Question

If the concentration of $\left[\right.NH_{4}^{+}\left]\right.$ in a solution having $0.02 \, M \, NH_{3}$ and $0.005 \, M \, Ca\left(O H\right)_{2}$ is $a\times 10^{- 6} \, M$ , determine $a$ . $\left[\right.k_{b}\left(N H_{3}\right)=1.8\times \left(10\right)^{- 5}\left]\right.$

Answer 1

$NH_{4}OH\leftrightharpoons NH_{4}^{+}+OH^{-}$

$k_{b}=\frac{\left[N H_{4}^{+}\right] \, \left[O H^{-}\right]}{\left[N H_{4} O H\right]}$

$\Rightarrow \, \, 1.8\times 10^{- 5}=\frac{\left[N H_{4}^{+}\right] \, \left[0.01\right]}{\left[0.02\right]}$

{As $OH^{-}$ will mainly come from $Ca\left(O H_{2}\right)$ only}

$\Rightarrow \left[N H_{4}^{+}\right]=36\times 10^{- 6}=a\times 10^{- 6}$

$\Rightarrow \, \, a=36$