Question

Calculate $\left(H^{+}\right)$ and % dissociation of 0.1 M solution of $NH_{4}OH$ solution. The ionisation constant for $NH_{4}OH$ is, $K_{b}=2.0\times 10^{- 5}$

• A. $7.09\times 10^{- 12}M,3\%$
• B. $7.09\times 10^{- 12}M,1.4\%$
• C. $9.02\times 10^{- 12}M,2.4\%$
• D. $9.02\times 10^{- 12}M,3\%$

Answer 1

Answer: (B)

$K _{ b }=\frac{( C \alpha}{ C (1-\alpha)}=\frac{ C \alpha^{2}}{1-\alpha}$
negligible $\left(1 - \alpha \right)=1$
$K_{b}=C\alpha ^{2}$
$\alpha =\sqrt{\frac{k_{b}}{C}}=\sqrt{\frac{2 \times 1 0^{- 5}}{0.1}}=\sqrt{2 \times 1 0^{- 4}}$
$\alpha =1.41\times 10^{- 2}$
$\alpha =1.4\%$
$\left[O H^{-}\right]=C\alpha =0.1\times 1.4\times 10^{- 2}$
$=1.4\times 10^{- 3}M$
$\left[H^{-}\right]=\frac{K_{w}}{\left[O H^{-}\right]}=\frac{1 0^{- 14}}{1.4 \times 1 0^{- 3}}$
$=7.09\times 10^{- 12}M$