VIEWS: 21 PAGES: 59 POSTED ON: 2/28/2012
Chemistry Ionic equilibrium-III Session Objectives Session Objectives 1. Hydrolysis of salts 2. Buffer and buffer capacity Buffer solution Types of buffer solution a. Acidic buffer: CH3COOH and CH3COONa b. Basic buffer: NH4OH and NH4Cl c. Salt or neutral buffer: CH3COONH4 Animation of buffer Explanation Now, to find the pH, Let us consider CH3 COOH CH3 COO H H CH COO Ka 3 CH3COOH Since CH3COOH is feebly ionized, so most of CH3COO– will come from the salt. Hence, [CH3COO–] = [Salt] CH COOH H K 3 a CH COO 3 Explanation Taking log of both sides log H logK log [Acid] a [Salt] [Salt] pH pKa log [Acid] This is known as Henderson’s equations. Similarly, for basic buffer [Salt] pOH pKb log [Base] We know, pH + pOH = 14 Questions Illustrative example 1 Calculate the pH of a buffer solution containing 0.1 M each of acetic acid and sodium acetate. (Ka = 1.8 × 10–5) What will be the change in pH on adding (a) 0.01 moles of HCl to 1.0 L of solution? (b) 0.01 moles of NaOH to 1.0 L of solution? (Assume that no change in volume occurs on the addition of HCl or NaOH) Solution: We know pH pKa log [Salt] [Acid] or pH log 1.8 105 log 0.1 0.1 pH = 4.745 + 0 = 4.745 Solution (a) 0.01 moles of HCl give 0.01 moles of H+ which reacts with 0.01 moles of CH3COO– to form CH3COOH. Therefore, CH COO 0.1 0.01 0.09 M 3 CH3COOH 0.1 0.01 0.11 M [0.09] pH 4.745 log 4.745 0.087 4.658 [0.11] Change in pH = 4.745 – 4.658 = 0.087 units Hence, pH will decrease. Solution (b) On adding 0.01 moles of NaOH to a litre of solution, 0.01 moles of OH– will react with 0.01 moles of CH3COOH. Therefore, [CH3COOH] = 0.1 – 0.01 = 0.09 M [CH3COO–] = 0.1 + 0.11 = 0.11 M [0.11] pH 4.745 log 4.745 0.086 4.831 [0.09] Change in pH = 4.831 – 4.745 = 0.086 units Hence, pH will increase. Illustrative example 2 Calculate the amount of NH3 and NH4Cl required to prepare a buffer solution of pH 9.0 when total concentration of buffering reagents is 0.6 mol L-1. pKb for NH3 = 4.7, log 2 = 0.30 ? Solution: NH3 and NH4CL form a basic buffer Let NH3 xM NH4Cl (0.6 x)M NH4Cl pOH pKb log NH3 Solution 0.6 x 14 pH 4.7 log x 0.6 x antilog 0.3 2 x 0.6 – x = 2x 3x = 0.6 0.6 x 0.2 3 NH3 0.2M NH4Cl 0.4M Buffer Capacity Buffer capacity is defined quantitatively as number of moles of acid or base added in 1 L of solution as to change the pH by unity, i.e. Buffer capacity Number of moles of acid or base added to 1 litre = Change in pH Questions Illustrative example 3 The pH of a buffer is 4.75. When 0.01 mole of NaOH is added to 1 L of it, the pH become 4.83. Calculate its buffer capacity. Solution: Moles of base added to 1 litre solution Buffer capacity = Change in pH 0.01 0.125 4.83 4.75 Illustrative example 4 10 ml. of 10-5 M solution of sodium hydroxide is diluted to one litre. The pH of the