Question

If the concentration of $[N{H}_4^{+}]$ in a solution having $0.02MN{H}_3$ and $0.005MCa{\left(OH\right)}_2$ is $a\times 10^{-6}M$ , determine $a$ . $[k_b\left(N{H}_3\right)=1.8\times {\left(10\right)}^{-5}]$

Answer 1

Answer: 36

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

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

$\Rightarrow 1.8\times 10^{-5}=\frac{\left[N{H}_4^{+}\right]\left[0.01\right]}{\left[0.02\right]}$

{As $O{H}^{-}$ will mainly come from $Ca\left(O{H}_2\right)$ only}

$\Rightarrow \left[N{H}_4^{+}\right]=36\times 10^{-6}=a\times 10^{-6}$

$\Rightarrow a=36$