Titrations of Weak Acids

Question

Calculate the pH for each case in the titration of 50.0 mL of 0.220 M HClO(aq)

with 0.220 M KOH(aq).

Use the ionization constant for HClO.

What is the pH before addition of any KOH?

pH=?

What is the pH after addition of 25.0 mL KOH?

pH= 7.54

What is the pH after addition of 40.0 mL KOH?

What is the pH after addition of 50.0 mL KOH?

What is the pH after addition of 60.0 mL KOH?

Answer 1

1. pH before adding KOH
2. at this point there is only water in equilibrium with HClO with is a weak acid. The balanced equation is as follow
3. HClO + H2O -> H3O+ + ClO-
4. Prepare ICE table (Initial, Change, Equilibrium)
5. initial HClO concentration = 50mL / 1000mL * 0.220M = 0.011 moles HClO
6. I: HClO = 0.011moles; H3O+ = 0; ClO- = 0
7. C: HClO = -x; H3O+ = +x; ClO- = +x
8. E: Ka HClO = 3.8 x 10^-8 = (H3O+)( ClO-)/HClO = X²/(0.011-x)
9. NOTE! this Ka value is found via google! if your give Ka is different, it will result in slightly different answers!
10. assuming the change in HClO is negligible due to properties of being a weak acid & general simplicity of calculation
11. Ka HClO = 3.8 x 10^-8 = (H3O)( ClO- )/ HClO = X²/(0.011)
12. x = H3O+ = ClO- = 1.8 x 10^-5 = H+
13. -log H+ = -log(1.8x 10^-5) = pH = 4.7
14. Given 25mL of KOH
15. balanced equation as given
16. HClO + KOH -> KClO- + H2O
17. Prepare ICE table same as previous; note there are both HClO and KOH now
18. I: HClO = 0.011moles (as previously calculated)
19. KOH 25mL of KOH = 0.0055 moles = 25mL / 1000mL * 0.220M
20. KClO- = 0
21. C: HClO = 0.011 moles -0.0055; KOH = 0.0055 - 0.0055; KClO = 0 + 0.0055
22. note KOH is a strong base so it will completely dissociate thus the change in moles will be equal to moles of KOH present
23. E: HClO = 0.0055; KOH = 0; KClO- = 0.0055
24. use Henderson Hasselbach equation to calculate the pH in this established buffer system as we have now identified dissociation constant, [A-] and [HA]
25. pH = pKa + log ([A-] / [HA])
26. pH = -log(3.8x10^-8) + log ([KClO]- /[HClO])
27. pH = -log(3.8x10^-8) + log (0.0055/0.0055)
28. pH = ~7.4
29. This problem follows the same principle as previous, i will answer in less depth for sake of you working through similar problems
30. 40mL of KOH = 0.0088 moles
31. equilibrium moles of:
32. HClO = 0.011 - 0.0088; KOH = 0.0088 - 0.0088; KClO- = 0+ 0.0088
33. HClO = 0.0022
34. KCLO- = 0.0088
35. pH = 7.4 + log (0.0088/0.0022) = ~8
36. 50mL of KOH; at this point we have reached the equivalence point of the titration and there are no acids of base remaining
37. the only major species is the salt KClO that is also the conjugate base ClO-
38. moles of KOH = (50mL * 0.220M) / 100mL (this is the volume of acid plus base for total volume at play) = 0.011 moles or 0.11M
39. hydrolysis of the weak base
40. ClO- + H2O -> HClO + OH-
41. set up ice table
42. initial: ClO = 0.11moles; rest are 0
43. change ClO = 0.11moles -x; HClO = +x; OH- = +x
44. equilibrium = Kb = Kw/Ka = 1x10^-14/3.8x10^-8 = 2.6 x 10^.7= x² / 0.11
45. again assume change in ClO is negligible and solve for x
46. Kw is a value to typically remember or know off the topic of your head; along with relationship between Ka, Kb, and Kw
47. our x or change in HClO or OH- = 1.7*10^-4
48. -log OH- = pOH = -log (1.7*10^-4) = ~3.8
49. pH = 14-pOH = 10.2
50. 60mL of KOH; we have far passed the equivalence point of the titration; so the excess strong base will entirely determine the pH at this point
51. 60mL KOH; 0.22M * 0.06L = 0.0132moles
52. 0.011moles of the KOH will be used to neutralize the HClO acid which leaves us with 0.0022 moles remaining
53. 0.0022 moles / 0.11L (note this is the cumulative volume of 60mL + 50mL) = 0.02M of KOH
54. the remaining KOH fully dissociates into OH- ions
55. -log (0.02) = 1.7
56. pH = 14-pOH = 12.3