Question

The dissociation constant for aniline, acetic acid and water at ${25}^{o}C$ are $3.83\times {10}^{-10}$, $1.75\times {10}^{-5}$ and $1.008\times {10}^{-14}$ respectively. What is the degree of hydrolysis of aniline acetate in a deci normal solution and $pH$?

• A. $h=54.95$%; $pH=6.4683$
• B. $h=45.95$%; $pH=4.6683$
• C. $h=54.95$%; $pH=4.6683$
• D. None of these

Answer 1

The correct option is C $h=54.95$%; $pH=4.6683$

$\underset{cc(1-h)}{{Aniline}^{+}}+\underset{cc(1-h)}{{Acetate}^{-}}+H_{2}O\rightleftharpoons \underset{0ch}{Aniline}+\underset{0ch}{AceticAcid}$
Let concentration of salt be $cmol{litre}^{-1}$

$\therefore K_H=\frac{\left[Aniline\right]\left[AceticAcid\right]}{\left[{Aniline}^{+}\right]\left[{Acetate}^{-}\right]}=\frac{ch\cdot ch}{c(1-h)c(1-h)}$

$K_H=\frac{h^2}{{(1-h)}^2}$

or $\frac{h}{1-h}=\sqrt{K_H}=\sqrt{\frac{K_w}{K_a\times K_b}}=\sqrt{\frac{1.008\times {10}^{-14}}{1.75\times {10}^{-5}\times 3.83\times {10}^{-10}}}$

$\therefore h\approx 54.95$%

Also,
$pH=\frac{1}{2}(p{K}_w-p{K}_a-p{K}_b)$

$=\frac{1}{2}[-log{K}_w+\log K_a+\log K_b]=\frac{1}{2}(\log [1.008\times {10}^{-14}]-\log [1.75\times {10}^{-5}]-\log [3.83\times {10}^{-10}])$

$=4.6683$

If $K_H=h^2$ is assumed (assuming $1-h\approx 1$), the value of $h$ comes greater than $1$ which is not possible and thus $1-h$ should not be neglected.