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

NTA AbhyasNTA Abhyas 2020Equilibrium

Solution:

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

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

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

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

$=4\text{log} 10 - \text{log} ⁡ 5$

$=4-0.6989$

$=3.3010$

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