1. Chemistry

2. Ionic equilibrium-II

3. Session Objectives

4. 1.pH of weak acids 2.pH of mixture of two strong acids 3.pH of mixture of strong and weak acids 4.Dissociation of polybasic acids 5.pH of mixture of two weak acids Session Objectives

5. For mixture of two strong acids Let us consider a mixture containing 200 ml 0.1 M HCl and 500 ml 0.2 M H 2 SO 4. Since both are strong electrolytes, 200 × 0.1×10 –3 0.02 mole = 0.02 mole 500 × 0.2 × 10 –3 2 × 0.1 mole = 0.1 mole

6. For mixture of two strong acids Note: For mixture of any two acids or bases, degree of ionization of water is taken negligible because of common ion effect. Total ion concentration,

7. Mixture of strong and weak acids Let us consider a mixture of strong acid (HA) and a weak acid (HX) of concentration C 1 and C 2 respectively. For weak acid, Dissociation constant of weak acid, Now, for strong acid

8. Mixture of strong acid and weak acid

9. Question

10. Illustrative example 1 A solution contains 0.10 M H 2 S and 0.3 M HCl. Calculate the concentration of [HS – ] and [S –2 ] ions in the solution. For H 2 S, K a1 = 1.0 x 10 -7 K a2 = 1.3 x 10 -13 Solution: H 2 S H + + HS –

11. Solution HS – H + + S –2

12. Solution Considering [HS – ] dissociates to a very small extent

13. Dissociation of polybasic acids Acids giving more than one hydrogen ion per molecule are ‘polybasic’ or ‘polyprotic’ acids. Examples are H 2 C 2 O 4, H 2 CO 3, H 2 S, H 3 PO 4, H 3 AsO 4, etc. These dissociate in stages. For example,

14. Dissociation of polybasic acids Normally K 2 << K 1. To calculate hydrogen ion concentration, only the first step should be considered as the H + obtained from successive dissociation can be neglected, but to calculate the concentration of then both the equilibria have to be considered.

15. pH of mixture of two weak acids Let HA and HB are two weak acids.

16. pH of mixture of two weak acids Dividing (1) by (2)

17. Question

18. Illustrative example 2 A solution is prepared by mixing 0.2M HCOOH with 0.5 M CH 3 COOH. Given K a CH 3 COOH=1.8 x 10 –5, K a HCOOH =2.1x10 -4 Calculate [HCOO – ], [CH 3 COO – ] and pH of the solution. Solution:

19. Solution From (2)

20. Solution

22. Hydrolysis of Salts A. Hydrolysis of a salt of weak acid and strong base The hydrolysis reaction is At eqm. where C = concentration of salt h = degree of hydrolysis.

23. Hydrolysis of Salts Hydrolysis constant

24. Hydrolysis of Salts

25. where K b = Dissociation constant of weak base. B. Salt hydrolysis of strong acid and weak base

26. Hydrolysis of Salts C. Hydrolysis for a salt of weak acid and weak base At eqm.

27. Hydrolysis of Salts

28. For pH, Now, to calculate the pH

29. Questions

30. Illustrative example 3 Calculate the pH at the equivalence point of the titration between 0.1 M CH 3 COOH (50 ml) with 0.05 M NaOH. Ka(CH 3 COOH) = 1.8 × 10 –5. At the equivalence point, Let V ml NaOH is required to reach the equivalence point. Solution: At the equivalence point,

31. Solution V = 100 ml

32. Solution

33. Illustrative example 4 Calculate the percentage hydrolysis of decinormal solution of ammonium acetate, given K a = 1.75 × 10 –5, K b = 1.80 × 10 –5 and K w = 1 × 10 –14. What will be the change in degree of hydrolysis when 2 L of water is added to 1 L of the above solution? Since CH 3 COONH 4 is a salt of weak acid and weak base Solution:

34. Solution Since degree of hydrolysis is not affected by the concentration in this case. So on dilution there will be no change in degree of hydrolysis.

35. Illustrative example 5 Hydrolysis constant of Zn +2 is 1 × 10 –9 (a) Calculate pH of a 0.001 M solution of ZnCl 2. (b) What is the basic dissociation constant of Zn(OH) + ? Solution:

36. Solution

37. Illustrative example 6 When 0.20 M acetic acid is neutralised with 0.20 M NaOH in 0.50 litre of water, the resulting solution is slightly alkaline, calculate the pH of the resulting solution (K a for acetic acid = 1.8 x 10 -5 ) Solution:

38. Solution

39. Illustrative example 7 Calcium lactate is a salt of a weak organic acid and represented as Ca(Lac) 2. A saturated solution of Ca (Lac) 2 contains 0.13 mol of this salt in 0.50 litre solution. The pOH of this solution is 5.60. Assuming a complete dissociation of the salt, calculate K a of lactic acid. Solution:

40. Solution

41. Class exercise

42. Class exercise 1 The hydrolysis constant for FeCl 2 will be FeCl 2 is the salt of strong acid and weak base. Solution:

43. Solution Hence, the answer is (b)

44. Class exercise 2 pH of a solution produced when an aqueous solution of pH 6 is mixed with an equal volume of an aqueous solution of pH 3, will be (a) 4.5(b) 4.3 (c) 4.0(d) 3.3 pH = 6 pH = 3 Solution:

45. Solution = 4 – log 5.005 = 3.3 Hence, the answer is (d).

46. Class exercise 3 Which one of the following is true for any diprotic acid, H 2 X? (a) Ka 2 > Ka 1 (b) Ka 1 > Ka 2 (c) Ka 1 = Ka 2 H 2 X being a diprotic acid, Due to the ‘common ion effect’ dissociation of HX – will be less. Solution:

47. Class exercise 4 Ka (CH 3 COOH) = 1.7 × 10 –5 and [H + ] = 3.4 × 10 –4. Then initial concentrations of CH 3 COOH is (a) 3.4 × 10 –4 (b) 6.8 × 10 –3 (c) 3.4 × 10 –3 (d) 6.8 × 10 –2 Solution:

48. Solution Hence, the answer is (b).

49. Class exercise 5 0.001 M HCl is mixed with 0.01 M HCOOH at 25° C. If K a HCOOH = 1.7 × 10 –4, find the pH of the resulting solution. Solution:

50. Solution

51. Class exercise 6 What is the percentage hydrolysis of NaCN solution when the K a HCN = 1.3 × 10 –9, K 2 = 1 × 10 –14 ? (a) 2.48(b) 5.26 (c) 9.6(d) 8.2 Solution:

52. Solution = 2.48 × 10 –2

53. Class exercise 7 Calculate the pH of a solution obtained by mixing 100 ml of 0.1 M HCl with 9.9 ml of 1 M NaOH. Acid remaining = 1 × 10 –4 moles Solution: pH = –log[H + ] = 4 – log 9.09 = 3.04

54. Class exercise 8 Calculate the percentage hydrolysis of 3 × 10 –3 M aqueous solution of NaOCN (K a HCON = 3.33 × 10 –4 M). Solution:

55. Solution

56. Thank you