Question

What is $\left[{NH}_4^{+}\right]$ in a solution containing 0.02M ${NH}_3$ ($K_b={1.8\times 10}^{-5}$) and 0.01M KOH?

• A. ${1.8\times 10}^{-5}$
• B. ${9\times 10}^{-6}$
• C. ${3.6\times 10}^{-5}$
• D. None of the above

Answer 1

Answer: (C)

${3.6\times 10}^{-5}$

Solution:

$N{H}_{4}OH\rightleftharpoons N{H}_4^{+}+O{H}^{-}$

$KOH\to K^{+}+O{H}^{-}$

$K_b=\frac{\left[N{H}_4^{+}\right][O{H}^{-}{]}^2}{\left[N{H}_{4}OH\right]}$

$1.8\times {10}^{-5}=\frac{\left[X\right]\left[{10}^{-2}\right]}{2\times {10}^{-2}}$

$\Rightarrow x=3.6\times {10}^{-5}$

Hence the correct option: C