Question

If the pH of an aqueous solution of ammonium chloride is 4.60, what is the $ {K_h} $ value of $ NH_4^ + $ ?

Answer 1

Hint: to solve the given question we need to know the values of $ {K_w},{K_a}\& {K_b} $ . By knowing the relationship between these three values, we can determine the value of $ {K_a} $ and hence of $ {K_h} $ consequently. Since the value of $ {K_b} $ is not given, we will assume the standard value.

Complete answer:
The value of $ {K_b} $ of ammonia is related to that of the $ {K_a} $ for ammonium ion $ NH_4^ + $ . The relationship between $ {K_w},{K_a}\& {K_b} $ can be given as: $ {K_a}.{K_b} = {K_w} = {10^{ - 14}} $
The $ {K_b} $ value of ammonia is: $ {K_b} = 1.8 \times {10^{ - 5}} $
The value of $ {K_w} = {10^{ - 14}} $
Therefore, $ {K_a} = \dfrac{{{K_w}}}{{{K_b}}} = \dfrac{{{{10}^{ - 14}}}}{{1.8 \times {{10}^{ - 5}}}} = 5.55 \times {10^{ - 10}} $
This is the value of $ {K_a} $ that is expected. Now let us find the same using the ICE table:
 $ {\text{NH}}_4^ + {\text{(aq) + }}{{\text{H}}_2}{\text{O(l) }} \rightleftharpoons {\text{ N}}{{\text{H}}_3}(aq){\text{ }} + {\text{ }}{H_3}{O^ + }(aq) $

T=0	0.1	-	--	-
T=equilibrium	$ 0.1 - x $	-	$ x $	$ x $

Hence, the value of $ {K_a} = \dfrac{{{x^2}}}{{0.1 - x}} $
We know that $ NH_4^ + $ is a weak acid since $ N{H_3} $ is a weak base. The value of $ {K_b} $ for the weak base itself is very small. Therefore, the value of $ {K_a} $ will be even smaller. In the above equation for calculating $ {K_a} $ will not ignore the value of x considering it to be small.
The concentration of hydronium ions can be found out by the value of pH; $ pH = - \log ({H^ + }) $
 $ [{H^ + }] = AL(pH) $
 $ [{H^ + }] = AL(4.60) = {10^{ - 4.60}} $
The value of $ {K_a} $ can be given as: $ {K_a} = \dfrac{{{{({{10}^{ - 4.60}})}^2}}}{{0.1 - ({{10}^{ - 4.60}})}} $
Therefore, $ {K_a} = 6.310 \times {10^{ - 9}} $
The relationship between $ {K_h}\& {K_a} $ can be given as $ {K_h} = \dfrac{{{K_w}}}{{{K_a}}} $ .
Substituting the values in the equation we have $ {K_h} = \dfrac{{{{10}^{ - 14}}}}{{6.310 \times {{10}^{ - 9}}}} = 1.5847 \times {10^{ - 6}} $
This is the required answer.

Note:
$ {K_h} $ is known as the Hydrolysis constant or the equilibrium constant for hydrolysis reaction and is given as $ {K_h} = \dfrac{{[HX][O{H^ - }]}}{{[{X^ - }]}} $ . The relationship between $ {K_h}\& {K_a} $ is given as: $ {K_h} = \dfrac{{{K_w}}}{{{K_a}}} $ . This shows that $ {K_h} $ is inversely proportional to $ {K_a} $ . Higher the value of $ {K_a} $ lower the value of $ {K_h} $ . The hydrolysis constant ‘h’ can be given as: $ h = \sqrt {\dfrac{{{K_h}}}{c}} = \sqrt {\dfrac{{{K_W}}}{{{K_a} \times c}}} $