Question

A 0.01 M ammonia solution is 5 % ionised, its pH will be

• A. 11.80
• B. 10.69
• C. 7.22
• D. 12.24

Answer 1

Answer: (B)

Given, aqueous solution of ${\mathrm{NH}}_3$ (weak base) $\mathrm{C}=0.01\mathrm{M}$
$\text{α}=5\text{\%}=\frac{5}{100}$

$\left[\text{O}{\text{H}}^{-}\right]=\text{C}\text{α}=0.01\times \frac{5}{100}=5\times 10^{-4}$

$\text{p}\text{O}\text{H}=-\text{log}[\text{O}{\text{H}}^{-}]$

$=\log \left(5\times 10^{-4}\right)$

$=4\text{log}10-\text{log}5$

$=4-0.6989$

$=3.3010$

$\therefore$ $\text{p}\text{H}=14-\text{p}\text{O}\text{H}=14-3.3010=10.6990$