Question

The $\mathrm{pH}$ value of $0.1\mathrm{M}$ solution of anilinium chloride is:$(K_a$ of $C_6{H}_{5}N{H}_3^{+}={10}^{-6})$

Answer 1

Correct Answer: 3.5

Explanation:
Given:The concentration of the solution of aniline chloride $=0.1$ MDissociation constant, Ka of $C_6{H}_{5}N{H}_3^{+}={10}^{-6}$We have to find the pH value of the solution.Dissociation of $C_6{H}_{5}N{H}_3^{+}$ is given as:$C_6{H}_{5}N{H}_3^{+}\leftrightharpoons C_6{H}_{5}N{H}_2^{+}+H^{+}$Considering Ostwald's dilution law, dissociation constant, $K_a$ is given as:$k_a=C{\alpha }^2$$\alpha =\sqrt{\frac{K_a}{C}}$$\alpha =3.16\times {10}^{-3}$Now, $\left[H^{+}\right]$ is given as:$\left[H^{+}\right]=C\times \alpha$.....(i)On substituting the values in equation (i), we get$\left[H^{+}\right]=0.1\times 3.16\times {10}^{-3}$$\left[H^{+}\right]=3.16\times {10}^{-4}$Also, $\left[H^{+}\right]$ is related to $pH$ as:$pH=-\log [H+]$...(ii)$pH=-\log (3.16\times {10}^{-4})$$pH=-\log (3.16)+4\log 10$$pH=-0.4997+4(As,\log 10=1)$$pH=3.5003$$PH\simeq 3.5$So, the correct answer is 3.50First Alternative Solution$pH$ of weak acid is given as:$pH=\frac{1}{2}(p{K}_a-\log C)$or,$pH=\frac{1}{2}(-\log K_a-\log C)$On substituting the values of $K_a$, we get$pH=\frac{1}{2}(-\log {10}^{-6}-\log 0.1)$$pH=\frac{1}{2}(6\log 10-\log 0.1)$$pH=\frac{1}{2}(6+1)$$pH=3.5$Hence, the correct answer is 3.5.