Questions & Answers

Question

Answers

\[

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 \\

\]

Answer

Verified

129k+ views

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).

\[

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, \[pH = \dfrac{1}{2}(p{K_{w\;\;}} - p{K_{a\;\;}} - p{K_{b\;\;}})\]

\[ = \dfrac{1}{2}\log \left[ {1.008 \times {{10}^{ - 14\;}}} \right] - \log \left[ {1.75 \times {{10}^{ - 5\;}}} \right] - \log \left[ {3.83 \times {{10}^{ - 10\;}}} \right]\]

\[ = 4.6683\]

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 )