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# The dissociation constant for aniline, acetic acid, and water at are $3.83 \times {10^{ - 10}},1.75 \times 1{O^{ - 5}},$and $1.008 \times {10^{ - 14}}$ respectively. What is the degree of hydrolysis of aniline acetate in a deci normal solution and pH?$A.\;\;\;\;\;h = 54.95\% {\text{ }};{\text{ }}pH = 6.4683 \\ B.\;\;\;\;\;h = 45.95\% ;{\text{ }}pH = 4.6683 \\ C.\;\;\;\;\;h = 54.955\% ;{\text{ }}pH = 4.6683 \\ D.\;\;\;\;\;none{\text{ }}of{\text{ }}these \\$

Last updated date: 29th May 2024
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Hint:
Degree of hydrolysis is the fraction (or percentage) of total salt which is hydrolysed at equilibrium whereas dissociation constant (ka) is the ratio of dissociated ions (products) to original acid (reactants).
Formula used:
$i)\,\,\,{K_h} = \dfrac{{{k_w}}}{{{k_a} \times {k_b}}}\; \\ ii)\;\;h = \dfrac{{\sqrt {{k_h}} }}{{1 + \sqrt {{k_h}} }} \\ iii)\,\,pH = \dfrac{1}{2}(p{K_{w\;\;}} - p{K_{a\;\;}} - p{K_{b\;\;}}) \\$

The expression for hydrolysis constant ${K_h} = \dfrac{{1 \times {{10}^{ - 14}}}}{{1.75 \times {{10}^{ - 5\;\;\;}} \times 3.83 \times {{10}^{ - 10}}}}$ (from formula i)
= 1.49
The expression for the degree of hydrolysis is ,$\;h = \dfrac{{\sqrt {{k_h}} }}{{1 + \sqrt {{k_h}} }}$
Therefore, $h = \dfrac{{\sqrt 1 .49}}{{1 + \sqrt 1 .49}}$
= 0.5495
Hence, percentage of hydrolysis $= {\text{ }}100X{\text{ }}0.5459$
$= {\text{ }}54.95\%$
Further, $pH = \dfrac{1}{2}(p{K_{w\;\;}} - p{K_{a\;\;}} - p{K_{b\;\;}})$
$= \dfrac{1}{2}\log \left[ {1.008 \times {{10}^{ - 14\;}}} \right] - \log \left[ {1.75 \times {{10}^{ - 5\;}}} \right] - \log \left[ {3.83 \times {{10}^{ - 10\;}}} \right]$
$= 4.6683$

Hence, option c is correct.

Note:

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. (from${K_{h\;\;\;\; = \;\;\;}}\dfrac{{{h^2}}}{{1 - h}}$ , which is another method )