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The $\mathrm{pH}$ of a $0.1 \mathrm{M}$ solution of $\mathrm{NH}_{4} \mathrm{OH}$ (having $K_{b}=1.0 \times 10^{-5}$ ) is equal to
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11
$\begin{aligned}\left[\mathrm{OH}^{-}\right] &=\sqrt{K_{b} \times C} \\ &=\sqrt{1 \times 10^{-5} \times 10^{-1}}=\sqrt{10^{-6}}=10^{-3} \\ K_{w} &=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right] \\ 10^{-14} &=\left[\mathrm{H}^{+}\right]\left[10^{-3}\right] \\\left[\mathrm{H}^{+}\right] &=10^{-11} \\ \text { Hence, } \mathrm{pH} &=-\log \mathrm{H}^{+} \\ &=-\log \left(1 \times 10^{-11}\right)=11 \end{aligned}$
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