SolutionHard
Question
An aqueous solution of 0.01 M CH3COOH has Van′t Hoff factor 1.01. If pH = - log [H+], pH of 0.01 M CH3COOH solution would be :
Options
A.2
B.3
C.4
D.5
Solution
CH3COOH ⇋ CH3COO- + H+
C 0 0
i = [1 + (y - 1) x] = 1 + x = 1.01
x = 0.01
[H+] = Cx = 0.01 × 0.01 = 1 × 10-4 M
∴ pH = 4.
C 0 0
i = [1 + (y - 1) x] = 1 + x = 1.01
x = 0.01
[H+] = Cx = 0.01 × 0.01 = 1 × 10-4 M
∴ pH = 4.
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