Question

The $pH$ of $0.1M$ solution of cyanic acid $\left(HCNO\right)$ is $2.34$. Calculate the ionization constant of the acid and its degree of ionization in the solution.
Calculate the degree of ionization of $0.05M$ acetic acid if its $p{K}_A$ value is $4.74$. How is the degree of dissociation affected when its solution is also (a) $0.01M$ and (b) $0.1M$ in hydrochloric acid?

Answer 1

$HCNO\rightleftharpoons H^{+}+{CNO}^{-}$

$C=0.1M$ $0$ $0$

$C(1-\alpha )$ $C\alpha$ $C\alpha$

$PH=-log\left[H^{+}\right]=2.34$

$\left[H^{+}\right]=4.6\times {10}^{-3}$

also known $\left[H^{+}\right]=C\alpha$

$C\alpha =4.6\times {10}^{-3}$

$\alpha =4.6\times {10}^{-2}$

$K=\frac{C\alpha \times l\alpha }{C(1-\alpha )}\Rightarrow \frac{{C\alpha }^2}{1-\alpha }\approx {C\alpha }^2$ ($\because$ $HCN$ weak acid)

$\Rightarrow 0.1\times {(4.6\times {10}^{-2})}^2$

$\Rightarrow 2.1160\times {10}^{-4}$

Given Acetic acid (weak acid)

${PK}_a=4.74\Rightarrow K_a={10}^{-4.74}=0.182\times {10}^{-4}$

we know $K_a={C\alpha }^2\Rightarrow {\alpha }^2=\frac{0.182\times {10}^{-4}}{0.05}={\left(0.0191\right)}^2$

$\alpha =0.0191$

In $0.01M$ $HCl$ (strong acid)

So, $\left[H^{+}\right]=0.01$

${HCH}_{3}COO\rightleftharpoons {CH}_3{COO}^{-}+H^{+}$

$C$ $0$ $0$ ($\alpha$ is small because coming from weak acid)

$C-C\alpha$ $C\alpha$ $0.01+C\alpha$

$K_a=\frac{C\alpha }{C(1-\alpha )}\times 0.01\approx \alpha \times 0.01$

$\alpha \approx K_a\times {10}^2\approx 0.182\times {10}^{-2}$