Question

What is the pH of a 0.987 M CH₃COOH aqueous solution at 25^oC? K_a(CH₃COOH)=1.8 x 10^–5

1.92

2.38

11.62

6.98

5.22

Answer 1

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