resulting solution will be nearly equal to: (a) 6 (b) 7 (c) 8 (d) 9 Solution: Moles of NaOH in 10ml 105 10 103 107 Solution Moles of NaOH in 103 ml diluted solution 7 103 10 105 10 NaOH Na OH 105 105 [OH ] 105 pOH 5 pH 14 5 9 Hence, answer is (d). Illustrative example 5 Calculate the pH value of the mixture containing 50 c.c M - HCl and 30 c.c. M-NaOH solution assuming both to be completely ionised Solution: HCl NaOH NaCl H2O 3 50110 301103 moles moles pH of the solution depends on the conc. of the remaining subs tan ce Solution (50 103 30 103 ) [HCl] re m ainin g 80 103 20 103 0.25 [H ] 0.08 pH log[H ] log0.25 0.602 Solubility and solubility product Any sparingly soluble salt will dissolve in very small amount and whatever dissolves will ionize into respective ions. At certain temperature solubility of a salt is fixed. Thus AgCl (s) Ag+ + Cl Ag Cl K [AgCl (s)] [AgCl] may be taken as constant, thus K·[AgCl] = [Ag+]·[Cl–] Solubility and solubility product or Ksp = [Ag+]·[Cl–] Where, Ksp = Solubility product in saturated solution. In saturated solution, the ionic product, Ki = Ksp If Ki < Ksp the solution is unsaturated. If Ki > Ksp the solution is super saturated and precipitation occurs. Different cases of solubility product a. Electrolyte of the type AB2/A2B AB2 s A 2 2B s 2s Where s =solubility of the salt in mole/L 2 Ksp A 2 B s (2s)2 4s3 PbCl2 Pb2 2Cl 2 Ksp Pb2 Cl = s × (2s)2 = 4s3 b. Electrolyte of the type AB3 AB3 s A 3 3B s 3s 3 Ksp A 3 B s(3s)3 27s4 Ksp s 4 27 AlI3 s Al3 3I 3 Ksp Al3 I = s × (3s)3 = 27s4 c. Mixture of electrolytes AB and A`B A 'B s A' B s (s s ') Ksp A B 1 Total s(s s ') (I) A 'B s A ' B s' s ' s Ksp A ' B 2 Total s '(s s ') (II) Comparing (i) and (ii) Now solving these three equations Ksp s we can find s and if K sp1 and Ksp2 1 (III) Ksp s' 2 Question Illustrative example 6 The concentration of Ag+ ion in a saturated solution of Ag2CrO4 at 20° C is 1.5 × 10–4 M. Determine the solubility product of Ag2CrO4 at 20°C. Solution: Ag2CrO4 2Ag CrO42 Ag 1.5 104 M CrO 2 1.5 104 4 0.75 104 M 2 2 Ksp Ag CrO42 (1.5 104 )2 0.75 104 1.69 1012 Solubility of a salt in presence of common ion Let us find out the solubility of Ag2CrO4 (Ksp =1.9 x 10–12) in 0.1 M Ag NO3 solution. In water Ag2 CrO4 2Ag CrO42 S 2S S Ksp (2S)2 S 4S3 1.9 1012 1.9 1012 S 3 0.78 104 mol3 / L3 4 In AgNO3 AgNO3 Ag NO3 0.1 Solubility of a salt in presence of common ion Ag2CrO4 2Ag CrO42 2S` S` Ksp [Ag ]2[CrO4 ] 1.9 1012 (2s` 0.1)2S ` S` 0.1 3 2 S` and S` can be neglected 1.9 1012 102 x S` S 1.9 1010 ` Solubility decreases in presence of common ion Question Illustrative example 7 The solubility of Mg(OH)2 in pure water is 9.57 × 103 gL–1. Calculate its solubility in gL–1 in 0.02 M Mg(NO3)2 solution. Solution: Solubility of Mg(OH)2 in pure water = 9.57 × 10–3 gL–1 9.57 103 