1 / 59

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:.

calliope
Télécharger la présentation

Chemistry

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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).

More Related