ion  Bronsted lowery concept does not differ appreciably from the Arrhenius theory for aqueous solution only.  Autoionisation or Autoprotolysis or Self ionisation H2O + H2O  H3O+ + OH– acid base NH3 + NH3  NH4+ + NH2– acid base H2SO4 + H2SO4  H3SO4+ + HSO4– acid  base A limitation of the Bronsted Lowery theory is that the extent to which a dissolved substance can act as an acid or a base depends largly on the solvent. (a) HClO4 + HF ‡ˆ ˆˆ ˆ† ˆˆ Acid (b) Base Acid ( HClO4 acts as a acid in HF ) Base HNO3 behaves as base in HClO4 and HF HNO3 + 4HF ‡ˆ ˆˆ ˆ† ˆˆ (base) (c) H2F+ + ClO4– H3O+ + NO2+ + 2HF2 – (acid) Urea is weak acidic in liquid NH3 NH2CONH2 + NH3  NH4+ + NH2CONH– Acid Base Acid Base Note: H2SO4 also acts as base in HF solvent. Classification of Bronsted - Lowery Acids and Bases Bronsted - Lowery acids and bases can be (i) Molecular (ii) Cationic and (iii) Anionic IONIC EQUILIBRIUM Table - 1 Type Type Acid Molecular HCl, HNO3 , HClO4 , H2 SO4 , H3 PO4 , H2 O etc. + + + Cationic NH4 , N2 H5 , PH4 , + Na , Ba 2+ cid Base 2+ [Fe(H2 O)5 OH] 2+ (All cations) 3+ Base NH3 , N2 H4 , Amines, H2 O, Alcohol, Ethers etc. [Al(H2 O)5 OH] etc. 3+ [Fe(H2 O)6 ] , [Al(H2 O)6 ] etc. Anionic – – – HS , HSO3 , H2 PO4 ,HSO4 – – 2– – – 2– HCO3 , HPO4 etc. all amphiprotic anions – – – Cl , Br , OH , HSO4 , CN , 2– – – CO3 ,SO4 , NH2 , CH3 COO etc. all anions. Reactions in Non-Aqueous Solvents (i) Solvents like C6H6, CCl4, THF (Tetrahydrofuran), DMF (N, N-dimethyl formamide) etc. are used in organic chemistry. In inorganic chemistry reactions are generally studied in water. However a large number of non–aqueous solvents (such as Glacial acetic acid, Hydrogen halides, SO2 etc.) have been introduced in inorganic chemistry. (ii) The physical properties of a solvent such as M.P., B.P., Dipole moment and Dielectric constant are of importance in deciding its behaviour. Classification of Solvents There are two types of solvents (i) Protonic (protic) and (ii) Aprotic (i) Protonic or Protic Solvents (i) They are characterized by the presence of a transferable hydrogen and the formation of "Onium" ions Autoionisation taking place in them. Ex. + – (a) H2O + H2O ‡ˆ ˆˆ ˆ† ˆˆ H3O + OH + – (b) NH3 + NH3 ‡ˆ ˆˆ ˆ† ˆˆ NH4 + NH2 – (c) 3HX ‡ˆ ˆˆ ˆ† ˆˆ H2X + HX2 + – (d) 2H2SO4 ‡ˆ ˆˆ ˆ† ˆˆ H3SO4 + HSO4 (ii) Protonic solvents may be (a) Acidic (Anhydrous sulphuric acid, liquid HF, Glacial acetic acid etc.) (b) Basic (liquid NH3) (c) Amphiprotic (H2O, proton containing anions) (ii) Aprotic Solvents Such solvents do not have replaceable hydrogen in them. These can be classified into three categories (a) Non polar or very weakly polar, nondissociated liquids, which do not solvate strongly. Ex. CCl4, hydrocarbons, C6H6, C6H12 etc. (b) Non-ionised but strongly solvating, generally polar solvents. Ex. Acetonitrile CH3CN, DMF, DMSO (dimethyl sulfoxide), THF and SO2. (c) Highly polar, autoionising solvents. Ex. Inter halogen compounds (BrF3, IF5 and trichloro phosphine) (a) 2 BrF3 ‡ˆ ˆˆ ˆ† ˆˆ (b) 2 IF5 ‡ˆ ˆˆ ˆ† ˆˆ (c) 2Cl3PO ‡ˆ ˆˆ ˆ† ˆˆ BrF2+ + BrF4– IF4+ + IF6– Cl2PO+ + Cl4PO– Levelling Solvents (i) The Bronsted - Lowery theory can be extended to acid - base reactions in non-aqueous solvents. It can be used in differentiating the acid strength of a particular acid and in titration of weak bases. (ii) In water solvent, mineral acids appear to be equally strong because of their complete ionisation, water is called here a levelling solvent because it levels all the acids to the same strength. (iii) If instead of water solvent, we take mineral acids in pure acetic acid solvent (which is poor proton acceptor as compared to water) it is found acids become weak and can be differentiated. Ex. ˆˆ HCl + CH3COOH ‡ˆ ˆˆ ˆ† Cl– + CH3COOH2+ Acid Base Base Acid In above example acetic acid and Cl– ions both compete for protons and the former being a poor proton acceptor does it much less effectively than water. Thus HCl in acetic acid solvent appears to be a much weaker acid than that in water. (iv) Mineral acids in acetic acid solvent follow the following order of their strengths. HNO3 < HCl < H2SO4 < HBr < HClO4 (v) A weak base like acetamide or acetanilide in aqueous medium can not be titrated with acids. If how ever, the weak base is taken in glacial acetic acid solvent, the former behaves as a strong base and can be titrated. This is because acetic acid (which acts as a better proton donor) exerts a levelling effect on the base. Lux - Flood Concept (1939 & 1947)* (i) The proton plays an important role in explaining the acid-base behaviour in the Bronsted-Lowery concept. Lux observed that acid - base reactions are also feasible in oxide systems without the aid of protons. (ii) Above approach was extended by Flood and applied to non-protonic systems, which were not covered by the Bronsted Lowery concept. (iii) According to this concept a base (like CaO, BaO or Na2O) is an oxide ion (O2–) donor and an acid (like SiO2, CO2 or P4O10) is an oxide ion (O2–) acceptor. Ex. Base (a) CaO (b) 6Na2O + P4O10  4Na3PO4 (iv) Substances are termed amphoteric if they show a tendency of losing as well as accepting an oxide ion. Ex. ZnO, Al2O3, BeO, Ga2O3 Acid + SiO2  CaSiO3 IONIC EQUILIBRIUM KEY POINTS (i) The concept is particularly applicable to reactions which take place at high temperature i.e. in metallurgical operations or during the manufacture of ceramics and glass. The approach can be extended to include other negative ion systems (like halides, sulphide or carbanion). (ii) Example based on : Bronsted - Lowery Concept Ex. Select the aprotic solvent from the following is : (A) H2O (B) C6H6 Sol. (B) according to the classification of solvents. Ex. The strongest conjugate base is (A) Cl– (B) CH3COO– (C) HF (D) NH3 (C) SO42– (D) NO2– Sol. (B) CH3COOH is weakest acid among HCl, CH3COOH, HSO4– and HNO2. Ex. Which is the strongest Bronsted base in the following anions : (A) ClO– (B) ClO2– (C) ClO3– Sol. (A), HClO is weakest acid among HClO, HClO2, HClO3 and HClO4. Ex. Give appropriate equation and label acid and base. Sol. NH2CONH2 + NH3  NH4+ + NH2CONH– Acid base Acid (D) ClO4– base In liquid NH3 solution urea can show weak acidic nature. Ex. Ammonium ion is(A) A Lewis acid Sol. (B) Lewis base (C) Bronsted acid (D) Bronsted base Ans. (C) (D) OH– Ans. (A) Correct answer is (C) Self Practice Problem 1. 2. (a) Write conjugate acids of 2– – SO4 , RNH2, NH2 , F – (b) Write conjugate base of HNO2, OH–, H2CO3, HClO4. (c) Write conjugate acids and conjugate base of amphoteric species. – HS–, NH3, H2O, HSO4 Which of the following is the strongest base– (A) NH2 (B) CH3COO– (C) C2H5O– Lewis Concept (Electronic Concept)  Ex.  Ex. An acid is a molecule/ion which can accept an electron pair with the formation of a coordinate bond. Acid e– pair acceptor Electron deficient molecules : BF3, AlCl3 Cations : H+, Fe2+, Na+ Molecules with vacant orbitals : SF4, PF3 A base is any molecule/ion which has a lone pair of electrons which can be donated. Base  (One electron pair donate) Molecules with lone pairs : NH3, PH3, H2O, CH3OH Ex. In which of the following reaction does NH3 act as an acid ? + (A) NH3 + H+ NH4 Sol. Ex. (B) NH3 + Na NaNH2 + 1 H 2 2 (C) NH3 + HCl NH4Cl (D) none of these (B) In the following reaction, NH3 changes of NaNH2 which contains NH2 ion. This means that NH3 has donated a proton to Na and hence acts as an acid. Sulphanilic acid is a/an (A) Arrhenius acid (B) Lewis base (C) Neither (A) or (B) (D) Both (A) and (B) SO3H Sol. (D) Sulphanilic acid is . NH2 The SO3H group is capable of donating H+ and hence it acts as arrhenius acid while nitrogen in the NH2 group contains lone pair of electrons and hence can act as lewis base by donating it. Properties of Water Amphoteric (Amphiprotic) Acid/Base Nature Water - an acid as well as base according to Bronsted - Lowry theory but according to Lewis concept it can only be taken as base only. In pure water [H+] = [OH–] so it is neutral. Molar Concentration / Molarity of Water 1000 g / litre Molarity = No. of moles/litre = 18g / mole = 55.55 mole/litre = 55.55 M (density = 1 g/cc) Ionic Product of Water According to arrhenius concept + – + – –14 H2O ‡ˆ ˆˆ ˆ† ˆˆ H + OH so, ionic product of water, kw = [H ][OH ] = 10 at 25° (exp.) Dissociation of water is endothermic, so on increasing temperature Keq increases. Kw increases with increase in temperature.  Ionic product of water is always a constant whatever has been dissolved in water since its an equilibrium constant so will be dependent only on temperature.  Degree of Dissociation of Water H2O ‡ˆ ˆˆ ˆ† ˆˆ H+ + OH–  = =  no. of moles dissociated Total no. of moles initially taken 10 – 7 = 18 × 10–10 or 1.8 × 10–7 % 55.55 Absolute Dissociation Constant of Water H2O ‡ˆ ˆˆ ˆ† ˆˆ H+ + OH– So, K a = Kb = [H + ][OH – ] 10 –7 ×10 –7 = = 1.8 × 10–16 [H 2O] 55.55 pKa = pKb = – log(1.8 × 10–16) = 16 – log1.8 = 15.74 IONIC EQUILIBRIUM Acidity and pH Scale + +  Acidic strength means the tendency of an acid to give H3O or H ions in water.   So greater then tendency to give H+, more will be the acidic strength of the substance. Basic strength means the tendency of a base to give OH– ions in water. So greater the tendency to give OH– ions, more will be basic strength of the substance. The concentration of H+ ions is written in a simplified form introduced by Sorenson known as pH scale. pH is defined as negative logarithm of activity of H+ ions. C activity (a) = . C o C  unitless, Co  standard concentration  pH = – log a H  (where a H  is the activity of H+ ions)     4 5 6 7 8 9 10 11 12 13 14 Bleach 3 Neutral Pure water 2 HCl Battery acid 1 Basic Stength increases Ammonia Acidic Stength increases Activity of H+ ions is the concentration of free H+ ions or H3O+ ions in a solution. The pH scale was marked from 0 to 14 with central point at 7 at 25 °C taking water as solvent. If the temperature and the solvent are changed, the pH range of the scale will also change. For example 0 - 14 at 25 °C (Kw = 10–14) Neutral point, pH = 7 –13 0 - 13 at 80 °C (Kw = 10 ) Neutral point, pH = 6.5 pH can also be negative or > 14 – pH Calculation of Different Types of Solutions (a) Strong acid Solution (i) If concentration is greater than 10–6 M. In this case H+ ions coming from water can be neglected, so [H+] = normality of strong acid solution (ii) If concentration is less than 10–6 M In this case H+ ions coming from water cannot be neglected. So [H+] = normality of strong acid + H+ ions coming from water in presence of this strong acid – Solved Examples Ex. Calculate pH of 10–8 M HCl solution. Sol. H 2O ‡ˆ ˆˆ ˆ† ˆˆ H+ OH– + 10–8+x + x – kw = [H ][OH ] 10–14 = x(x + 10–8)  x2 + x × 10–8 – 10–14 = 0 –10 –8 ± 10 –16 + 4 × 10 –14 –10 x= = 2 –8 + 10 –7 4 + 2 1 –8 100 = ( 401 –1)10 = 0.95 × 10–7 2 [H+] = 10.5 × 10–8 = 1.05 × 10–7 pH = –log [H+] pH = 7 – log 1.05 6.98 10–9 M HCl pH 7 10–16 M HCl pH 7 (b) Strong Base Solution : Calculate the [OH–] which will be equal to normality of the strong base solution and then use Kw = [H+] × [OH–] = 10–14, to calculate [H+] Ex. Calculate pH of 10–7 M of NaOH solution Sol. [OH–] from NaOH = 10–7 [OH–] from water = x < 10–7 M H2O ‡ˆ ˆˆ ˆ† ˆˆ OH– (due to common ion effect) + (x + 10–7) – + – H+ x –14 Kw = [H ] [OH ] = 10 = x (x + 10–7) x2 + 10–7x – 10–14 = 0  x= 5 1 × 10–7 = 0.618 × 10–7 2 [OH–] = 10–7 + 0.618 × 10–7 = 1.618 × 10–7 pOH = 7 – log(1.618) = 6.79 pH = 14 – 6.79 = 7.21 Self Practice Problem 1. Calculate pH of a KOH solution having (a) 5.6 g of KOH mixed in 50mL water (b) if it is further diluted to make 100 mL. Ans. (a) 14.3 (b) 14 ( 5 = 2.236) IONIC EQUILIBRIUM (c) pH of Mxture of Two Strong Acids : If V1 volume of a strong acid solution of normality N1 is mixed with V2 volume of another strong acid solution of normality N2, then Number of H+ ions from I-solution = N1V1 Number of H+ ions from II-solution = N2V2 If final normality is N and final volume is V, then NV = N1V1 + N2V2 [dissociation equilibrium of none of these acids will be disturbed as both are strong acid] [H+] = N = (d) N1V1 + N 2 V 2 V1 + V2 pH of Mixture of Two Strong Bases similar to above calculation [OH–] = N = Ex. Calculate pH of mixture of (400 mL, Sol. Normality Method N 1V 1 = N1V1 + N 2 V 2 V1 + V2 [H+] = 10 –14 [OH – ] 1 1 M H2SO4) + (400 mL, M HCl) + (200 mL of water) 200 1 00 1 4 00 4 4 × = , N 2V 2 = , H+ ions from water will be neglected 1 00 0 100 1 00 0 1 00 0 N1V1 + N2V2 = 8 × 10–3 [H+] = 8  10 3 = 8 × 10–3 1 pH = 3 – log8 = 2.1 Molarity Method :   H2SO4 2m mole 2H+ + SO42– 4m mole   HCl 4m mole H+ + Cl– 4m mole 8m mole  H     8  103 total 1000mL pH = –log [H+] = – log [8 × 10–3] = 2.1 Ans. Ex. 500 mL of 10–5 M NaOH is mixed with 500 mL of 2.5 × 10–5 M of Ba(OH)2. To the resulting solution 99 L water is added. Calculate pH. Sol. [OH–] = 500 ×10 –5 + 500 × 2 × 2.5 ×10 –5 1000 = 3 × 10–5 M M1 = 3 × 10–5 M V2 + V1 = 1 L V1 = 100 L no. of moles of [OH–] initially = no. of moles of [OH–] 3 × 10–5 = M2 × 100  M2 = 3 × 10–7 < 10–6 + H2O ‡ˆ ˆˆ ˆ† ˆˆ H + x OH– (x + 3 × 10–7) Kw = x (x+3 × 10–7) = 10–14  13 –  x=  2  3 –7  × 10  x = 0.302 × 10–7   3+ 13  13 – 3  [OH–]Net = 3 + 2  × 10–7 =  2  × 10–7 = 3.302 × 10–7     1. Self Practice Problem Calculate the pH for- (a) 50 mL of 0.1 M HCl, 25 mL of 0.1M H2SO4, 25 mL of 0.2 M HNO3 + 100 mL of H2O (b) 50 mL of 0.2 M NaOH + 100 mL of 0.1 M RbOH the resulting solution is diluted by 350 mL H2O. Ans. (a) 1.123 (e) pH of Mixture of a Strong Acid and a Strong Base    (b) 12.6 Acid Base neutralisation reaction will take place. The solution will be acidic or basic depending on which component has been taken in excess. If V1 volume of a strong acid solution of normality N1 is mixed with V2 volume of a strong base solution of normality N2, then Number of H+ ions from I-solution = N1V1 Number of OH– ions from II-solution = N2V2 If N1V1 > N2V2 [H+] = N = If N2V2 > N1V1 N1V1 – N 2V2 V1 + V2 [OH –] = N = Solution will Solution will be acidic in nature be basic in nature [H+] = Ex. Sol. N1V1 – N 2V2 V1 + V2 1 10 –14 [OH – ] 1 Calculate pH of mixture of (400 mL, M Ba(OH)2) + (400 mL, M HCl) + (200 mL of water) 50 200 1 1 400 × – 400× ×2 50 200 [H+] = = 4 × 10–3, so pH = 3 – 2 log2 = 2.4 1000 294 IONIC EQUILIBRIUM Ex. Sol. What will be the resultant pH when 150 mL of an aqueous solution of HCl (pH = 2.0) is mixed with 350 mL of an aqueous solution of NaOH (pH = 12.0) ? pH of HCl = 2  [HCl] = 10–2 M pH of NaOH = 12, pOH = 2  [NaOH] = 10–2 M HCl + NaOH  NaCl + H2O Meq. initial 150×10–2 350×10–2 0 0 = 1.5 = 3.5 Meq. final 0 0.2 1.5 1.5  [OH–] from NaOH = 2 = 4 × 10–3 M 500 pOH = – log[OH–] = –log (4 × 10–3)  pOH = 2.3979  pH = 14 – pOH = 14 – 2.3979 = 11.6021 Self Practice Problem Ans. (f) 1. Calculate pH of mixture 200 mL of 0.2 M H2SO4 + 300 mL of 0.2 M NaOH + 200 mL of 0.1 M KOH. 2. Calculate the pH when 200 mL of 0.25 M H2SO4 is mixed with 200 mL of 0.2 M Ba(OH)2 1. 7 2. 1.30 pH of a Weak Acid (Monoprotic) Solution  Weak acid does not dissociated 100 % therefore we have to calculate the percentage dissociation using K a dissociation constant of the acid.  We have to use Ostwald's Dilution law (as have been derived earlier) + – HA ‡ˆ ˆˆ ˆ† ˆˆ H + A t=0 C teq C(1 – ) 0 0 CC If  << 1 (1 – ) 1 Ka C2  = [H+] = C= C Ka = Cα 2 [H + ][A – ] = 1– α [HA] Ka (is valid if  < 0.1 or 10%) C Ka 1 = K a × C So pH = (pKa – logC) valid only if  <<< 1. C 2 on increasing the dilution C= and [H+]pH Ex. Calculate pH of (a) 10–1 M CH3COOH Take Ka = 2 × 10–5 Sol. – + (a) CH3COOH ‡ˆ ˆˆ ˆ† ˆˆ CH3COO + H C 0 0 C(1 – ) C C Ka = Cα 2 = 1– α So, [H+] = 10–1 × Ka = C 2 ×10 –5 10 –1 (b) 10–3 M CH3COOH = –2 2 × 10 pH = 3 – 2 ×10 –4 (<< 0.1) 1 log 2 = 2.85 2 (c) 10–6 M CH3COOH (b)  = Ka  = C Ka = C 2 ×10 –5 10 –3 = 2 × 10 – 2 ( > 0.1) So we have to do the exact calculations Ka = 10 –3 × α 2 Cα 2 2 × 10–5 =  = 13.14 % 1– α 1– α [H+] = 10–3 × 0.1314 = 1.314 × 10–4 pH = 4 – log(1.314) 3.8 (c) If approximation is used the,  = 2 ×10 – 5 10 – 6 = 2 0 > 1, so we have to do the exact calculation, 2 × 10–5 = 10–6 2 0.95 or 95% 1  [H+] = 0.95 × 10–6 = 9.5 × 10–7 pH = 7 – log(9.5) = 6.022  At very low concentration (at infinite dilution) weak electrolyte will be almost 100% dissociate, so behave as strong electrolyte (pH) of 10–6 M HCl ; pH of 10–6 M CH3COOH ; 6) (g) pH of a Mixture of Weak Acid (Monoprotic) and a Strong Acid Solution  Weak acid and Strong acid both will contribute H+ ion.  For the first approximation we can neglect the H+ ions coming from the weak acid solution and calculate the pH of the solution from the concentration of the strong acid only.  To calculate exact pH, we have to take the effect of presence of strong acid on the dissociation equilibrium of the weak acid.  If [SA] = C1 and [WA] = C2 , then [H+] from SA = C1 the weak acid will dissociate as follows. HA ‡ˆ ˆˆ ˆ† ˆˆ H+ + A– C2 0 0 C2(1 – ) C2+C1 C2 Ka = (C2 α + C 1 )C2α C2 (1– α) ( <<< 1) (The weak acids dissociation will be further suppressed because of presence of strong acid, common ion effect) Ka = (C2 + C1) + Total H ion concentration = C1 + C2  If the total [H+] from the acid is more than 10–6 M, then contribution from the water can be neglected, if not then we have to take [H+] from the water also. IONIC EQUILIBRIUM Relative Strength of Weak Acids and Bases The relative strength of weak acids and bases are generally determined by their dissociation constants K a and Kb respectively. For weak acid, i.e. CH3COOH – + CH3COOH ‡ˆ ˆˆ ˆ† ˆˆ CH3COO + H C C(1 – ) Ka = 0 C 0 C Cα 2 Cα.Cα =  Ka = C2 C(1– α) (1– α) (if  << 1) Similarly, for weak base, i.e. NH4OH NH4OH ‡ˆ ˆˆ ˆ† ˆˆ NH4+ + OH– C 0 0 C(1 – ) C C 2 Kb = C Ka and Kb are just the equilibrium constants and hence depends only on temperature. Greater the value of dissociation constant of the acid (Ka), more is the strength of the acid and similarly greater the value of dissociation constant of the base, more is the strength of the base. For two acids of different concentrations. K a1 .c1 Strength of acid (I) = Strength of acid (II) K a 2 .c 2 Similarly for bases, K b1 .c1 Strength of base (I) = K b2 .c 2 Strength of base (II) The modern method is to convert Ka as a power of 10 and express acid strength by power of 10 with sign changed and call this new unit pKa. Thus, if Ka for acid is equal to 10–4, pKa = 4. So higher pKa value means lower acid strength, that is, pKa = – log Ka Also, pKb = – log Kb Total [H+] in a mixture of two weak Acids. [H  ]  K a1 c1  K a 2 c 2 Similarly for two weak bases [OH  ]  K b1 c1  K b2 c 2 (h) pH of a Mixture of Two Weak Acid (Both Monoprotic) Solution Both acids will dissociate partially.  Let the acid are HA1 & HA2 and their final concentrations are C1 & C2 respectively, then  HA1 ‡ˆ ˆˆ ˆ† ˆˆ t=0 C1 At eq. C1(1 – ) H+ + – A1 0 0 C11 + C22 C11 HA2 ‡ˆ ˆˆ ˆ† ˆˆ C2 C2(1 – ) Ka = C 1 1 (C 1 1  C 2 2 ) 1 C 1 (1  1 ) (Since 1, 2 both are small in comparision to unity) H+ + – A2 0 0 C22 + C11 C22 Ka 2 = (C 2  2  C1 1 )C 2  2 C 2 (1  2 ) Ka1 = (C11 + C22)1 ; K a 2 = (C11 + C22)2  [H+] = C11 + C22 = C1K a1 C1K a1 + C 2 K a 2 + C2K a 2 Ka1 Ka2 C1K a1 + C 2 K a 2 = 1 2 [H+] = C1K a1 + C2 K a 2 If the dissociation constant of one of the acid is very much greater than that of the second acid then contribution from the second acid can be neglected.  So, [H+] = C11 + C22 C11 Ex. Ka for acid HA is 2.5 × 10–8 calculate for its decimolar solution at 25°C. (i) % dissociation (ii) pH (iii) OH– ion concentration Sol. HA ‡ˆ ˆˆ ˆ† ˆˆ H+ + A– C 0 0 C(1–)   Ka = Cα2 [H + ][A – ] Cα.Cα Ka = =  C2 [HA] C(1– α) (1– α) (i) = Ka = C 10 8 (C = 1/10 M) 1 / 10 = 5 × 10–4 = 0.05% 1  5 1 0 4 = 5 × 10–5 mol/L  So pH = 5 – log 5 = 4.30 10 (ii) [H+] = C = (iii) [H+] [OH–] = 1 × 10–14  [OH–] = 10 –14 = 2 × 10–10 mol/L 5×10 –5 Ex. Determine the degree of dissociation of 0.05 M NH3 at 25°C in a solution of pH = 10. Sol. NH4OH ‡ˆ ˆˆ ˆ† ˆˆ C NH4+ + OH– 0 0 Given, pH = 10 [H+] = 10–10 [H+] [OH–] = 1 × 10–14  [OH–] = = 1×10 –14 = 10–4 = C 10 –10 [OH – ] 10 –4 = = 2 × 10–3 or 0.2 % C 0.05 IONIC EQUILIBRIUM Ex. Sol. Two weak monobasic organic acids HA and HB have dissociation constants as 1.6 × 10 –5 and 0.4 × 10–5 respectively at 25° C. If 500 mL of 1 M solutions of each of these two acids are mixed to produce 1 litre of mixed solution, what is the pH of the resulting solution ? In such cases, we have to consider H+ from both HA and HB simultaneously. The concentration of HA and HB in the mixture = 0.5 M [equal volumes are mixed] = say 'c' HA  H+ + A– HB  H+ + B– Let, x = [H+] from HA and y = [H+] from HB  [H+]final = x + y K HA = (x + y)y (x + y)x and K HB = c c [H + ]HA x = = [H + ]HB y k HA × C k HB × C x 1.6 ×10 –5 = =2 y 0.4 ×10 –5 x = 2y  y = x 2 Substitute for y = x x 2 + xy in K HA = 2 c 2x 2 + 3x 2 2×0.5 2 –5 3x = 1.6 × 10  x2 = 5.33 × 10–6 x = 2.30 × 10–3 M, y = 1.15 × 10–3 M [H+]Final = x + y = 2.30 × 10–3 + 1.15 × 10–3 = 3.45 × 10–3 M pH = –log10 (3.45 × 10–3) pH = 2.462 1.6 × 10–5 = Ex. Saccharin (Ka = 2 × 10–12) is a weak acid represented by formula HSaC. A 8 × 10–4 mole amount of saccharin is dissolved in 400 cm 3 water of pH = 3. Assuming no change in volume, calculate the concentration of SaC– ions in the resulting solution at equilibrium. Sol. [HSaC] = mole 8×10–4 = = 2 × 10–3 M litre 400 /1000 The dissociation of HSaC takes place in presence of [H+] = 10–3 + – HSaC ‡ˆ ˆˆ ˆ† ˆˆ H + SaC Conc. Before dissociation 2×10–3 10–3 0 + In presence of H , the dissociation of HSaC is almost negligible because of common ion effect. Thus, at equilibrium [HSaC] = 2 × 10–3 ; [H+] = 10–3 M Ka = [H + ][SaC – ] [HSaC] [10 –3 ][SaC –1 ] 2×10 –3 –  [SaC ] = 4 × 10–12 M  2 × 10–12 = Ex. Sol. A solution contains 0.08 M HCl, 0.08 M CHCl2COOH and 0.1 M CH3COOH. The pH of this solution is 1. If Ka for acetic acid is 10–5, calculate Ka for CHCl2COOH. pH will be decided by [H+] furnished by HCl and CHCl2COOH. CHCl2COOH ‡ˆ ˆˆ ˆ† ˆˆ CHCl2COO– + H+ Initial conc. 0.08 0 Final conc. (0.08 – x) x 0.08 (from HCl) (0.08 + x) +  [H ] = 0.08 + x ; but pH = 1,  [H+] = 10–1 = 0.1  0.08 + x = 0.1 M  x = 0.02 Ka for CHCl2COOH can be given as Ka = 0.1×0.02 [H + ][CHCl2COO – ] = = 3.33 × 10–2 (0.08 – 0.02) [CHCl 2COOH] ISOHYDRIC SOLUTIONS (i) Solutions of electrolytes are said to be isohydric if the concentration of the common ion present in them is the same and on mixing such solutions, there occurs no change in the degree of dissociation of either of the electrolyte. (ii) Let the isohydric solution is made by HA1 and HA2 acids, then [H+] of both acids should be equal i.e. K a1 C  2 Ka2 C1 Example Based on : Dissociation of Acid and Base and pH Calculation K a1 C1 = K 2 C2 Ex. Sol. or The degree of dissociation of pure water at 25°C is found to be 1.8 × 10–9. Find KW and Kd at 25°C. (A) 3.24 × 10–18 ; 5.83 × 10–20 (B) 1 × 10–14 ; 1.8 × 10–16 (C) 1.8 × 10–16 ; 1 × 10–14 (D) 1 × 10–14 ; 1 × 10–14 (B) Since  = 1.8 × 10–9 1000 = 55.56 18 [H+] = [OH–] = c = 55.56 × 1.8 × 10–9 = 1 × 10–7 M and for water c = KW = [H+] × [OH–] = (1 × 10–7)2 = 10–14 and Kd = Ex. Sol. 10¬14 [H + ][OH – ] K W = = = 1.8 × 10–16 [H 2 O] [H 2 O] 55.56 The concentration of [H+] and [OH–] of the 10–1 M aqueous solution of 2% ionised weak acid is : (A) 2 × 10–3 M and 5 × 10–12 M (B) 1 × 10–3 M and 3 × 10–11 M –4 –11 (C) 2 × 10 M and 5 × 10 M (D) 3 × 10–2 M and 4 × 10–13 M (A) [H+] = C = 2 × 10–3 M 10 –14 or [OH ] = = 5 × 10–12 M [H + ] – IONIC EQUILIBRIUM Ex. Sol. When a 0.1 N solution of an acid at 25°C has a degree of ionisation of 4%, the concentration of OH– present is : (A) 2.5 × 10–3 (B) 2.5 × 10–11 (C) 2.5 × 10–12 (D) 2.5 × 10–13 (C) [H+] = C = 0.1 × 4 × 10–2 = 4 × 10–3 M or [OH–] = Ex. Sol. 10 –14 = 2.5 × 10–12 N [H + ] Calculate the molar concentration of a solution of acetic acid (HOAc) that has a pH of 4.00. (Ka = 1.8 × 10–5) : (A) 1.0 × 10–3 (B) 1.0 × 10–6 (C) 0.057 × 10–2 (D) 0.010 (C) Ka = [H + ][OAc – ] [HOAc] (10 –4 ) 2 [HOAc] or [HOAc] = 5.56 × 10–4 M or 1.8 × 10–5 = Ex. Sol. Ex. Select the correct option from the following ? (A) pKW increases with increase of temperature (B) pKW decreases with increase of temperature (C) pKW = 14 at all tempera tures (D) pKW = pH at all temperatures (B) How much water must be added to 200 mL of 0.2 M solution of CH3COOH for the degree of dissociation of the acid to double ? Ka for the acetic acid = 1.8 × 10–5: 2 Sol.  α1  C C11 = C22 or C2 = C1  α  = 1 4  2 2 so 2 M1 = 0.2 M ; M2 = V1 = 200 mL, V2 = ? M1V1 = M2V2 Ex. Sol. 0.2 M 4 M1V1 0.2 × 200 × 4 = = 800 mL 0.2 M2 or V2 = so 800 – 200 = 600 mL water should be added. The degree of dissociation of acetic acid in a 0.1 M solution is 1.32 × 10–2. Calculate dissociation constant of acid and its pKa value : CH COOH ‡ˆ ˆˆ ˆ† CH COO– + H+ ˆˆ 3 Initially at equilibrium Ka = [CH 3COO – ][H + ] [CH3COOH] 0.1 0.1(1–0.0132) = 3 0 0.1× 0.0132 0.1× 0.0132 × 0.1× 0.0132 = 1.76 × 10–5 0.1(1 – 0.0132) pKa = – log Ka = –log (1.76 × 10–5) = 4.75 0 0.1× 0.0132 Ex. A solution having pH = 13, calculate the no. of H+ ions present in 1 mL of this solution : Sol. pH = 13 so [H+] = 10–13 M moles of H+ in one mL = 10 –13 = 10–16 mol. 103  No. of H+ ions = 10–16 × 6.022 × 1023 = 6.022 × 107 Ex. Calculate pH of solution obtained by mixing equal vol. of 0.02 M HOCl & 0.2 M CH3COOH solution given that –4 Ka1 (HOCl) = 2 × 10 Ka2 (CH3COOH) = 2 × 10–5 Sol. also calculate OH–, OCl–, CH3COO– Final solution volume become double C1 = 0.01, C2 = 0.1 + [H ] = = 2 C2 2 ×10 – 4 × 0.01 + 2 ×10 – 5 × 0.1 = 2 ×10 –6 + 2 ×10 –6 = 2 × 10–3 pH = 3 – log2 = 3 – 0.3010 = 2.69 1 = 2 ×10 –4 2 ×10 –3 HOCl = 10–1 2 = ‡ˆ ˆˆ ˆ† ˆˆ C1(1 – 1) 2 ×10 –5 2 ×10 –3 H+ + C11 + C22 = 10–2 OCl– CH3COOH C11 C2(1 – 2) [OCl–] = C11 ‡ˆ ˆˆ ˆ† ˆˆ C11 + C22 + CH3COO– C22 [CH3COO–] = C22 = 0.01 × 10–1 = 0.1 × 10–2 = 1 × 10–3 = 1 × 10–3 [OH–] = H+ 10 –14 = = 0.5 × 10–11 [H + ] 2 ×10 –3 Kw = 5 × 10–12 M [HOCl] = 10–2 (1 – 0.1) = 9 × 10–3 M [CH3COOH] = 10–1(1 – 0.01) 10–1 (i) pH of a Solution of a Polyprotic Weak Acid  Diprotic acid is the one, which is capable of giving 2 protons per molecule in water. Let us take a weak diprotic acid (H2A) in water whose concentration is c M. In an aqueous solution, following equilbria exist. If 1 = degree of ionization of H2A in presence of HA– – 2 = degree of ionisation of HA in presence of H2A Ka1 = first ionisation constant of H2A K a = second ionisation constant of H2A 2 IONIC EQUILIBRIUM – H2A + H2O ‡ˆ ˆˆ ˆ† ˆˆ HA at eq. C – x + x–y (Keq) [H2O] = 1 HA– + H2O ‡ˆ ˆˆ ˆ† ˆˆ A2– x–y y H 3O + x+ y at eq. [H3O + ][HA – ] = Ka 1 [H 2 A] (Keq) [H2O] = 2 (x – y) (x + y) (C – x)  Ka1 = [H3O + ][A 2– ] Ka 2 = [HA – ] + H3O+ x+y = Ka 2 (y) (x + y) (x – y) Approximation For diprotic acids, Ka 2 << K a 1 and y would be even smaller than x.  y < < < x  x – y  x and x + y  x Thus, equation (i) can be reduced to K a1  x2 , K a2  y Cx This is expression similar to the expression for a weak monoprotic acid. +  Hence, for a diprotic acid (or a polyprotic acid) the [H3O ] can be calculated from its first equilibrium constant expression alone provided K a 2 << Ka1 Ex. Sol. Calculate pH of [HS–], [S2–], [Cl–] in a solution which is 0.1 M HCl & 0.1 M H2S given that K a 1 (H2S) = 10–7, Ka2 (H2S) = 10–14 also calculate 1 & 2. HCl + H2S 0.1 0.1 C1 = C2 = 0.1 pH = 1 (most of [H+] comes from HCl] ‡ˆ ˆˆ ˆ† ˆˆ H 2S 0.1(1 – 1) Ka1 = H+ + 10–1 HS– C1 = 0.1 1 Cα1 × 10 –1 10 –7 = –1 = 1 C(1– α1 ) 10 1 = 10–6 HS– ‡ˆ ˆˆ ˆ† ˆˆ S2– + H+ C1(1 – 2) C12 10–14 = 0.1 × 2  2 = 10–13 2– [S ] = C12 = 10–6 × 10–1 × 10–13 = 10–20 M (j) 0.1 pH of a Mixture of a Polyprotic Weak Acid and a Strong Acid  pH can be calculated by taking the concentration of strong acid only (for first approximation)  For precise calculation we should take only the first dissociation constant of the weak polyprotic acid. (As can be predicted from the equations we have presented so far for different cases.)  All these steps can be followed for the calculation of pOH for weak base and their mixtures (we just have to replace Ka with Kb) (k) pH of a Mixture of a Weak Acid/Weak Base with Weak/Strong Base/Acid Respectively. For this type of mixtures there can be two cases in general, (i) if the acids and bases are mixed in equal amounts (equivalents) (ii) if the acids and bases are mixed in different amounts (equivalents) First case will lead to phenomenon of Salt hydrolysis and second case will lead to formation of Buffer Solution. Relation Between Ka and Kb for Conjugate Acid Base Pair For conjugate-acid base pairs, the acid dissociation constant Ka and base ionizsation constnat Kb are related by the following equations : Ka . Kb = Kw where Kw is the autoionization constant pKa + pKb = 14 at 25°C Weak acids, generically abbreviated as HA, donate H+ (or proton) to water to form the conjugate base A– and H3O+ : HA(aq) + H2O(l) ƒ H3O+(aq) + A–(aq) acid base acid base Similarly, a base (abbreviated as B) will accept a proton in water to form the conjugate acid, HB+, and OH– : B(aq) + H2O(l) ƒ HB+ (aq)+ + OH– (aq) base acid acid base For a weak acid or base, the equilibrium constant for theionization reaction quantities the relative amount of each species. In this ariticle, we will discuss the relationship between the equilibrium constants K a and Kb for a conjugate acid-base pair. Let’s look more closely at the dissociation reaction for a monoprotic weak acid HA : HA(aq) + H2O(l) ƒ H3O+(aq) + A–(aq) The products of this reversible reaction are A–, the conjugate base of HA, and H3O+. We can write the following expressin for the equilibrium constant Ka : Ka = [H 3O  ][A  ] [HA] Finding Kb for A– Reacting as a Base Since A– is a base, we can also write the reversible reaction for A– acting as a base by accepting a proton from water : A–(aq) + H2O(l) ƒ HA (aq) + OH–(aq) The products of this reaction are HA and OH–. We can write out the equilibrium constnat Kb for the reaction where A– acts as a base : Kb = [HA][OH  ] [A  ] Even though this almost looks like the reverse of HA acting as an acid, they are actually very different reactions. When HA acts as an acid, one of the products is H3O+. When the conjugate base A– acts as a base, one of the products is OH–. Relationship between Ka and Kb for conjugate acid-base pair If we multiply Ka for HA with the Kb of its conjugate base A–, that gives : Ka . Kb  [H 3O  ][A  ]  [HA][OH  ]   =  [HA]  [A  ]    = [H3O+] [OH–] = Kw IONIC EQUILIBRIUM where Kw is the water dissociation constant. This relationship is very useful for reating Ka and Kb for a conjugate acidbase pair. We can also use the value of Kw at 25°C to derive other handy equations. Ka . Kb = Kw = 1.0 × 10–14 at 25°C (Eq. 1) If we take the negativbe log10 of both sides of the Eq. 1, we get : pKa + pKb = 14 at 25°C (Eq. 2) We can use these equations to determine Kb (or pKb) of a weak base given Ka of the conjugate acid. We can also calculate the Ka (or Pka) of a weak acid given Kb of the conjugate base. An inportant thing to remember is that these equations only work for conjugate acid-base pairs. SALTS (i) Salts are the ionic compounds formed when its positive part (Cation) come from a base and its negative part (Anion) come from an acid. (ii) Salts may taste salty, bitter, astringent or sweet or tasteless. (iii) Solution of salts may be acidic, basic or neutral. (iv) Fused salts and their aqueous solutions conduct electricity and undergo electrolysis. (v) The salts are generally crystalline solids. 1. Classification of Salts The salts may be classified into four categories. 1.1 Simple Salts The salts formed by the neutralisation process between acid and base. These are of three types. (i) Normal Salt (i) The salt formed by the loss of all possible protons (replaceable H+ ions) Ex. NaCl, NaNO3, K2SO4, Ca3(PO4)2, Na3BO3, Na2HPO3, NaH2PO2 etc. (ii) Acid Salts (i) Salts formed by incomplete neutralisation of polybasic acids. Such salts contain one or more replaceable H atom. Ex. NaHCO3, NaHSO4, NaH2PO4, Na2HPO4 etc. (these are salts containing amphiprotic ions) (ii) Above salts when neutralized by base form normal salts. (iii) Basic Salts (i) Salts formed by in complete neutralisation of poly acidic bases are called basic salts. These salt contain one or more hydroxyl groups. Ex. Zn(OH)Cl, Mg(OH)Cl, Fe(OH)2 Cl, Bi(OH)2Cl etc. (ii) Above salts when neutralised by acids form normal salts. 