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  4. The pH of a 0.1 M solution of NH4OH (having dissociation constant Kb = 1.0 × 10-5) is equal to

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Q. The pH of a 0.1 M solution of $NH_4OH$ (having dissociation constant $K_b = 1.0 \times 10^{-5})$ is equal to

Equilibrium

Solution:

$[OH^{-}]=\sqrt{K_{b}\times\left(NH_{4}OH\right)}$
$=(10^{-5}\times10^{-1})^{1/2}=1\times10^{-3}$
$\left[H^{+}\right]=\frac{10^{-14}}{10^{-13}}=10^{-11}$
Hence $pH=-log\,10^{-11}=11.0$