Question

${K_{a1}},{K_{a2}}~and~{K_{a3}}$ are the respective ionization constant for the following reactions.
${H_2}S{H^ + } + H{S^ - }$
$H{S^ - }{H^ + } + {S^{2 - }}$
${H_2}S2{H^ + } + {S^{2 - }}$

The correct relationship between ${K_{a1}},{K_{a2}}~and~{K_{a3}}$ is
(A) ${K_{a3}} = {K_{a1}} \times {K_{a2}}$
(B) ${K_{a3}} = {K_{a1}} + {K_{a2}}$
(C) ${K_{a3}} = {K_{a1}} - {K_{a2}}$
(D) ${K_{a3}} = {K_{a1}} \div {K_{a2}}$

Answer 1

We can calculate the relation of ${K_{a1}},{K_{a2}}~and~{K_{a3}}$ by finding the ionisation or dissociation constant of each reaction and further comparing it. Dissociation constant is given by ratio of product of concentration of the product by product of concentration of the reactants.

Complete Solution :
Let us first understand what a ${K_a}$ is. ${K_a}$ is the Dissociation constant of an acid. It is the equilibrium constant for dissociation reaction of an acid. From this value, we can say how much acid or base has been dissociated. We can measure quantitatively and know about how much of an acid or base will be present in a solution.

- Let us consider an example of an acid, say HA. This acid when it is dissolved in water it will dissociate into hydronium ion and conjugate base. The reaction is given below:
$HA + {H_2}O \to {H_3}{O^ + } + {A^ - }$……. (1)

The dissociation constant of the above reaction is given as
${K_a} = \dfrac{{[{A^ - }][{H_3}{O^ + }]}}{{[HA][{H_2}O]}}$……... (2)

The dissociation constant ${K_a}$ is given as the ratio of product of concentration of the product by product of concentration of the reactants. It is also known as the ionization constant.

In the reaction, ${H_2}S{H^ + } + H{S^ - }$, hydrogen sulphide dissociate into ${H^ + }$and $H{S^ - }$ions. The dissociation constant is
${K_{a1}} = \dfrac{{[{H^ + }][H{S^ - }]}}{{[{H_2}S]}}$…. (3)

In the reaction $H{S^ - }{H^ + } + {S^{2 - }}$, $H{S^ - }$ will dissociate to give ${H^ + }$ and ${S^{2 - }}$ ions. The dissociation constant is
${K_{a2}} = \dfrac{{[{H^ + }][{S^{2 - }}]}}{{[H{S^ - }]}}$…… (4)

In the reaction, ${H_2}S2{H^ + } + {S^{2 - }}$, ${H_2}S$ will be dissociated into $2{H^ + }$ and ${S^{2 - }}$ ions. The dissociation constant is
${K_{a3}} = \dfrac{{{{[{H^ + }]}^2}[{S^{2 - }}]}}{{[{H_2}S]}}$…... (5)
Multiply the equation (3) and (4)

Then we get:
${K_{a1}} \times {K_{a2}} = \dfrac{{[{H^ + }][H{S^ - }]}}{{[{H_2}S]}} \times \dfrac{{[{H^ + }][{S^{2 - }}]}}{{[H{S^ - }]}}$
${K_{a1}} \times {K_{a2}} = \dfrac{{{{[{H^ + }]}^2}[{S^{2 - }}]}}{{[{H_2}S]}}$ which is similar to equation (5)
Therefore, ${K_{a1}} \times {K_{a2}} = {K_{a3}}$
The correct relationship between ${K_{a1}},{K_{a2}}~and~{K_{a3}}$ is ${K_{a1}} \times {K_{a2}} = {K_{a3}}$
So, the correct answer is “Option A”.

The product of molar concentration of hydrogen ion, ${H^ + }$ and hydroxide ion $O{H^ - }$ will give the ionic product of water. The ionic product of water is given by the equation:
${K_W} = [{H^ + }][O{H^ - }]$

Note: Weak acids are the substances that will dissociate only partially into ${H^ + }$ ions, but strong acids will ionise completely into ${H^ + }$ ions. In this reaction, we have used hydrogen sulphide which is a weak acid. Hence the dissociation of ${H^ + }$ ions is very partial.