1.2 Double Salts (i) The addition compounds formed by the combination of two simple salts are termed as double salts. Ex. FeSO4 (NH4)2SO4 . 6H2O (Ferrous ammonium sulphate), K2SO4 Al2 (SO4)3.24H2O (Alum) and other alums. (ii) Above salts are stable in solid state only. (iii) When dissolved in water, it furnishes all the ions present in the simple salt from which it has been constituted. (iv) The solution of double salt shows the properties of the simple salts from which it has been constituted. 1.3 Complex Salts (i) These are formed by combination of simple salts or molecular compounds. Ex. K4Fe(CN)6, Co(NH3)6 SO4 etc. (ii) FeSO 4 + 6KCN  K 4 Fe(CN)6  K 2 SO 4 1 4 4 2 4 43 complex salt simple salt (iii) CoSO4 + Simple 6NH3  Co(NH3)6SO4 Molecular salt compound complex salt (iv) These are stable in solid states as well as in solutions. (v) On dissolving in water, if furnishes a complex ion. H O K4Fe(CN)6 ‡ˆ ˆˆ ˆˆ ˆˆ†ˆ 2 4K+ + [Fe(CN)6]4– complex ion (vi)The properties of the solution are different from the properties of the substance from which it has been constituted. 1.4 Mixed Salts (i) The salt which furnishes more than one cation or more than one anion when dissolved in water is called mixed salt. Ex. CaOCl2, NaKSO4, NaNH4HPO4 etc. Ca OCl Cl , Na K Na SO4 , NH4 H PO4 Hydrolysis of Salt Salt Hydrolysis Salt hydrolysis is defined as the process in which water reacts with salt to form acid & base. ˆˆ Acid + Base Water + Salt ‡ˆ ˆˆ ˆ† H = +ve It is always an endothermic process because it is reverse of acid – base neutralization reaction which is always exothermic. Hydrolysis constant Kh =  Acid   Base  Salt  Here H2O is a solvent (in excess) so active mass of H2O is 1. – Types of Salt Hydrolysis (1) Hydrolysis of Strong Acid - Weak Base [SA - WB] Type Salt – Ex. CaSO4, NH4Cl, (NH4)2SO4, Ca(NO3)2, ZnCl2, CuCl2, CaCl2 ˆˆ NH4Cl + H2O ‡ˆ ˆˆ ˆ† NH4OH + HCl ˆˆ NH4++Cl– + H2O ‡ˆ ˆˆ ˆ† ˆˆ NH4+ + H2O ‡ˆ ˆˆ ˆ† NH4OH + H+ + Cl– NH4OH + H+ IONIC EQUILIBRIUM (1) (2) (3) (a) In this type of salt hydrolysis, cation reacts with H2O, therefore called as cationic hydrolysis. Solution is acidic in nature (SA WB) as [H+] is increased. pH of the solution is less than 7. Relation between Kh , KW & Kb ˆˆ NH4+ + H2O ‡ˆ ˆˆ ˆ† Hydrolysis constant Kh  NH4OH   H+  Kh = [NH +4 ] ˆˆ NH4OH ‡ˆ ˆˆ ˆ† For weak Base NH4OH + H+ ..... (1) NH4+ + OH–  NH +4   OH –    Kb =   NH 4OH  ˆˆ H+ + OH– H2O ‡ˆ ˆˆ ˆ† – Kw = [OH ] [H+] Now multiplying Eq. (1) & (2) = Eq. (3) ..... (2) For water ..... (3)  NH 4OH   H +   NH +4  OH –        × = [H+] [OH–]  NH 4OH   NH +4      i.e. Kh × K b = K w Kh = Kw Kb (b) Degree of hydrolysis – Represented by h ˆˆ NH4+ + H2O ‡ˆ ˆˆ ˆ† C C–x nx = a 1x = Ch x = Ch C – Ch Kh = =  NH4OH  H +   NH +4    Ch 2 C2 h 2 = C 1 – h  1 – h  Since h <<<< 1 then (1 – h) 1 Kh = Ch2 NH4OH + H+ 0 0 x x Ch = Ch × Ch C – Ch Ch (initial concentration at equilibrium) h2 = Kh C  h = Kw Kb  K h =  h = Κh C Kw Kb C h=  Kw Kb × C (c) pH of the solution : pH = – log [H+] Kw Kb × C [H+] = Ch = C  [H+] = Kw × C Kb On taking – log on both sides Kw × C  Kw × C   pH = – log  – log [H ] = – log Kb  Kb  1 2 + 1 [ log Kw + logC – log Kb ] 2 1 1 1 pH =  logKw  logC  (–log Kb) 2 2 2 1 1 1 pH = pK w  log C  pK b 2 2 2 pH =  pH = 7  1 1 pK b  log C 2 2 KEY POINTS Summary (1) K h = Kw Kb  + (3)  H  = Ch = (2) h = Kw × C Kb Kh = C Kw Kb × C (4) pH = – log [H+] 1 1 pH = 7 – pK b – logC 2 2 Ex. Sol. Find out the Kh of centi normal [10–2 N]solution of NH4Cl (SA - WB) if dissociation constant of NH4OH is 10–6 and Kw = 10–14. Find out degree of hydrolysis and also find [H+] and pH of solution ? Given : Kw = 10–14 ; Kb = 10–6 Kw 10 –14 (1) K h = K = = 10–8 b 10 –6 (2) h = Kh = C 10 –8 10 –2 (3) [H+] = Ch = 10–2 × 10–3 = 10–5 = 10 –6 = 10–3 IONIC EQUILIBRIUM (4) pH = – log [H+] = – log [10–5] = + 5 log10 =+5×1 =5 Ex. Find out the K h at 363 K (90°C) of a salt of [Strong Acid – Weak Base] if the value of K b is 10 –5 [At 90°C Kw = 10–12] Sol. Kh = Ex. How many grams of NH4Cl should be dissolved per litre of solution to have a pH of 5.13 ? Kb for NH3 is 1.8 × 10–5. NH4Cl is a salt of strong acid and weak base for solutions of such salts. Sol. pH = Kw 10 –12 = = 10–7 Kb 10 –5 1 [pKW – log C – pKb] 2  10.26 = 14 – log C – 4.74  log C = 9.26 – 10.26 = –1.0 C = 10–1 M [NH4Cl] = 10–1 M WNH 4 NO 3 = 10–1 × 53.5 gL–1 = 5.35 gL–1 Ex. Sol. What is the pH of 0.4 M aqueous NaCN solution? (Given pKb of CN – = 4.70) pKa for HCN = 14 – 4.7 = 9.30 NaCN + H2O ‡ˆ ˆˆ ˆ† ˆˆ C C(1–h)  [OH–] = Ch = C pOH = NaOH + HCN Kh = C 0 Ch 0 Ch KhC = KwC Ka 1 1 (pKw – pKa – log C) = [14 – 9.30 – log (0.4)] = 2.548 2 2  pH = 14 – 2.5 = 11.45 Ex. The acid ionization constant for + + Zn2+ + H2O ‡ˆ ˆˆ ˆ† ˆˆ Zn(OH) + H is 1.0 × 10–9. Calculate the pH of 0.10 M solution of ZnCl2. Also calculate basic dissociation constant of Zn(OH)+. Sol. + + Zn2+ + H2O ‡ˆ ˆˆ ˆ† ˆˆ Zn(OH) + H  [H+] = C.h = C = K wC Kb Kh = KhC C    where K b is basic dissociation constant of Zn(OH) +    i.e.Zn(OH) + ‡ˆ ˆˆ ˆ†ˆˆ Zn 2+ + OH – We know Zn2+ and Zn(OH)+ are conjugate acid and base.  Ka × Kb = 10–14 or Kb = Now, [H+] = 1 0 14 = 10–5 1 0 9 10 –14 × 0.1 = 10 –6 = 10–3 10 –9 pH = 3 (2) Hydrolysis of [WA – SB] Type Salt – Ex. KCN, NaCN, K2CO3, BaCO3, K3PO4 NaCN +H2O ‡ˆ ˆˆ ˆ† ˆˆ NaOH + HCN Na+ + CN– + H2O ‡ˆ ˆˆ ˆ† ˆˆ CN– + H2O ‡ˆ ˆˆ ˆ† ˆˆ Na+ + OH– + HCN HCN + OH– (1) In this type of salt hydrolysis anion reacts with water therefore called as anionic hydrolysis. (2) Solution is basic in nature as [OH–] increases. (3) pH of the solution is greater than 7. (a) Relation between Kh, Kw, Ka CN– + H2O ‡ˆ ˆˆ ˆ† ˆˆ Kh = For weak acid HCN + OH–  HCN   OH –  ..... (1) CN –    HCN ‡ˆ ˆˆ ˆ† ˆˆ CN– + H+ CN-   H +    Ka =   HCN  For water H2O ‡ˆ ˆˆ ˆ† ˆˆ ..... (2) H+ + OH– Kw = [OH–] [H+] ..... (3) eq. (1) × eq. (2) = eq. (3)  HCN [OH – ] × [CN – ][H + ] [CN – ] Kh × Ka = K w Kh = Kw Ka [HCN] = [H+][OH–] IONIC EQUILIBRIUM (b) Degree of hydrolysis : CN– + H2O ‡ˆ ˆˆ ˆ† ˆˆ HCN + OH– C C–x 0 x 0 x Ch Ch Initial concentration at equlibrium nx = a 1x = Ch x = Ch C – Ch Kh = Kh =  HCN  OH –  CN –    = C2 h 2 Ch × Ch = C 1 – h  C – Ch Ch 2 1 – h  Since h <<<< 1 therefore (1 – h) 1  Kh = Ch2 h2 = h= Κh C h= Kw Ka × C Kh  C (c) pH of the solution [OH–] = Ch [OH–] = C × OH –  =   Kw Ka × C Kw × C Ka  Kw = [OH–] [H+]  H+  = K w    OH –    H+  =   Kw Kw × C Ka H+  =   K w × Ka C On taking – log on both sides – log [H+] = – log K w × Ka C  K w × Ka  pH = – log   C   pH =  pH =  pH = 1 2 1 [ log Kw + logKa – logC ] 2 1 1 1 logKw  logKa  logC 2 2 2 1 1 1 pKw+ pKa + logC 2 2 2 pH = 7 + 1 1 pKa + logC 2 2 KEY POINTS Summary Kw (1) K h = K a (2) h = Kw × C Ka (3) [OH–] = Ch = (4) [H+] = Kh = C Kw Ka × C K w × Ka C (5) pH = – log [H+] pH = 7 + 1 1 pKa + log C 2 2 Ex. Find out pH, h and [OH–] of milli molar solution of KCN 10–3 M, if the dissociation constant of HCN is 10–7. Sol. (1) pH = 7 + =7 (2) h = 1 1 1 1 pKa + log C = 7   7  log10 3 2 2 2 2 14  7  3 21  3 7 3 18 = = 9  log10 = 2 2 2 2 2 Kh = C (3) [OH–] = Kw = Ka × C 10 –14 10 –7 ×10 –3 = 10 –14 ×1010 = 10 –4 = 10–2 Kw × C 10 –14 × 10 –3 = = 10 –17 ×10+7 = 10 –10 = 10–5 Ka 10 –7 IONIC EQUILIBRIUM Ex. Calculate the pH and degree of hydrolysis of 0.01 M solution of NaCN, Ka for HCN is 6.2 × 10–12. Sol. NaCN is a salt of strong base NaOH and weak acid HCN. K+ does not react with water whereas CN– reacts with water as here under – CN– + H2O ‡ˆ ˆˆ ˆ† ˆˆ HCN + OH Kh = [HCN][OH – ] K w 10 –14 = = = 1.6 × 10–3 [CN ] K a 6.2 ×10 –12 Let, x moles of salt undergo hydrolysis then concentrations of various species would be [CN–] = (0.01 – x) 0.01, [HCN] = x [OH–] = x  Kh = x.x = 1.6 × 10–3 0.01 x2 = 1.6 × 10–5 x = 4 × 10–3 [OH–] = x = 4 × 10–3 M [H3O+] = 10 –14 KW = 0.25 × 10–11 – = [OH ] 4 ×10 –3 pH = – log(0.25 × 10–11) = 11.6020 x 4 ×10 –3 = = 4 × 10–11 0.01 0.01 Degree of hydrolysis = Ex. Sol. Calculate for 0.01 N solution of sodium acetate (i) Hydrolysis constant (ii) Degree of hydrolysis Given Ka of CH3COOH = 1.9 × 10–5. For CH3COONa + Initial After H 2O ‡ˆ ˆˆ ˆ† ˆˆ C C(1–h) CH3COOH 0 Ch (iii) pH + NaOH 0 Ch 10 –14 Kw = = 5.26 × 10–10 K a 1.9 ×10 –5 (i) Kh = (ii) h= (iii) [OH–] from NaOH, a strong base = Ch = 0.01 × 2.29 × 10–4 = 2.29 × 10–6 M pOH = 5.64  pH = 14 – 5.64 = 8.36 Kh = C 5.26 ×10 –10 = 2.29 × 10–6 M 0.01 3. Hydrolysis of (WA - WB) Type Salt Ex. NH4CN, CaCO3, (NH4)2 CO3, ZnHPO3 NH4CN + H2O ‡ˆ ˆˆ ˆ† ˆˆ NH4OH + HCN NH4+ + CN– + H2O ‡ˆ ˆˆ ˆ† ˆˆ (1) NH4OH + HCN Solution is almost neutral but it may be acidic or basic depending upon the nature of acid & base & pH of the solution is near to 7. (a) Relation between Kh, Kw, Ka & Kb NH + + CN— + H O ‡ˆ ˆˆ ˆ† ˆˆ NH OH + HCN 4 2 4 Kh =  NH 4OH   HCN  For weak base  NH +4    NH OH ‡ˆ ˆˆ ˆ† ˆˆ For weak acid  NH +4   OH –    Kb =   NH 4 OH    + ˆ ˆ ˆ† HCN ‡ ˆ ˆˆ H + CN– 4 ..... (1) CN –    NH4+ + OH– ..... (2)  H +  CN –   Ka =     HCN  ..... (3) H2O ‡ˆ ˆˆ ˆ† ˆˆ H++ OH– Kw = [OH–] [H+] Multiply Eq. (1) × Eq. (2) × Eq. (3) = Eq. (4) For water  NH 4OH   HCN   NH +4   CN –      NH +4   OH –   H +   CN –      = [H+] [OH–] × ×    NH 4OH  HCN     Kh × Kb × K a = K w Kh = (2) ..... (4) Kw Ka × Kb Degree of Hydrolysis – NH4+ + CN– + H2O ‡ˆ ˆˆ ˆ† ˆˆ C C C–x C–x  nx = a  x = Ch Kh = C – Ch  NH 4OH   HCN   NH +4  CN –     Since h <<<< 1 Then (1 – h) 1  Kh = h2 = C – Ch NH4OH + HCN 0 0 x x Ch Ch Ch × Ch C2 h 2 h2 = =  C – Ch  C – Ch  C 1 – h  × C 1 – h  1 – h 2 Initial concentration at equilibrium IONIC EQUILIBRIUM or h2 = KW Ka × Kb h= KW Ka × Kb ..... (5) (b) pH of the solution From eq. (3)  H +  CN –   Ka =     HCN   H +  = K a ×  HCN     CN –    K ×h  H +  = K a × Ch = a   C – Ch 1– h Since h <<<< 1 (1 – h) 1 [H+] = Ka × h = Ka × [Now put the value of h from eq. (5)] Kw Ka × Kb [H + ] = K w × Ka Kb On taking – log on both sides  K × Ka  – log [H ] = – log  w   Kb  + 1 2 pH = – 1 [ log (Kw × Ka) – log Kb ] 2 pH = – 1 [ log Kw + log Ka – log Kb ] 2 pH = – 1 1 1 [ log Kw] – [log Ka] – [ – log Kb ] 2 2 2 pH = + 1 1 1 pKw + pKa – pKb 2 2 2 pH = 7 + 1 1 pKa – pKb 2 2 314 KEY POINTS Summary (1) K h  Kw Ka  K b (2) h = K h = K w × Ka = Ka.h Kb (3) [H+] = pH = 7 + Kw Ka × Kb (4) pH = – log [H+] 1 1 pKa – pKb 2 2 Note : Degree of hydrolysis of [WA – WB] type salt does not depend on the concentration of salt. Ex. Salt of weak acid and weak base CalculatepH of themixture(25 mL of 0.1 M NH4OH + 25 mL of 0.1 M CH3COOH). (i) Sol. Given that Ka : 1.8 × 10–5, and Kb = 1.8 × 10–5 NH4OH + CH3COOH  CH3COONH4 + H2O Initial milli moles 25 × 0.1 25× 0.1 0 0 = 2.5 = 2.5 – Final milli moles 0 0 2.5 2.5 As salt is formed (salt of weak acid and weak base) and pH will be decided by salt hydrolysis pH = Ex. pK w + pK a – pK b 1 = (–log 10–14 – log 1.8 × 10–5 + log 1.8 × 10–5) = 7 2 2 In the following which one has highest / maximum degree of hydrolysis. (1) 0.01 M NH4Cl (2) 0.1 M NH4Cl (3) 0.001 M NH4Cl (4) Same Sol.  [3]  h =  Ex. In the following which one has lowest value of degree of hydrolysis. Kh C  if C decreases, h increases   (1) 0.01 M CH3COONH4 (2) 0.1 M CH3COONH4 (3) 0.001 M CH3COONH4 (4) Same Sol. [4] Ex. Sol. Find out the concentration of [H+] in 0.1M CH3COONa solution (Ka = 10–5) Salt is [WA – SB] type  [H+] = K w × Ka = C ×10 –5 10 –1 = 10 –19 × 10+1 = 10 –18 = 10–9 IONIC EQUILIBRIUM Ex. Calculate the degree of hydrolysis of a mixture containing 0.1N NH4OH and 0.1N HCN Ka = 10–5 Sol. Kb = 10–5 Salt is [WA – WB] h= (4) & Kw = Ka × Kb 10 –14 10 –5 ×10 –5 = 10 –14 ×10+10 = 10 –4 = 10–2 Hydrolysis of [SA – SB] Type Salt – Ex. NaCl, BaCl2, Na2SO4, KClO4 etc. NaCl + H2O ‡ˆ ˆˆ ˆ† ˆˆ NaOH + HCl Na+ + Cl– + H2O ‡ˆ ˆˆ ˆ† ˆˆ H2O ‡ˆ ˆˆ ˆ† ˆˆ Na+ + OH– + H+ + Cl– H+ + OH– (It is not salt hydrolysis) (1) Hydrolysis of salt of [SA – SB] is not possible (2) Solution is neutral in nature (pH = pOH = 7) (3) pH of the solution is 7 Hydrolysis of Polyvalent Anions or Cations     The hydrolysis of these species will take place in steps (just like dissociation of weak acids). Out of different steps generally first step hydrolysis dominants mainly because of two reasons The hydrolysis constant of second and further steps is generally negligible in comparison to first step hydrolysis constant. The second and further step hydrolysis will be suppressed in presence of ions produced due to first step hydrolysis. For a polyprotic acid (H2S, H3PO4, H2CO3, H2C2O4) we already know that the disscociation always takes place in steps, so for example for H3PO4 + – H3PO4 ‡ˆ ˆˆ ˆ† ˆˆ H + H2PO4 Ka 1 = –2 – + H2PO4 ‡ˆ ˆˆ ˆ† ˆˆ H + HPO4 Ka 2 = –3 2– + HPO4 ‡ˆ ˆˆ ˆ† ˆˆ H + PO4 Ka 3 = [H + ][H 2 PO 4– ] [H3PO4 ] ..... (1) [H + ][HPO 42– ] ..... (2) [H + ][PO3– 4 ] ..... (3) [H 2 PO4– ] [HPO 2– 4 ] For all acids we always have Ka >> K a 2 >> K a 3 1 pH of the solution can be calculated from Ist step only because [H+] from IInd & IIIrd step can be neglected as (a) Ka >> K a 2 >> K a 3 1 (b) [H+] from Ist dissociation will suppress the dissociation of IInd & IIIrd step. Now for the hydrolysis of polyvalent ions of salts (like K3PO4, Na2C2O4, ZnSO4, FeCl3, (NH4)2C2O4 or ions like 3– 2– PO4 , C2O4 , Zn2+, Fe3+ etc). Consider the hydrolysis in step 3– 2– – PO4 + H2O ‡ˆ ˆˆ ˆ† ˆˆ HPO4 + OH C 0 0 K h1 = [OH – ][HPO2– 4 ] 3– [PO 4 ] ..... (4) 2– – – HPO4 + H2O ‡ˆ ˆˆ ˆ† ˆˆ H2PO4 + OH Kh 2 = [OH – ][H 2 PO 4– ] [HPO 2– 4 ] ..... (5) – – H2PO4 + H2O ‡ˆ ˆˆ ˆ† ˆˆ H3PO4 + OH Kh 3 = [OH – ][H 3PO 4 ] [H 2 PO 4– ] ..... (6) C(1 – h) Ch Ch + – H2O ‡ˆ ˆˆ ˆ† ˆˆ H + OH From above equations we get. Kw = [H+] [OH–] ..... (7) K a 1 × K h 3 = Kw K a 2 × K h 2 = Kw K a 3 × K h 1 = Kw Genarally pH is calculated only using the first step hydrolysis Ch 2 ChCh = Ch2 C(1– h) 1– h K h1 = h= K h1 C [OH–] = Ch = K h1 × C [H+] = Kw – [OH ] = Kw Ka3 KwC = k ×K a 3 C 1 [pKw + pK a 3 + logC] 2 What is the pH of 1.0 M Na3PO4 in aqueous solution ? So pH = Ex. 3– 2– – –2 PO4 + H2O ‡ˆ ˆˆ ˆ† ˆˆ HPO4 + OH ; Kb = 2.4 × 10 Sol. 10 –14 2– Ka(HPO4 ) = 2.4 ×10 –2 pKa = – logKa = 12.38 or pH = 7 + = 4.17 × 10–13 1 1 1 1 pKa + logC = 7 + (12.38) + log(1) = 13.19 2 2 2 2 Hydrolysis of Amphiprotic Anion (Cation is not Hydrolysed) NaHCO3, NaHS, etc., can undergo ionisation to from H+ ion and can undergo hydrolysis to from OH– (Na+ ion is not hydrolysed) ionisation + (a) (i) HCO3 + H2O ‡ˆ ˆˆ ˆˆ ˆˆ ˆˆ ˆˆ ˆ† ˆˆˆˆ CO3 + H3O (acid) – 2– hydrolysis – (ii) HCO3 + H2O ‡ˆ ˆˆ ˆˆ ˆˆ ˆˆ ˆˆ ˆˆ†ˆˆ H2CO3 + OH– (base)  pK a1  pKa 2 – pH(HCO3 ) =  2     IONIC EQUILIBRIUM Similarly for H2PO4– and HPO42– amphiprotic anions. (b) pH (H – 2 PO 4 )  pK a 2 + pK a 3   pK a1 + pK a 2  =  =    and pH (HPO 2– 4 ) 2 2     Cation is Also Hydrolysed (i) Salts like NH4HCO3, NH4HS in which HCO3– and HS– are amphiprotic respectively but NH4+ will also hydrolysed. (ii) Equilibria in such solutions will be : (Hydrolysis of anion) – HCO3– + H2O ‡ˆ ˆˆ ˆ† ˆˆ H2CO3 + OH (Hydrolysis of cation) + NH4+ + H2O ‡ˆ ˆˆ ˆ† ˆˆ NH4OH + H so, K  [H+] = K a1  w – K a 2   Kb  Example Based on : Salt Hydrolysis Ex. Select the compound whose 0.1 M solution is basic : (A) ammonium chloride (B) ammonium acetate (C) ammonium sulphate (D) sodium acetate Sol. (D), since sodium acetate is salt of (WA + SB) so its pH > 7. Ex. If one equivalent of a strong acid is added to one equivalent of a weak base, the resulting solution will be. (A) neutral (B) acidic (C) alkaline Sol. (B), since after neutralisation salt of (SA + WB) will form and its pH < 7. Ex. Which is the correct option for hydrolysis constant of NH4CN ? (A) Sol. Kw Ka Kw (B) K × K a b Kb Kc Ka (D) K b (B), Since NH4CN is a salt of (WA + WB). Equilibrium constant of hydrolysis of WA + WB is Ex. (C) (D) none of these Kw Ka × Kb Increasing order of pH of 0.1 M solution of the following salts is : (A) NaCl < NH4Cl < NaCN (B) NH4Cl < NaCl < NaCN (C) NaCN < NH4Cl < NaCl (D) NaCl < NaCN < NH4Cl Sol. (B), Since NH4Cl is the salt of (WB + SA) so pH < 7, NaCl is salt of (SA + SB) so pH = 7 and NaCN is salt of (WA + SB) so pH < 7. Ex. When a solution of 0.01 M CH3COOH is titrated with a solution of 0.01 M NaOH. Calculate the pH at the equivalence point. (pKa of CH3COOH is 4.74) Sol. CH3COOH + NaOH ‡ˆ ˆˆ ˆ† ˆˆ CH3COONa + H2O Let acid be = V mL V mL of 0.01 M CH3COOH will require V mL of 0.01 M NaOH. But CH3COONa formed will make solution alkaline due to hydrolysis CH3COONa + H2O ‡ˆ ˆˆ ˆ† ˆˆ CH3COOH + NaOH 0.01 = 0.005 M 2 for pH of salt of weak acid and strong base [CH3COONa] = pH = 7 + 4.74 log 0.005 + = 8.22 2 2 =7+ Ex. log C pK a + 2 2 Calculate the pH of 0.5 M Na3PO4 in aqueous solution ? 2– – –3 –2 PO43– + H2O ‡ˆ ˆˆ ˆ† ˆˆ HPO4 + OH ; Kb (PO4 ) = 2.4 × 10 Sol. HPO42– and PO4–3 are conjugate acid and base so Ka × Kb = 10–14 Ka(HPO42–) = 10 –14 = 4.17 × 10–13 2.4 ×10 –2 pKa = – log Ka = 12.38 or pH = 7 + 1 1 pKa + log C 2 2 pH = 13.04 Ex. What is degree of hydrolysis, Kh and pH of 1 M urea hydrochloride solution in water. Kb (urea) = 1.5 × 10–14. Sol. NH2CONHCl is a salt of (SA + WB) so h = Kw 10 –14 = K b .C 1.5 ×10 –14 ×1 or h = 81.65 % Kh = Kw 10 –14 = = 6.667 × 10–1 K b 1.5 ×10 –14 pH = 7  =7 1 1 pKb – log C 2 2 1 1 (13.82) – log(1) or pH = 0.09 2 2 BUFFER SOLUTIONS A solution that resists change in pH value upon addition of small amount of strong acid or base (less than 1 %) or when solution is diluted is called buffer solution. The capacity of a solution to resist alteration in its pH value is known as buffer capacity and the mechanism of buffer solution is called buffer action. Types of Buffer Solutions (A) Simple buffer solution (B) Mixed buffer solution IONIC EQUILIBRIUM (A) Simple Buffer Solution A salt of weak acid and weak base in water e.g. CH3COONH4, HCOONH4, AgCN, NH4CN. Buffer Action of Simple Buffer Solution Consider a simple buffer solution of CH3COONH4 ,since it is a salt will dissociated completely. CH3COONH4  CH3COO + NH4+ If a strong acid such as HCl is added then HCl  H+ + Cl The H+ ions from the added acid (HCl) combine with CH3COO ions to form CH3COOH, which is a weak acid so will not further ionized.  Thus there is no rise in H+ ion concentration and the pH remains constant. CH3COO + H+ CH3COOH (Weak acid) If a strong base is added as NaOH NaOH  Na+ + OH NH4+ + OH NH4(OH) (Weak base)  Thus change in OH ion concentration is resisted by NH+ ions by forming NH4OH which is a weak base. So it will not 4 further ionized and pH remains constant. pH of a simple buffer solution :- 1 1 pH = 7 + pk a – pk b 2 2 (B) (a) Mixed Buffer Solutions Acidic Buffer Solution An acidic buffer solution consists of solution of a weak acid and its salt with strong base. The best known example is a mixture of solution of acetic acid and its salt with strong base (CH3COONa). Other example : HCN + KCN, (H2CO3 + NaHCO3) blood CH3COOH CH3COO + H+ (Weakly ionised) CH3COONa CH3COO + Na+ (Highly ionised) When a few drops of an acid (HCl) are added to it, the H+ ions from the added acid (HCl) combine with the CH3COO ions to form CH3COOH. Thus there is no rise in H+ ion concentration and the pH of solution remains constant. On the other hand, when a few drops of base(NaOH) are added, the OH of the added base reacts with acetic acid to form unionise water and acetate ions. CH3COOH + OH H2O + CH3COO. Thus there is no increase in OH ion concentration and hence the pH of the solution remains constant. pH of an Acidic Buffer Solution (Handerson Equation) Consider a buffer mixture (acidic buffer) HA + NaA (CH3COOH + CH3COONa) where A = CH3COO, HA H+ + A NaA  Na+ + A Applying law of mass action to dissociation equilibrium of HA Ka = taking log K a [HA] [H + ][A – ] ; so [H+] = [HA] [A – ] log [H+] = log Ka + log [HA] [A – ] A– = CH3COO– –log [H+] = – log Ka – log pH = pKa + log [HA] [A – ] [A – ] [HA] [A] = Initial concentration of salt as it is mainly comes from salt. [HA] = Initial concentration of the acid. pH = pKa + log [Salt] [Acid] (it is known as Handerson-Hasselbalch equation.) K EE YYPP OOI INN TTS S A solution can act as buffer only if ratio of concentration of salt to acid is between 0.1 to 10. CH3COOH : CH3COONa 1 10 pH = pKa  1 10 1 pH = pK a  1 Thus pH range of an acidic buffer solution is pKa  1 to pKa  1 pH range = pKa  1 Maximum buffer capacity when concentration of salt is equal to that of acid. [Salt] = [Acid] Maximum buffer action will be only when ratio of concentration of acid and salt is 1. So for maximum buffer action. pH = pKa + log1/1  pH = pKa Ex. Sol. Calculate the pH after the addition of 80 mL and 100 mL respectively of 0.1 N NaOH to 100 mL, 0.1 N CH 3COOH. (Given pKa for CH3COOH = 4.74) If 80 mL of 0.1 N NaOH is added to 100 mL of 0.1 N CH3COOH, acidic buffer will form as Initial Final H3CCOOH + NaOH  H3CCOONa + H2O 0.01 eq. 0.002 eq. 0.008 eq. 0 0 0 0.008 eq. [CH 3COO – ] 8 pH = pKa + log [CH COOH] = 4.74 + log = 4.74 + 0.6020 = 5.342 2 3 If 100 mL of 0.1 N NaOH is added is added to 100 mL of 0.1 N CH3COOH, complete neutralization takes place and the concentration of H3CCOONa = Now, pH = 7 + pH = 8.72 0.1 M = 0.05 M 2 1 1 pKa + log C 2 2 IONIC EQUILIBRIUM Ex. How much volume of 0.2 M solution of acetic acid should be added to 100 mL of 0.2 M solution of sodium acetate to prepare a buffer solution of pH = 6.00? (pKa for acetic acid is 4.74) Sol. pH = pKa + log log [Salt] [Acid] [Salt] [Salt] = pH – pKa = 6.00 – 4.74 = 1.26  = 18.2 [Acid] [Acid] Moles of CH3COONa in solution 100 × 0.2 = 0.02 1000 Let, volume of 0.2 acetic acid added = V mL  Moles of acetic acid =  V × 0.2 1000 0.02 = 18.2 0.2 V× 1000  V = 5.49 mL Ex. Calculate the pH of a solution when 0.20 moles of HCl is added to one litre solution containing ? (a) 1 M each of acetic acid and acetate ion ? (b) 0.1 M each of acetic acid and acetate ion ? –5 Given Ka for acetic acid is 1.8 × 10 . Sol. (a) Initially [Acetic acid] =1 M [Acetate] = 1 M Now 0.2 moles of HCl are added to it. HCl Mole before reaction CH3COO–  CH3COOH + 0.2 Mole after reaction 0 Cl– + 1 1 0 0.8 1.2 0.2 –  New [CH3COOH] = 1.2 ; [CH3COO ] = 0.8  pH = pka + log [conjugate] [acid]  pH = – log 1.8 × 10–5 + log (b) 0.8 = 4.5686 1.2 In II case initially [Acetic acid] = 0.1 M [Acetate] = 0.1 M Now 0.2 mole of HCl are added to it HCl + CH3COO–  CH3COOH + Cl– Mole before reaction 0.2 0.1 0.1 0 Mole after reaction 0.1 0 0.2 0.1 + –1  [H ] from free HCl = 0.1 = 10 M  pH = 1 Note : CH3COOH no doubt gives H+ but being weak acid as well as in presence of HCl does not dissociate appreciably and thus, H+ from CH3COOH may be neglected. Ex. Calculate the ratio of pH of a solution containing 1 mole of CH3COONa + 1 mole of HCl per litre and of other solution containing 1 mole CH3COONa + 1 mole of acetic acid per litre. Sol. Case I : pH when 1 mole CH3COONa and 1 mole HCl are present. CH3COONa + HCl  CH3COOH + NaCl Before reaction 1 1 0 0 After reaction 0 0 1 1  [CH3COO] = 1 M  Ka   [H+] = C. = C   =  C  (K a .C) = (K a ) C=1 1  pH1 = – logK a 2 Case II : pH when 1 mole CH3COONa and 1 mole of CH3COOH ; a buffer solution  pH2 = – log Ka + log [salt] [acid]  [Salt] = 1M pH2 = – log Ka  (b)  [Acid] = 1M pH1 1 pH 2 = 2 Basic Buffer Solution A basic buffer solution consists of a mixture of a weak base and its salt with strong acid. The best known example is a mixture of NH4OH and NH4Cl. NH4OH NH4Cl NH4+ + OH   NH + Cl + 4  (Weakly ionised) (Highly ionised) When a few drops of a base (NaOH) are added, the OH ions from NaOH combine with NH4+ ions to form feebly ionised NH4OH thus there is no rise in the concentration of OHions and hence the pH value remains constant. NH4+ + OH  NH4OH If a few drops of a acid (HCl) are added the H+ from acid combine with NH4OH to form H2O and NH4+ ions. NH4OH + H+  NH4+ + H2O Thus the addition of acid does not increase the H+ ion concentration and hence pH remains unchanged. Ex. Calculate the amount of NH3 and NH4Cl required to prepare a buffer solution of pH = 9 when total concentration of buffering reagents is 0.3 mol L–1. (pKb for NH3 = 4.7, log 2 = 0.30) Sol. pOH = – log Kb + log 5 = 4.7 + log  a = 2b [salt] [Base] a a  =2 b b IONIC EQUILIBRIUM Given a + b = 0.3 2b + b = 0.3  3b = 0.3 or b = 0.1 mole/L Amount of base = 0.1 × 17 = 1.7 g/L a = 0.2 mole/L  Amount of salt = 0.2 × 53.5 = 10.7 g/L Thus, [Salt] = 0.2 M and [Base] = 0.1 M • pH of Basic Buffer Solution NH4OH NH4+ + OH NH4Cl  NH4+ + Cl  NH +4  OH –    Kb =   NH4OH  OH –  =   K b  NH 4 OH  K  Base   = b  NH +4  Salt    ( NH +4 mainly comes from salt) taking –log on both side • –log OH– = –log K b  Base   Base pOH = –log Kb – log Salt   Salt  pOH = pKb + log Salt   Base  pH = 14 – pOH pOH Range A solution can act as buffer solution only if ratio of concentration of salt to base is from 0.1 to 10. NH4OH : NH4Cl 1 10 pOH = pKb + 1 10 1 pOH = pKb – 1 So pOH range is pKb ± 1 • Condition for Maximum Buffer Action [NH4OH] : [NH4Cl] 1 1 pOH = pK b + log pOH = pKb i.e. [Salt] = [Base] 1 1 and pH = 14 – pKb Maximum buffer action because pH remains constant. Ex. Sol. A solution of weak base LiOH was titrated with 0.1 N HCl. The pH of the solution was found to be 10.04 and 9.14 after the addition of 5 mL and 20 mL of the acid respectively. Find the dissociation constant of the base. Case I : LiOH + HCl  LiCl + H2O Millimole before reaction a 0.1× 5=0.5 0 0 Millimole after reaction (a–0.5) 0 0.5 0.5 [LiCl] [LiOH] pOH = – log Kb + log pH = 10.04 so  3.96 = – log Kb + log .........(i) pOH = 3.96 0.5 (a – 0.5) Case II : Millimole before reaction Millimole after reaction LiOH a (a–2) .........(ii) + HCl  0.1× 20=2 0 [LiCl] [LiOH] pOH = – log Kb + log pH = 9.14 LiCl 0 2 + H 2O 0 2 .........(iii) pOH = 4.86  4.86 = – log Kb + log 2 (a – 2) .........(iv) Solving Eqs. (ii) and (iv), Kb = 1.81 × 10–5 Ex. An organic base B has Kb value equal to 1 × 10–8. In what amounts should 0.01 M HCl and 0.01 M solution of B be mixed to prepare 1 L of a buffer solution having pH = 7.0 ? Sol. + – B + H2O ‡ˆ ˆˆ ˆ† ˆˆ BH + OH [XH + ][OH – ] Kb = = 1 × 10–8 [B] pOH = pKb + log [BH + ] [B]  7 = – log (10–8) + log log [BH + ] [BH + ]  7 = 8 + log [B] [B] [BH + ] = –1 [B] [BH + ]  = 10–1 = 0.1 [B] Let, volume of HCl taken = xL  Volume of base taken = (1 – x) L After the reaction, millimole of BH+ formed = 0.01 × (x) Millimoles of base left = 0.01 (1 – 2x)  [BH  ] x = = 0.1 [B] [1 – 2x]  x = 0.083 L = Volume of HCl  Volume of base = 0.0917 L IONIC EQUILIBRIUM BUFFER CAPACITY It is defined as the number of moles of acid (or base) require by one litre of a buffer solution for changing its pH by one unit. Buffer capacity = No.of moles of acid or bases added per litre changein pH Buffer capacity gives the tendency of buffer to resist change in its pH. Higher is the buffer capacity, smaller will be the change in pH and more efficient will be the buffer. Ex. When 2 moles of HCl is added to 1 lit. of an acidic buffer solution, its pH changes from 3.4 to 3.9. Find its buffer capacity. Sol. B.C. = 2 =4 0.5 EXAMPLE BASED ON : Buffer solutions Ex. Which of the following buffers containing NH4OH and NH4Cl show the lowest pH value? Conc. of NH4Cl (mol L–1) 0.10 0.50 (C) 0.50 1.50 (D) 0.50 0.10 (B) pOH = pkb + log (A) 0.50 (B) Sol. Conc. of NH4OH (mol L–1) 0.50 [salt] for NH4Cl = 0.5 and NH4OH = 0.1 [base] pOH will be maximum and so pH will be minimum. Ex. pH of a mixture containing 0.2 M X– (base) and 0.4 M HX with pKb (X–) = 4 is : (A) 4 + log 2 Sol. (D). HX ‡ˆ ˆˆ ˆ† ˆˆ [H+] = Ka or Ex. (B) 4 – log 2 H+ + X– , Ka = (D) 10 – log 2 10 –14 = 10–10 Kb  HX  10 –10 × 0.4 = [X – ] 0.2 pH = 10 – log 2 pH of a mixture of 1 M benzoic acid (pKa = 4.2) and 1 M sodium benzoate is 4.5, in 150 mL buffer, benzoic acid is : (A) 200 mL Sol. (C) 10 + log 2 (D) pH = pKa + log (B) 150 mL (C) 100 mL [salt] [Acid]  (150 – x)  4.5 = 4.2 + log   where x is the volume of benzoic acid  x  (150 – x)   (150 – x)  0.3 = log   or 2.0 =    x x or x = 50 mL (D) 50 mL Ex. Buffering action of a mixture of CH3COOH and CH3COONa is maximum when the ratio of salt to acid is equal to : (A) 1.0 (B) 100.0 (C) 10.0 (D) 0.1 Sol. (A). The buffer action of a buffer mixture is effective in the pH range pKa ± 1. It is maximum when pH = pKa. Ex. What amount of sodium propanoate should be added to one litre of an aqueous solution containing 0.02 mole of propanoic acid to obtain a buffer solution of pH 4.74 ? What will be the pH of 0.01 mol of HCl is dissolved in the above buffer solution ? Compare the last pH value with the pH of 0.01 molar HCl solution. Dissociation constant of propanoic acid at 25°C is 1.34 × 10–5. Sol. Using Henderson's expression pH = pKa + log [salt] [acid] We get 4.74 = – log (1.34 × 10–5) + log Which gives 4.74 = 4.87 + log [Salt] 0.02 [Salt] [Salt] or = 0.74 or [Salt] = 1.48 × 10–2 M 0.02 0.02 Hence, amount of sodium propanoate to be added = 1.48 × 10–2 × 96 g = 1.42 g The addition of 0.01 mol of HCl converts the equivalent amount of sodium propanoate into propanoic acid. Hence, we will have pH = 4.87 + log (0 .01 48 2  0.01)molL1 (0.0 2  0.0 1)molL1 pH = 4.87 + log (0.160) = 4.87 – 0.79 = 4.08 (The pH of 0.01 molar HCl solution would be pH = – log (0.01) = 2) Ex. A solution of a weak acid was titrated with NaOH the equivalence point was reached when 25.06 mL of 0.1 N NaOH have been added. Now 12.53 mL of 0.1 N HCl were added to the titrated solution, the pH was found to be 4.92. What is Ka of the acid. Sol. For complete neutralisation, meq. of acid = meq. of NaOH = 25.06 × 2.506 = 3.612 meq. HA + NaOH  NaA + H2O Initial 2.506 At equivalence point 2.506 0 0 0 0 2.506 2.506 Now 1.253 meq. of HCl are added so, NaA + HCl  NaCl + HA 2.506 1.253 0 0 (Before) 1.253 0 1.253 1.253 (After) Now mixture contains HA and NaA so it will be Acidic buffer, pH = pKa + log [Salt]  1.253  = pKa + log    1.253  [Acid] or pH = pKa or Ka = antilog (–4.92) or Ka = 1.2 × 10–5 IONIC EQUILIBRIUM Solubility (s) and Solubility Product (Ksp) Solubility : “Moles of a substance dissolved per unit volume of a solution.” Classification of Salts : If S > 0.1 M  Soluble Salts If 0.01 M < S < 0.1 M  Partial Soluble salts If S < 0.01 M  Sparingly soluble salts Note : All salts of alkali metals and NH4+ ion are generally water soluble. Examples of sparingly soluble salts are AgCl, PbCl2, Hg2Cl2, PbSO4, Ag2CO3, CaSO4, AgCN, etc. Process of Dissolution of Sparingly Soluble Salts Let AB   Sparingly Soluble Salt A(aq) + B(aq) Initially, rate of dissociation > rate of ppt.  more salt can be dissolved and solution is unsaturated. But, when Saturated solution rate of dissolution = rate of ppt ion In a saturated solution, whatever salt is dissolved will be present in the form of its ions and therefore, concentration of ions in a saturated solution will represent solubility of the salt. This is generally used for sparingly soluble salts. We will be dealing with the solubilities in the following type of solution Solubility Product (Ksp) For a saturated solution, AB ƒ A + B [A + ][B– ] Keq. = [Q concentration of solid is constant] [AB] + –  Keq. [AB] = [A ] [B ] Ksp = [A+] [B–] Solubility product (Ksp) is a type of equilibrium constant, so will be dependent only on temperature for a particular salt. Here different methods for writing Ksp for different types of salts are following : (a) AB Type Salt A+ (aq) + B– (aq) AB(s) ƒ s Ksp = [A ] [B–] = s2 s + s= K sp (b) A2B Type Salt 2A+(aq) A2B(s) ƒ B– (aq) + 2s s Ksp = [A+]2[B–] = [2s]2[s] = 4s3 Ksp = 4s3 (c) AB3 Type Salt A3+(aq) AB3(s) ƒ 3B–(aq) + s 3s Ksp = [s][3s]3 = 27s4 (d) A2B3 Type Salt A2B3(s) ƒ 2A3+(aq) 2s Ksp = [2s] [3s]3 + 3B2–(aq) 3s 2 Ksp = 108 s5 (e) AxBy Type Salt AxBy ƒ xAy+ + yBx– xs Ksp = (xs)x (ys)y ys Ksp = xx.yy.sx+y Following examples will illustrate the different type of solubilities and the effects of different factors or situation on solubility of a salt. Simple Solubility Let the salt is Ax By, in solution in water, let the solubility in H2O = 's' M, then y+ –x AxBy ‡ˆ ˆˆ ˆ† ˆˆ xA + yB – xs ys Ex. Calculate ksp of Li3Na3[AlF6]2 Sol. ksp = 33.33.22.(s)8 = 36.4 (s)8 = 2916 s8 Ex. Sol. Self Practice Problem Calculate ksp of Mg3(PO4)2 108 s5 ksp = (xs)x (ys)y = xx, yy.