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Question: The dissociation constants of acetic acid and aniline are \(1.8 \times {10^{ - 5}}\)and \(4.2 \times...

The dissociation constants of acetic acid and aniline are 1.8×10−51.8 \times {10^{ - 5}}and 4.2×10−104.2 \times {10^{ - 10}}respectively. What is the degree of hydrolysis of aqueous aniline acetate? What is the pH of the solution?

Explanation

Solution

As we know that degree of hydrolysis is the fraction of total salt which is hydrolysed or we can say that it is the ratio of the concentrations of products to the concentration of reactants of a salt and the pH can be calculated using the negative algorithm of dissociation constant of acid and base.

Complete solution:
As we know that degree of hydrolysis is the ratio of the concentrations of products to the concentration of reactants of a salt. We also know that Acetic acid is a weak acid so acetate ion will be hydrolysed to given acetic acid, similarly the Aniline is a weak base so aniline ions will be hydrolysed in water to give the aniline. We can represent this in an equation for hydrolysis of salts of weak acid and weak base as:
C6H5NH2++CH3COO−+H2O⇌C6H5NH2+CH3COOH{C_6}{H_5}NH_2^ + + C{H_3}CO{O^ - } + {H_2}O \rightleftharpoons {C_6}{H_5}N{H_2} + C{H_3}COOH
Let the concentration of Let the concentration of these compounds be CC at initial time and CαC\alpha at equilibrium.
C6H5NH2++CH3COO−+H2O⇌C6H5NH2+CH3COOH{C_6}{H_5}NH_2^ + + C{H_3}CO{O^ - } + {H_2}O \rightleftharpoons {C_6}{H_5}N{H_2} + C{H_3}COOH

At initial timeCCCC0000
At equilibriumC(1−α)C(1 - \alpha )C(1−α)C(1 - \alpha )CαC\alpha CαC\alpha

Now, we know that hydrolysis constant KH{K_H} is given as the ratio of concentrations of product to the concentrations of the reactants. So we will get:
KH=[C6H5NH2][CH3COOH][C6H5NH2+][CH3COO−]{K_H} = \dfrac{{[{C_6}{H_5}N{H_2}][C{H_3}COOH]}}{{[{C_6}{H_5}NH_2^ + ][C{H_3}CO{O^ - }]}}
Putting the equilibrium values we will get:
KH=[Cα][Cα][C(1−α)][C(1−α)]{K_H} = \dfrac{{[C\alpha ][C\alpha ]}}{{[C(1 - \alpha )][C(1 - \alpha )]}}
KH=α1−α\sqrt {{K_H}} = \dfrac{\alpha }{{1 - \alpha }}
Where α\alpha is the degree of hydrolysis, we also know that, KH=KWKa×Kb{K_H} = \dfrac{{{K_W}}}{{{K_a} \times {K_b}}}, where Kw=10−14{K_w} = {10^{ - 14}}which is the ionic product of water.
Now we can put this value in the above formula and calculate the degree of hydrolysis.
KwKa×Kb=α1−α\sqrt {\dfrac{{{K_w}}}{{{K_a} \times {K_b}}}} = \dfrac{\alpha }{{1 - \alpha }}
10−141.8×10−5×4.2×10−10=α1−α\sqrt {\dfrac{{{{10}^{ - 14}}}}{{1.8 \times {{10}^{ - 5}} \times 4.2 \times {{10}^{ - 10}}}}} = \dfrac{\alpha }{{1 - \alpha }}
α=0.53\alpha = 0.53
Now we know that pH for hydrolysis of weak acid and weak bases can be easily calculated using the formula:
pH=7+12pKa−12pKbpH = 7 + \dfrac{1}{2}p{K_a} - \dfrac{1}{2}p{K_b}
So we need to calculate the values of PkaP{k_a}and PkbP{k_b}which are given by the negative logarithm of dissociation constants of acid and base. So we will get:
pKa=−log⁡(1.8×10−5)p{K_a} = - \log (1.8 \times {10^{ - 5}})
pKa=4.74p{K_a} = 4.74
Similarly, pKb=−log⁡(4.2×10−10)p{K_b} = - \log (4.2 \times {10^{ - 10}})
pKb=9.37p{K_b} = 9.37
Thus putting these values in the pH formula we will get:
pH=7+12×4.74−12×9.37pH = 7 + \dfrac{1}{2} \times 4.74 - \dfrac{1}{2} \times 9.37
pH=7+(2.37−4.685) pH=4.68  pH = 7 + (2.37 - 4.685) \\\ pH = 4.68 \\\
Therefore from the above solution we can say that the degree of hydrolysis is 0.530.53and pH of the solution is 4.684.68.

So the correct option is D.D.

Note: Remember that degree of hydrolysis is used in quantitative estimation of salts of strong acids, strong bases, weak acids and weak bases. The degree of hydrolysis increases with increases in dilution for the salts of strong acid and weak base but when salt of weak acid and weak base are taken into account then degree of hydrolysis is independent of dilution.