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Chemistry

Chemistry. Ionic equilibrium-III. Session Objectives. Session Objectives. Hydrolysis of salts Buffer and buffer capacity. Buffer solution. Types of buffer solution. a. Acidic buffer:. CH 3 COOH and CH 3 COONa. b. Basic buffer:. NH 4 OH and NH 4 Cl. c. Salt or neutral buffer:.

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Chemistry

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  1. Chemistry

  2. Ionic equilibrium-III

  3. Session Objectives

  4. Session Objectives • Hydrolysis of salts • Buffer and buffer capacity

  5. Buffer solution Types of buffer solution a. Acidic buffer: CH3COOH and CH3COONa b. Basic buffer: NH4OH and NH4Cl c. Salt or neutral buffer: CH3COONH4

  6. Animation of buffer

  7. Explanation Now, to find the pH, Let us consider Since CH3COOH is feebly ionized, so most of CH3COO– will come from the salt. Hence, [CH3COO–] = [Salt]

  8. Explanation Taking log of both sides This is known as Henderson’s equations. Similarly, for basic buffer We know, pH + pOH = 14

  9. Questions

  10. 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 = 4.745 + 0 = 4.745

  11. 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, Change in pH = 4.745 – 4.658 = 0.087 units Hence, pH will decrease.

  12. 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 Change in pH = 4.831 – 4.745 = 0.086 units Hence, pH will increase.

  13. 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

  14. Solution 0.6 – x = 2x 3x = 0.6

  15. 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.

  16. Questions

  17. 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:

  18. 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

  19. Solution Hence, answer is (d).

  20. 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:

  21. Solution

  22. 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] may be taken as constant, thus K·[AgCl] = [Ag+]·[Cl–]

  23. 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.

  24. Different cases of solubility product a. Electrolyte of the type AB2/A2B Where s =solubility of the salt in mole/L = s × (2s)2 = 4s3

  25. b. Electrolyte of the type AB3 = s × (3s)3 = 27s4

  26. Now solving these three equations we can find s and if and c. Mixture of electrolytes AB and A`B Comparing (i) and (ii)

  27. Question

  28. 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:

  29. 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 In AgNO3

  30. Solubility of a salt in presence of common ion

  31. Question

  32. 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

  33. Solution Let the solubility ‘s’ of Mg(OH)2 in presence of Mg(NO3)2

  34. Solution Solubility in gL–1 = 14.99 × 10–6 × 58 = 8.69 × 10–4 gL–1

  35. Solubility of a salt forming complex Let the solubility of AgCl is s mol L–1 Kf=Eqm. Constant for the formation of complex Solubility of salt increases due to complex formation.

  36. Question

  37. 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 Illustrative example 8 Solution: To make AgCl soluble in solution, maximum conc. of Ag+in the solution,

  38. Solution So, rest of Ag+ forms complex with NH3

  39. 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

  40. Selective precipitation 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

  41. Selective precipitation We can find out % of Cl– precipipated when Ag2SO4 starts precipitating Then

  42. Question

  43. Illustrative example 9 A solution has Zn+2and Cu+2each At 0.2M.The Ksp of ZnS and H2S are 1 x 10–22 andof CuS is 8 x 10–37. If the solution is made1M in H3O+ and H2S gas is passed until thesolution is saturated, should a precipitate form? Also report the ionic concentration product for ZnS and CuS in solution whenfirst ion starts precipitating. Solution:

  44. Solution

  45. Solution

  46. Class exercise

  47. 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: Hence, the answer is (d).

  48. 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:

  49. 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).

  50. 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: pH = 14 – pOH = 10 + log 2 Hence, the answer is (d).

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