Question

The pH of a 0.1 M solution of $N{H}_{4}OH$ (having dissociation constant $K_b=1.0\times 10^{-5})$ is equal to

• A. 10
• B. 6
• C. 11
• D. 12

Answer 1

Answer: (C)

$[O{H}^{-}]=\sqrt{K_b\times \left(N{H}_{4}OH\right)}$
$=(10^{-5}\times 10^{-1}{)}^{1/2}=1\times 10^{-3}$
$\left[H^{+}\right]=\frac{10^{-14}}{10^{-13}}=10^{-11}$
Hence $pH=-log10^{-11}=11.0$