Question

Calculate $\left(H^{+}\right)$ and % dissociation of 0.1 M solution of $N{H}_{4}OH$ solution. The ionisation constant for $N{H}_{4}OH$ 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 $\left(1-\alpha \right)=1$

$k_b=C{\alpha }^2$

$\alpha =\sqrt{\frac{k_b}{C}}=\sqrt{\frac{2\times 10^{-5}}{0.1}}=\sqrt{2\times 10^{-4}}$

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

$\alpha =1.4\%$

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

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

$\left[H^{-}\right]=\frac{K_w}{\left[O{H}^{-}\right]}=\frac{10^{-14}}{1.4\times 10^{-3}}$

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