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

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: $h = \sqrt {\dfrac{{{K_h}}}{c}} = \sqrt {\dfrac{{{K_W}}}{{{K_a} \times c}}}$