If the Ka of a monoprotic weak acid is 1.6 × 10-6, what is the pH of a 0.12 M solution of this acid?
Dear friend your answer should be by taking acetic acid because acetic acid is a weak monoprotic acid. only weak acids can have acid dissociation constant Ka
CH3COOH -----> CH3COO- +
Ka = [H+][CH3COO-] / [CH3COOH] = 1.6 x 10-6
Let x = the amount of CH3COOH that dissociates.
[H+] = x
[CH3COO-] = x
[CH3COOH] = 0.12 - x = 0.12 (assuming x << 0.12)
1.6 x 10-6 = (x) (x) / (0.12)
x = 0.00043817 = [H+]
pH = - log (0.00043817) = 3.3583
PH of a given monoprotic acid is 3.3583
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