Question

The dissociation constant of acetic acid at a given temperature is 1.69 × 10–5. The degree of dissociation of 0.01 M acetic acid in the presence of 0.01 M HCl is equal to

• A. 0.41
• B. 0.13
• C. 1.69 × 10–3
• D. 0.013.

Answer 1

Answer: (C)

1.69 × 10–3

Answer 2

CH3COOH (aq) $\rightleftharpoons$ H+ (aq) + CH3COO– (aq)

t = 0 0.01

t = eq 0.01 – x x x

[H+] = x + 0.01 $\approx$ 0.01 M

$\therefore$ Ka = $\Rightarrow$ 1.69 × 10–5

= $\frac { 0.01 \times \left[ \mathrm { CH } _ { 3 } \mathrm { COO } ^ { - } \right] } { 0.01 }$

$\therefore$ [CH3COO–] = 1.69 × 10–5 M

So, degree of dissociation of CH3COOH = $\frac { 1.69 \times 10 ^ { - 5 } } { 0.01 }$

= 1.69 × 10–3