(s)x+y IONIC EQUILIBRIUM Solubility in Different Solutions Solubility in Presence of Common - Ion Because of the presence of common ion the solubility of the salt decreases Ex. Calculate solubility of silver oxalate in 10–2M Potassium oxalate solution given that ksp of silver oxalate = 10–10 Sol. [oxalate] = 10 –2 + x, where x is the solubility of silver oxalate, this can be neglected in comparison to 10–2. so ksp = 10–10 = 10–2 × (2x)2  10 –8 = x2 x = 5 × 10–5 2× 2 Self Practice Problem Ex. Sol. Calculate the solubility of BaCl2 in presence of 'c' mol/litre NaCl in terms of Ksp(BaCl2). Ksp/c2 • • Calculation of Simultaneous Solubility When two sparingly soluble salts are added in water simultaneously, there will be two simultaneous equilibria in the solution. Let simultaneous solubility of AB be x mol L–1 A+(aq) + B–(aq) AB(s) ƒ x+y x Ksp1 = x(x + y) Simultaneously solubility of AC be y mol L–1 A+(aq) + C–(aq) AC(s) ƒ x+y y Ksp2 = y(x+y) Solving x+y= K sp1 + K sp2 and x K sp1 = y K sp2 Self Practice Problem Ex. Sol. Ex. Sol. Calculate solubility of BaSO 4 when BaSO 4 and CaSO 4 are dissolved in water simultaneously KspCaSO4 = p, Ksp BaSO4 = q and solubility of CaSO4 is 'b' mol/litre. bq/p Calculate simultaneous solubility of silverthiocyanate and sliver bromide in water given that ksp of silver thiocyanate = 10–12 and ksp of silver bromide = 5 × 10–13 respectively. Let the solubility of AgSCN be x and that of AgBr is y, then + – AgSCN ‡ˆ ˆˆ ˆ† ˆˆ Ag + SCN x+y + AgBr ‡ˆ ˆˆ ˆ† ˆˆ Ag x –12 x+y 10 = x (x + y) ...... (i) –13 ...... (ii) 5 × 10 = y(x + y) On solving we get, x = 2y So y = 4.08 × 10–7 and x = 8.16 × 10–7 + Br– y Ex. Sol. 50 mL of 0.02 M solution of Ca(NO3)2 is added to 150 mL of 0.08 M solution of (NH4)2SO4. Predict whether CaSO4 will beprecipitated or not, Ksp(CaSO4) = 4 × 10–5. Calculation of Ca2+ concentration, M1V1 = M2V2 0.02 × 50 = M2 × 200  [Ca(NO3)2] after mixing = 5 × 10–3 mol L–1 Since Ca(NO3)2 is completely ionized, [Ca2+] = [Ca(NO3)2] = 5 × 10–3 mol L–1 Calculation of SO42– ion concentration Applying M1' V1' = M2'V2'  0.08 × 150 = M2' × 200  M2' = 0.08×150 = 6 × 10–2 M 200  [(NH4)2SO4] is completely ionized, [SO42–] = [(NH4)2SO4) = 6 × 10–2 mol/L Ionic product = [Ca2+] [SO42–] = [5 × 10–3] [6 × 10–2] = 3 × 10–4 Since Ionic product (3 × 10–4) is greater than solubility product of CaSO4 (4 × 10–5), hence precipitate of CaSO4 will be formed. Ex. What is the maximum volume of water required to dissolve 1 g of calcium sulphate at 25°C. For calcium sulphate, Ksp= 9.0 × 10–6. Sol. 2+ 2– CaSO4 (aq) ‡ˆ ˆˆ ˆ† ˆˆ Ca (aq) + SO4 (aq) If S is the solubility of CaSO4 in moles L–1 Ksp= [Ca2+] × [SO42–] = S2 S= K sp = 9.0 ×10 –6 = 3 × 10–3 mol L–1 = 3 × 10–3 × 136 g L–1 = 0.408 gL–1 For dissolving 0.408 g of CaSO4 water required = 1 L  For dissolving 1g CaSO4 water required = 1 L = 2.45 L 0.408 Ex. A weak acid HA after treatment with 12mL of 0.1M strong base BOH has a pH of 3. At the end point, the volume of same base required is 26.6 mL. Calculate Ka of acid. Sol. For neutralization : Total Meq. of acid = Meq. of base = 26.6 × 0.1 = 2.66 Now for partial neutralization of acid HA + BOH  BA Meq. before reaction 2.66 1.2  0 Meq. after reaction 1.46 0  1.2 + H 2O 0 1.2 The resultant mixture acts as a buffer and [HA] and [BA] may be placed in terms of Meq. since volume of mixture is constant.  pH = – log Ka + log  Ka = 8.219 × 10–4 [Salt] [Acid] or 3 = – log Ka + log [1.2] [1.46] IONIC EQUILIBRIUM Ex. Calculate simultaneous solubility of AgCNS and AgBr in a solution of water. Ksp of AgBr = 25 × 10–13 and Ksp of AgCNS = 5 × 10–12. Sol. Let solubility of AgCNS and AgBr in a solution be a and b mol litre–1 respectively. ‡ˆ ˆˆ ˆ† ˆˆ AgCNS (s) Ag+ + CNS– a ‡ˆ ˆˆ ˆ† ˆˆ AgBr (s) a Ag+ + b Br– b K sp AgCNS  [Ag  ][CNS  ]  For AgCNS 5 × 10–12 = (a + b) (a) ..............(1) K sp AgBr  [Ag  ][Br  ] For AgBr 25 × 10–13 = (a + b) (b) ..............(2) By Eqs. (1) and (2),  a 5 ×10 –12 = =2 b 25 ×10 –13  By Eq. (1), (2b + b) (2b) = 1 × 10–12 or a = 2b 6b2 = 5 × 10–12 b = 0.912 × 10–6 mol litre–1 = 9.12 × 10–5 M By Eq. (1), (a + a/2) (a) = 5 × 10–12 a = 1.82 × 10–6 mol litre–1 Ex. Equal volumes of 0.04 M CaCl2 and 0.0008 M Na2SO4 are mixed. Will a precipitate form? Ksp for CaSO4 = 2.4 × 10–5 Sol. CaCl2 Millimole added 0.04 V + Na2SO4  CaSO4 0.0008×V Suppose V mL of both are mixed   [Ca2+] = 0.04 V 2V [SO2– ]= 4 0.0008 V 2V [Ca2+] [SO2– ]= 4 0.04 V 0.0008 V × = 8 × 10–6 2V 2V Thus, [Ca2+] [SO2– ] in solution < Ksp 4 8 × 10–6 < 2.4 × 10–5  CaSO4 will not precipitate. 0 + 2NaCl 0 Ex. Calculate the [OH–] of a solution after 50 mL of 0.2 M MgCl2 is added to 500 mL of 0.4 M NaOH. Ksp of Mg(OH)2 is 1.2 × 10–11. Sol. MgCl2 + 2NaOH  Mg(OH)2 + 2NaCl m mole before reaction 10 20  0 0 0 0 10 20 2  10   20  Thus, 10 m mole of Mg(OH)2 are formed. The product of [Mg2+] [OH–]2 is therefore  × = 4 × 10–3 which 100  100  is more than Ksp of Mg(OH)2. Now solubility (S) of Mg(OH)2 can be derived by Ksp = 4S3 K sp = 3 1.2 ×10 –11 = 1.4 × 10–4 M  S=  [OH–] = 2S = 2.8 × 10–4 M 3 Ex. Will a precipitate of Mg(OH)2 be formed in a 0.002 M solution of Mg(NO3)2, if the pH of solution is adjusted to 9 ? Ksp of Mg(OH)2 = 8.9 × 10–12. Sol. pH = 9  [H+] = 10–9 M or [OH–] = 10–5 M Now if Mg(NO3)2 is present in a solution of [OH–] = 10–5 M, then, Product of ionic conc. = [Mg+2] [OH–]2 = [0.002] [[10–5]2 = 2 × 10–13 lesser than Ksp of Mg(OH)2 i.e, 8.9 × 10–12    Mg(OH)2 will not precipitate. Condition of Precipitation For precipitation ionic product [IP] should be greater than solubility product ksp Ex. You are given 10–5 M NaCl solution and 10–8 M AgNO3 solution, they are mixed in 1 : 1 volume ratio, predict whether AgCl will be precipitated or not, if solubility product of AgCl in 10–2 M AgNO3 is = 10–10 mole per litre. Sol. Ionic product = 10 –8 10 –5 × = 25 × 10–15 < ksp 2 2 Hence no precipitation will take place. Solubility in Appropriate Buffer Solutions Appropriate buffer means that the components of buffer should not interfere with the salt or only H+ or OH– ions should be interacting with the ions of the salt. For example in AgCN since, a buffer solution resist change in its pH, H+ ion concentration remains same and hence, solubility of AgCN in buffer solution can be calculated as K sp Ka = s2 . [H ] buffer + Selective Precipitation: When the ksp values differ then one of the salt can be selectively precipitated. IONIC EQUILIBRIUM Ex. What [H+] must be maintained in saturated H2S(0.1 M) to precipitate CdS but not ZnS, if [Cd2+] = [Zn2+] = 0.1 initially ? Ksp = (ZnS) = 1 × 10–21 Ka = (H2S) = 1.1 × 10–21 Sol. In order to prevent precipitation of ZnS [Zn2+] [S2–] < Ksp(ZnS) = 1 × 10–21 (ionic product) or (0.1) [S2–] < 1 × 10–21 or [S2–] < 1 × 10–20 This is the maximum value of [S 2–] before ZnS will precipitate. Let [H+] to maintain this [S 2–] be x. Thus for H2S ‡ˆ ˆˆ ˆ† ˆˆ 2H+ + S2– Ka = [H + ]2 [S2 – ] x 2 (1×10 –20 ) = [H 2S] 0.1 = 1.1 × 10–21 or x = [H+] = 0.1 M No ZnS will precipitate at a concentration of H+ greater than 0.1 M  Effect on Solubility Because of Complex Formation Taking AgCN as an example we can explain the solubility AgCN in different mediums as follows : Acidic medium > water > basic Ligand [Cu(NH3)4]2+ central metal atom No. of ligands bonded (coordination number) Formation of [Ag(NH3)2]+ Ag+(aq) + NH3(aq) K1 + Ag(NH3) + NH3(aq) Ag+(aq) + 2NH3(aq) ƒ K2 Ag(NH3)+(aq) K1  [Ag(NH 3 ) + ] [Ag + ][NH 3 ] + K2 = [Ag(NH 3 ) 2 + ] [Ag(NH 3 )+ ][NH 3 ] [Ag(NH3)2] (aq) [Ag(NH3)2]+  formation constant / stability constant Kstab / Kf   Kf = K 1 × K 2 [Cu(NH3)4]2+ NH3 NH3 Cu2+ NH3 NH3 2+ Ex. Cu2+(aq) + 4NH3(aq) Kstab / Kf = Kf [Cu(NH3)4]2+ [Cu(NH 3 ) 4 2+ ] [Cu 2+ ][NH 3 ]4 Higher is the value of Kf /Kstab, more stable will be the complex and vice-versa. Kinstant [Cu(NH3)4]2+  Kstab / Kf = Cu2+ + 4NH3 1 K instant The ion which undergo complex formation will be more stable in aqueous solution as compared to that of parent ion. Therefore, the salts whose ions (both cation or anion) undergo complex formation will have more solubility as compared to solubility in pure water. Ex. S1  AgCl in H2O S2  AgCl in NaCl  common ion effect S3  AgCl in aq. NH3 or NH4OH  complex formed S3 > S1 > S2 Ex. What must be the concentration of aq. NH3(eq.) which must be added to a solution containing 4 × 10–3 M Ag+ and 0.001 M NaCl, to prevent the precipitation of AgCl. Given that Ksp(AgCl) = 1.8 × 10–10 and the formation constant of 8 [Ag(NH3)2]+ is Kformation = 10 . 6 Sol. Calculate silver ion concentration which can be allowed to remain in the solution, 1.8 × 10–10 = [Ag+][Cl–] [Ag+] = 1.8 ×10 –10 = 1.8 × 10–7 M, 0.001 This quantity is so small that almost all the Ag+ ion will be consumed. Ag+ + 4 × 10–3 b (ii) Sol. [Ag(NH3)2]+ K= 0 108 6 4 ×10 –3 108 = b = 0.0445 6 1.8 ×10 –7 × (b – 8 ×10 –3 )2 The solubility of Mg(OH)2 is increased by addition of NH4+ ion. Find (i) Kc for the reaction 1.8 × 10–7 Ex. 2NH3 ‡ˆ ˆˆ ˆ† ˆˆ (i) (b – 8 × 10–3) 4 × 10–3 K= 2+ Mg(OH)2 + 2NH4+ ‡ˆ ˆˆ ˆ† ˆˆ 2NH3 + 2H2O + Mg Calculate solubility of Mg(OH)2 in a solution containing 0.5 M NH4Cl ( K sp [Mg(OH )2 ] = 1.0 × 10–11, K b (NH 3 ) = 1.8 × 10–5) For the reaction 2+ Mg(OH)2 + 2NH4+ ‡ˆ ˆˆ ˆ† ˆˆ 2NH3 + 2H2O + Mg Kc = [NH 3 ]2 [Mg 2+ ] [NH +4 ]2 ..........(1) Also for the reaction 336 IONIC EQUILIBRIUM + – NH3 + H2O ‡ˆ ˆˆ ˆ† ˆˆ NH4 + OH + – K b(NH ) = [NH 4 ][OH ] 3 [NH 3 ] ..........(2) [As water is in large excess in both the cases] Now, Kc × Kb2 = + 2 – 2 [NH 3 ]2 [Mg 2+ ] [NH 4 ] [OH ] × [NH3 ]2 [NH +4 ]2 = [Mg+2] [OH–]2 = K sp [Mg (OH )]2 K sp Kc = (ii) Now, let us, assume that 'a' moles of Mg(OH)2 be dissolved in presence of 0.5 M NH4Cl.  Mg(OH)2 + Initial — Eqm. —  K 2b = 10 –11 = 3.08 × 10–2 (1.8 ×10 –5 ) 2  2NH4+ ‡ˆ ˆˆ ˆ† ˆˆ 0.5 2NH3 + 2H2O + Mg2+ 0 — 0 2a — a (0.5–2a) a × (2a) 2 4a 3 Kc = = 3.08 × 10–2 2  (0.5 – 2a) 0.25 or a = 0.124 M Ex. 0.10 mol sample of AgNO3 is dissolved in one litre of 2.00 M NH3. Is it possible AgCl(s) form the solution by adding 0.010 mol of NaCl ? ( K sp(AgCl) = 1.8 × 10–10, K f[Ag(NH Sol. Ag + + 0.10 M =0 = 1.6 × 107) + 2NH3 ‡ˆ ˆˆ ˆ† ˆˆ [Ag(NH3)2 ] 2.00 0.10–0.10 + 3 )2 ] (2–0.20 M) 0 0.10 M = 1.80 M It is assumed that all Ag+ ions have been complexed and only x amount is left Kf =  0.10 [Ag(NH 3 ) 2+ ] 7 + 2  1.6 × 10 = x(1.80) 2 [Ag ] [NH 3 ] x = 1.93 × 10–9 M = [Ag+] undisolved [Cl–] = 1.0 × 10–2 M  [Ag+] [Cl–] = 1.93 × 10–9 × 1.0 × 10–2 = 1.93 × 10–11 < 1.8 × 10–10 [Ksp(AgCl)] Hence, AgCl (s) will not precipitate. Ex. How many grams of CaC 2O 4 will dissolve in distilled water to make one litre of saturated solution (Ksp = 6.25 × 10–10 and its molecular mass is 128) : (A) 0.0064 g Sol. (B) 0.0128 g (C) 0.0032 g (D) 0.0640 g (C) [Ca2+] [C2O42–] = 6.25 × 10–10 so solubility = 2.5 × 10–5 mol/L = 2.5 × 128 × 10–5 = 3.2 × 10–3 g/L Ex Select the sulphides which has maximum solubility in water ? (A) CdS (Ksp = 36 × 10–30) Sol. (B) FeS (Ksp = 11 × 10–20) (C) HgS (Ksp = 32 × 10–54) (D) ZnS (Ksp = 11 × 10–22) (B) All salt are AB type so solubility will be Ex. K sp . Higher the value of Ksp, the maximum will be solubility.. If equal volumes of the following solutions are added, precipitation of AgCl (Ksp = 1.8 × 10–10) will occur only with. Sol. (A) 10–4 M (Ag+) and 10–4 M (Cl–) (B) 10–5 M (Ag+) and 