Question

Calculate $\left[{OH}^{-}\right]$ in 0.20 M solution of ${NH}_3$ if $K_b$ for ${NH}_3$ is $1.8\times {10}^{-5}$

Answer 1

$N{H}_3+H_{2}O\rightleftharpoons N{H}_4^{+}+O{H}^{-}$

$K_b=1.8\times {10}^{-5}$

$\alpha =\sqrt{\frac{K_b}{c}}=\sqrt{\frac{1.8\times {10}^{-5}}{0.2}}$

$=3\times {10}^{-3}\sqrt{10}\alpha 0.1$

$\therefore$ correct

$\left[O{H}^{-}\right]=C\alpha$

$=3\sqrt{10}\times {10}^{-3}\times 0.2$

$=6\times {10}^{-4}\times \sqrt{10}$

$=6\times {10}^{-7/2}$