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 $kb=2.0\times 10^{- 5}$

• 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)

$k_{b}=\frac{(C \alpha)^{2}}{C(1-\alpha)}=\frac{C \alpha^{2}}{1-\alpha}$

negligible $\left(1 - \alpha \right)=1$

$k_{b}=C\alpha ^{2}$

$\alpha =\sqrt{\frac{k_{b}}{C}}=\sqrt{\frac{2 \times 1 0^{- 5}}{0.1}}=\sqrt{2 \times 1 0^{- 4}}$

$\alpha =1.41\times 10^{- 2}$

$\alpha =1.4\%$

$\left[O H^{-}\right]=C\alpha =0.1\times 1.4\times 10^{- 2}$

$=1.4\times 10^{- 3}M$

$\left[H^{-}\right]=\frac{K_{w}}{\left[O H^{-}\right]}=\frac{1 0^{- 14}}{1.4 \times 1 0^{- 3}}$

$=7.09\times 10^{- 12}M$