What is the pH of a 0.987 M CH3COOH aqueous solution at 25oC? Ka(CH3COOH)=1.8 x 10–5
1.92
2.38
11.62
6.98
5.22
Acetic acid is weak acid dissociate in water as
CH3COOH + H2O
CH3COO- + H3O+
Ka = [CH3COO-] [H3O+] / [CH3COOH]
but [CH3COO-] = [H3O+]
consider [CH3COO- ] = [ H3O+] = X
then Ka = [X] [X] / [CH3COOH]
Ka = X2 / [CH3COOH]
X2 = Ka X [CH3COOH]
substitute the value
X2 = (1.8 X 10-5) X (0.987) = 1.7766 X 10-5
X = 0.004215 M
[CH3COO- ] = [H3O+] = X = 0.004215 M
[H3O+] = 0.004215 M
pH = -log [H3O+] = -log (0.004215) = 2.38
Ans = 2.38
What is the pH of a 0.987 M CH3COOH aqueous solution at 25oC? Ka(CH3COOH)=1.8 x 10–5...
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