10–5 M (Cl–) (C) 10–6 M (Ag+) and 10–6 M (Cl–) (D) 10–10 M (Ag+) and 10–10 M (Cl–) (A) One can calculate ionic product from given data and for precipitation Ionic product > Ksp. Ex. Sol. If hydrolysis of any one of the ions will occur, after the dissolution of a sparingly soluble salt, then (A) solubility of salt decreases. (B) solubility of salt increases (C) there will be no effect on solubility (D) question is absurd (B) Dissolution equilibria shift towards right side due to hydrolysis of cation or anion. Ex. What is the concentration of Ag+ ions in 0.01 M AgNO3 that is also 1.0 M NH3 ? Will AgCl precipitate from a solution that is 0.01 M AgNO3, 0.01 M NaCl and 1 M NH3? Kd(Ag[NH3]2+) = 5.88 × 10–8 ; Ksp(AgCl) = 1.8 × 10–10. Sol. Let us first assume that 0.01 M AgNO3 shall combine with 0.02 NH3 to form 0.01 M Ag(NH3)2+ and the consider its dissociation. AgNO3 + 2NH3  0.01 M Ag(NH3)2+ 1M 0 0 (1–0.02)=0.98M + 3 2 Ag(NH ) ‡ˆ ˆˆ ˆ† ˆˆ + Ag (0.01 – x) x = 0.01 M ...Initial conc. + 0.01 M ...at eq. conc. 2NH3 (0.98 + 2x)  0.98 M ....Equib. conc. Since x <<< 1 Kd = [Ag + ][NH 3 ]2 = 5.88 × 10–8 [Ag(NH 3 ) 2+ ]  [Ag+] = 5.88 ×10 –8 × 0.01 = 6.12 × 10–10 M (0.98)2 Further, ionic product of AgCl = [Ag+] [Cl–] = (6.12 × 10–10) (0.01) = 6.12 × 10–12 Because the ionic product is smaller than Ksp = 1.8 × 10–10, no precipitate should form. IONIC EQUILIBRIUM INDICATORS An indicator is a substance which changes its colour at the end point or neutral point of the acid-base titration i.e. the substance which is used to indicate neutral point of acid-base titration are called indicators. At End Point N1V1 = N2V2 Indicators are of two types (i) Acidic (i) (ii) Basic Phenolphthalein (HPh) :- HPh is acid indicator. It ionises in water to give colourless H+ ions and pink coloured anions. H+ HPh + Colourless If Ph Pink – [Hph] > [Ph ] Colourless [Hph] < [Ph–] Pink  In acidic medium the dissociation of HPh is almost nil so it gives no colour because acid suppress the ionisation of HPh due to the presence of common ion H+ and the solution remains colourless. HPh H+ + Ph– HCl H+ + Cl–  In alkaline medium, the OH ions combine with H+ ions of the indicator to form water. HPh H+ + Ph NaOH Na+ + OH– Thus Ph ions gives pink colour in alkaline medium. (ii) Methyl Orange (MeOH) :- It is a weak base and dissociates as :MeOH Yellow If Me+ + OH Red Colourless + [MeOH] > [Me ] Yellow [MeOH] < [Me+] Red MeOH is not dissociated in alkaline medium due to the presence of common ions OH  and the solution remains yellow. MeOH Me+ + OH– NaOH Na+ + OH– In acidic medium OH combine with H+ thus increase the ionisation of MeOH. Hence yellow colour of solution change to red colour. MeOH Me+ + OH– HCl H+ + Cl– Modern Quinonoid Theory According to this theory, (i) An acid-base indicator is a dynamic equilibrium mixture of two alternative tautometric forms ; ordinarily one form is benzenoid while the other is quinonoid. (ii) The two forms have different colours. (iii) Out of these one form exist in acidic solution, while the other in alkaline solution. (iv) The change in pH cause the transition of benzenoid form to quinonoid form and vice-versa and consequently change in colour. Ex. (a) For methyl orange Na + O3S + – + (Acid) + H N(CH3)2 ‡ˆ ˆˆ ˆˆ+ ˆˆ ˆˆ ˆ† ˆˆ –H (Alkali) N=N Yellow benzenoid form (in bases) Red quinonoid form (in acids) (b) For Phenolphthalein O OH HO C O  H ‡ˆ ˆˆˆHˆ ˆ† ˆˆ HO C COO– C= O Colourless benzenoid form (in Acid) Red quinonoid form (in Alkali) THEORIES OF INDICATORS (a) Ostwald Theory :- According to this theory (i) Indicators are organic, aromatic weak acids or weak bases. (ii) The colour change is due to ionisation of the acid - base indicator. The unionised form has different colour than the ionised form. (iii) Every indicator shows colour changes in opposite medium due to the conversion of unionized part into ionized part. For example phenolpthalein shows pink colour in basic medium and methyl orange shows red colour in acidic medium. Note : For oxidation reduction (Redox) reactions indicators are not used because these reactions are very fast. Indicators are not used in coloured solution also. TITRATION OF STRONG ACID AGAINST STRONG ALKALI The graph (A) shows how pH changes during the titration of 50 cm3 of 0.1 M HCl with 0.1 M NaOH. NaOH (aq) + HCl (aq)  NaCl (aq) + H2O () The pH of 0.1 M solution of HCl in the beginning would be 1. As alkali is added, the pH changes slowly in the beginning. However, at the equivalence point pH changes rapidly from about 3.5 to 10. It can be shown by simple calculations that pH of the solution is 3.7 when 49.8 cm3 of 0.1 M NaOH solution have been added. The pH suddenly changes to 10 after addition of 50.1 cm3 of the NaOH solution. Thus, any indicator having pH range between 3.5 to 10 will identify the equivalence point. This means that any one of phenolphthalein, methyl orange or bromothylmol blue could be used as an indicator. IONIC EQUILIBRIUM 14 14 10 Phenolphthalein (A) pH 7 10 Strong acid v/s strong base Methyl orange 4 (B) pH 7 0 50 100 3 Volume of base/cm 14 14 10 Phenolphthalein 10 Phenolphthalein pH 7 4 Methyl orange Weak acid v/s strong base Strong acid v/s weak base Methyl orange 4 0 100 50 Volume of base/cm3 (C) Phenolphthalein pH (D) 0 50 100 3 Volume of base/cm Weak acid v/s weak base 7 Methyl orange 4 0 50 100 3 Volume of base/cm TITRATION OF STRONG ACID AGAINST WEAK ALKALI The graph (B) shows how pH changes during titration of 50 cm3 of 0.1 M HCl with 0.1 M NH3. HCl (aq) + NH4OH (aq) NH4Cl (aq) + H2O () In this case, the pH changes rapidly from 3.5 to 7.0 at the equivalence point. Methyl orange, methyl red and bromocresol green are suitable indicators for this type of titration. Phenolphthalein is unsuitable because its pH range lies outside the vertical portion of the curve. TITRATION OF WEAK ACID AGAINST STRONG BASE The graph (C) shows how pH changes during titration of 50 cm3 of 0.1 M CH3COOH with 0.1 M NaOH. CH3COOH (aq) + NaOH (aq)  CH3COONa (aq) + H2O () The vertical portion of this titration curve lies between pH range 7 to 10.6. Phenolphthalein is suitable indicator for this titration. Methyl orange is not suitable for this titration because its pH range lies on the flat portion of the curve. TITRATION OF WEAK ACID AGAINST WEAK BASE The graph (D) represents the titration curve obtained for titration of 50 cm3 of 0.1 M CH3COOH with 0.1 M NH3. CH3COOH (aq) + NH4OH (aq)  CH3COONH4 (aq) + H2O () For this type of titration there is no sharp increase in pH at the equivalence point. No indicator is suitable for this type of titration. Indicator pH range Colour change pKa Neutral colour pH Ex. Methyl orange 3.2 – 4.5 Pink to yellow 3.7 Methyl red 4.4 – 6.5 Red to yellow 5.1 Litmus 5.5 – 7.5 Red to blue 7.0 Phenol red 6.8 – 8.4 Yellow to red 7.8 Phenolpthalein 8.3 – 10.5 Colourless to pink 9.6 Bromophenol blue is an indicator with a value of Ka = 6.84 × 10–6. At what pH it will work as an indicator? Also report the % of this indicator in its basic form at a pH of 5.84. Sol. + – HBPh ‡ˆ ˆˆ ˆ† ˆˆ H + BPh Ka = [H + ][BPh – ] , when BPh– = HBPh, indicator will work. Thus [HBPh] [H+] = 6.84 × 10–6  pH = 5.165 Also if pH = 5.84 or [H+] = 1.44 × 10–6, then Ka = 1.44 ×10 –6.Cα [H + ][BPh – ] or 6.84 × 10–6 = or  = 0.83 or 83 % C(1 – α) [HBPh] IONIC EQUILIBRIUM 1. A strong electrolyte is defined as a substance which dissociates almost completely into ions in aqueous solution and hence is a very good conductor of electricity Ex., NaOH, KOH, HCl, H2SO4, NaCl, KNO3 etc. 2. A weak electrolyte is defined as a substance which dissociates to a small extent in aqueous solution and hence conducts electricity also to a small extent e.g. NH4OH, CH3COOH etc. 3. Degree of dissociation :- The fraction of the total amount of an electrolyte which dissociates into ions is called the degree of dissociation (), i.e. = Number of moles dissociated Number of moles taken 4. According to Arrhenius concept of acids and bases, an acid is a substance which gives H+ ions in the aqueous solution whereas a base is a substance which gives OH– ions in the aqueous solution. 5. According to Bronsted-Lowry concept of acids and bases, an acid is a substance which can give a proton and a base is a substance which accepts a proton. 6. According to Lewis concept of acids and bases, an acid is a substance which can accept a lone pair of electrons whereas a base is a substance which can donate a lone pair of electrons. Types of Lewis Bases •• •• (i) Neutral molecules containing a lone pair of electrons on the central atom like : NH3, R O H , H 2O : etc. (ii) All •• negative ions like F–, Cl–, Br–, l–, OH– etc. Types of Lewis Acids (i) Molecules having central atom with incomplete octet e.g. BF3, AlCl3 etc. (ii) Simple cations e.g. Ag+ , Cu2+, Fe3+ etc. (iii) Molecules having central atom with empty d-orbitals e.g. SnCl4, SiF4, PF5 etc. (iv) Molecules containing multiple bonds between different atoms e.g. O = C = O. 7. According to Ostwald's dilution law, for the solution of a weak electrolyte with concentration C, mol L–1 and  as the degree of dissociation, Ka = Cα 2  Cα 2 1– α or  = Ka / C = Ka V K a1 Strength of acid HA1 = Strength of acid HA 2 Ka2 8. Relative strength of two weak acids is given by 9. Ionic product of water, Kw = [H+] [OH–] or [H3O+] [OH–]. Its value at 25°C = 10–14 10. pH = –log [H3O+], pOH = –log [OH–], pKa = –log Ka, pKb = – log Kb 11. As Kw = [H+] [OH–] = 10–14 therefore pKw = pH + pOH = 14. 12. Solubility product of a sparingly soluble salt Ax By is given by Ksp = [Ay+]x × [B x–]y Ex. 13. for AgCl, Ksp = [Ag+] [Cl–], for Ca3(PO4)2, Ksp = [Ca2+]3 [PO43–]2 etc. If two solutions are mixed in which ions can combine to form a precipitate, concentration of ions in the solution after mixing are calculated. Then ionic product is calculated using the same expression as for Ksp. If ionic product > solubility product, a precipitate is formed. 14. To calculate the solubility of a salt like AgCl in the presence of a strong electrolyte like NaCl, total [Cl–] is calculated (Cl– ions from AgCl being negligible). Knowing Ksp, [Ag+] can be calculated. 15. pH of an acidic buffer is given by Henderson equation viz pH = pKa + log 16. [Salt] [Acid] pH of a basic buffer is given by pOH = pKb + log [Salt] and then pH = 14 – pOH [Base] n No.of moles of the acid or base added to1litre of buffer = ΔpH Change in pH 17. Buffer capacity = 18. pH of boiling water is 6.5625. It does not mean that boiling water is not neutral. It is due to greater dissociation of H2O into H+ and OH–. 19. pH can be zero in 1 N HCl solution or it can be negative for more concentrated solution like 2N, 3N, 10 N etc. 20. The buffer system present in blood is H2CO3 + NaHCO3. IONIC EQUILIBRIUM IONIC EQUILIBRIUM