mol L1 1.65 104 mol L1 58 Mg(OH)2 Mg2 2OH 2 Ksp Mg2 OH 1.65 104 (2 1.65 104 )2 Solution 17.97 1012 Let the solubility ‘s’ of Mg(OH)2 in presence of Mg(NO3)2 Mg(NO3 )2 is completely ionized. Mg(NO3 )2 Mg2 2NO3 0.02 M 0.02 M Mg2 s ' 0.02 OH 2s ' 2 Ksp Mg2 OH Solution 2 s ' 0.02 2s ' s ' 0.02 4s '2 s ' 0.02 0.02 4 s '2 0.08 s '2 17.97 1012 17.97 1012 s' 0.08 14.99 106 M Solubility in gL–1 = 14.99 × 10–6 × 58 = 8.69 × 10–4 gL–1 Solubility of a salt forming complex Let the solubility of AgCl is s mol L–1 AgCl(s) Ag (aq.) Cl (aq.) Sx S Ag 2NH3 [Ag(NH3 )2 ] (aq.) S x C 2x x Ksp (S x)S x Kf (S x)(C 2x)2 Kf=Eqm. Constant for the formation of complex Solubility of salt increases due to complex formation. Question Illustrative example 8 A solution of Ag+ ion at a concentration of 4 x 10–3 M just fails to yield a precipitate of AgCl with a concentration of 1 x 10–3 M of Cl– ion when the concentration of NH3 in the solution is 2 x 10–2 M at equilibrium. Calculate the magnitude of the equilibrium constant for the reaction Ag(NH3 )2 Ag 2NH3 Ksp AgCl at 250 C 1 1010 Solution: To make AgCl soluble in solution, maximum conc. of Ag+ in the solution, Ksp 1010 [Ag ] 107 [Ag ]max 107M [Cl ] 3 10 Solution So, rest of Ag+ forms complex with NH3 Ag 2NH3 Ag(NH3 )2 4103 x 2102 2x x Now, 4 103 x 107 x 4 103 107 4 103 4 103 Kf 107 (2 102 8 103 )2 4 103 2.78 108 107 1.44 104 Eqm. cons tan t for the given reaction 1 K f` 0.36 108 3.6 109 Kf Selective precipitation Let us consider a solution containing 0.01M Cl– and 0.02 M SO4–2 AgNO3 is being added to this solution. Let us find out Ki=ionic product for AgCl and Ag2SO4 As Ki > Ksp precipitate forms Ksp AgCl = 10–10 Ksp Ag2SO4 = 10–13 AgCl(s) Ag Cl Now, [Ag ][Cl ] Ksp 1010 1010 [Ag ] 108M 102 Selective precipitation Ag2SO4 2Ag SO42 [Ag ]2[SO42 ] 1013 1013 [Ag ] 2.24 106 2 102 For precipitation of AgCl, lesser conc. of Ag+ is required. So it will precipitate first and Ag2SO4 gets precipitated only when [Ag+] becomes greater than 2.24 x 10–6 M Selective precipitation We can find out % of Cl– precipipated when Ag2SO4 starts precipitating Then Ksp AgCl [Cl ]remaining [Ag ] 1010 4.46 105 2.24 106 [Cl ] initial 102 M 4.46 105 % Cl remaining 2 100 0.446% 10 Cl precipitated 99.55% Question Illustrative example 9 A solution has Zn+2 and Cu+2 each At 0.2M. The Ksp of ZnS and H2S are 1 x 10–22 and of CuS is 8 x 10–37. If the solution is made 1M in H3O+ and H2S gas is passed until the solution is saturated, should a precipitate form? Also report the ionic concentration product for ZnS and CuS in solution when first ion starts precipitating. Solution: ZnS Zn2 S2 Ksp 1 x1022 CuS Cu2 S2 Ksp 8 x1037 H2S 2H S2 Ksp 1 x1022 Solution As 1M solution of H3O is saturated with H2S, Ksp 1022 [S2 ] 2 1022 [H ] 1 Now, Ki ZnS [Zn2 ][S2 ] 0.2 1022 2 1023 Ki Ksp No