The pH of a 0.00156 M solution of a certain weak acid, HA is found to be 3.64. What is the Ka of acid?
2.37 x 10-4 |
3.94 x 10-5 |
2.81 x 102 |
4.29 x 10-4 |
HA <-------------> H+ + A-
pH = 3.64
[H+] = 10-pH = 10-3.64 = 2.30 x 10-4 M
at equilibrium [H+] = [A-] = 2.30 x 10-4 M
[HA] = 0.00156 - 0.000230 = 0.00133 M
Ka = [H+] [A-] / [HA]
Ka = [2.30 x 10-4] [2.30 x 10-4] / [0.00133]
Ka = 3.9 x 10-5
answer = option 2 = 3.94 x 10-5
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