Question

The $H^{+}$ ion concentration in 0.001M acetic acid is $1.34\times {10}^{-4}gion{L}^{-1}$. The $H^{+}$ ion concentration of 0.164 g of $C{H}_{3}COONa$ is added to a litre of 0.001 M $C{H}_{3}COOH$ will be :-

• A. $9\times {10}^{-6}$
• B. $18\times {10}^{-6}$
• C. $4.5\times {10}^{-6}$
• D. $5\times {10}^{-6}$

Answer 1

The correct option is A $9\times {10}^{-6}$

$pH=-log(1.8\times {10}^{-5})+log\left(\frac{0.164/82}{0.001}\right)$

where $K_a=1.8\times {10}^{-5}$

$=4.7447+0.3010=5.0457$

$pH=-log\left[H^{+}\right]$

$=5.0457=-log\left[H^{+}\right]$

$=\left[H^{+}\right]=9\times {10}^{-6}$

Option A is correct.