precipitates of ZnS will form Ki CuS [Cu2 ][S2 ] 0.2 1022 2 1023 Solution Here, Ki Ksp CuS starts precipitating Ki ZnS 2 1023 Ki CuS Class exercise Class exercise 1 The solubility of A2X3 is s mol dm–3. Its solubility product is (a) 6s4 (b) 64s4 (c) 36s5 (d) 108 s5 Solution: A2X3 2A+3 + 3X-2 2s 3s 2 3 Ksp = 2s 3s =108 s5 Hence, the answer is (d). Class exercise 2 A compound M(OH)y has a Ksp of 4 × 10–6 and the solubility is 10–2 M. The value of y should be (a) 1 (b) 2 (c) 3 (d) 4 Solution: MOHy M+y + Y OH- s ys y Ksp = M y+ OH- Solution 4 × 10– 6 = 10–2 (y × 10– 4)y (y × 10–2)y = 4 × 10– 4 If we put y = 2, then only Ksp can be 4 × 10–6 Hence y = 2. Hence, the answer is (b). Class exercise 3 pH of a mixture containing 0.1 M and 0.2 M HX will be Given pKb(X–) = 4 (a) 4 + log 2 (b) 10 + log 2 (c) 4 – log 2 (d) 10 – log 2 Solution: [Salt] pOH=pKb +log [Base] 0.1 = 4+log 0.2 1 = 4+log = 4-log 2 pH = 14 – pOH = 10 + log 2 2 Hence, the answer is (d). Class exercise 4 For a basic indicator Kb is 1 × 10–10. It changes colour when the indicator is 10–5 M. The pH of solution at this point is (a) 10 (b) 4 (c) 2 (d) 8 Solution: Since colour change happen at the end point of titration and it is assumed that all indicator has been consumed during the process HpH H+ + pH- It is assumed that at the end point [Hph] = [Ph–] Solution [Ph- ] Hence, pOH= pKb +log [4Ph] pOH = 10 pH = 14 – 10 = 4 Hence, the answer is (b). Class exercise 5 For CH3COOH, pKa = 4.74 . A buffer solution of acetic acid 0.1 M and lead acetate (xM) has pH = 5.04. The value of x is (a) 0.2 (b) 0.1 (c) 0.02 (d) 0.05 Solution: For an acidic buffer, CH COO- 3 pH=pKa+log CH3COOH CH COO- 3 5.04 = 4.74+log 0.1 Solution CH COO- 3 = Antilog 0.3 0.1 CH COO- =0.199 0.2 3 For CH3COO2 Pb Pb+2 +2CH3COO- CH COO- 3 CH3COO2 Pb = = 0.1 2 Hence, the answer is (b). Class exercise 6 Saccharin (Ka = 210–12) is a weak acid represented by formula HSac. A 4 × 10–4 mole of saccharin is dissolved in 200 ml water of pH 3. Assuming no change in volume, calculate the concentration of Sac– ions in the resulting solution at equilibrium. Solution: HSac H+ +Sac– Intial conc. 0.002 0.001 0 conc. at eqm. (0.002–x) (0.001+x) x Solution Given [H+] = 0.001 M -4 4×10 ×1000 [HSac] = = 2×10-3M = 0.002 M 200 [H+] [Sac-] Ka = [HSac] (0.001+ x) (x) = (0.002 - x) (0.001+ x) x 2×10–12 = (0.002 - x) x = 410-12 M [Sac- ]eqm = 4×10-12M Class exercise 7 Calculate the amount of NH3 and NH4Cl required to prepare a buffer solution of pH 9.0 when total concentration of buffering reagents is 0.6 mol L–1. pKb for NH3 = 4.7, log 2 = 0.30. Solution: NH3 and NH4Cl forms a basic buffer NH+ =x 4 NH+ 4 pOH=pKb +log NH4OH x 5= 4.74+log 0.6 - x x = Antilog0.3=1.999 2 0.6 - x Solution x = 1.2 – 2x 1.2 x= = 0.4 3 NH+ =0.4 M (approx.) 4 NH OH =0.2 M (approx.) 4 Thank you