Question

Calculate the concentration of sulfate ions in $0.01M$ of ${H_2}S{O_4}$ , if ${K_{a1}} = 1 \times {10^{ - 2}}$ and ${K_{a2}} = 1 \times {10^{ - 6}}$
A.${10^{ - 2}}$
B.$0.01 \times {10^{ - 8}}$
C.$1 \times {10^{ - 6}}$
D.$0.01 \times {10^{ - 10}}$

Answer 1

Hint:We know that Sulphuric acid is represented by the molecular formula ${H_2}S{O_4}$ . It is a diprotic molecule that gives two protons per molecule. The dissociation of sulphuric acid occurs in two stages. ${K_a}$ is known as dissociation constant and it tells us the amount of dissociation that has occurred.

Formula Used:
$${K_a} = \dfrac{{[{H^ + }][{A^ - }]}}{{[HA]}}$$
${K_a} = $ Dissociation constant
$$[{H^ + }]$$$ = $ The concentration of ${H^ + }$ ions in the solution
$$[{A^ - }]$$$ = $ The concentration of conjugate base
$$[HA]$$$ = $ The concentration of acid in the solution

Complete step-by-step answer:We know that Sulphuric acid is represented by the molecular formula ${H_2}S{O_4}$ . When it is dissolved in water, dissociation occurs. As we know that sulphuric acid is a strong diprotic acid. This means it can donate two protons per molecule. Hence the dissociation of sulphuric acid occurs in two stages.
The first stage of dissociation is represented as
${H_2}S{O_4} + {H_2}O \to {H^ + } + HS{O_4}^ - $
The dissociation constant for this equation is represented as ${K_a}$ . It tells us the amount of dissociation that has occurred. For this equation ${K_{a1}} = 1 \times {10^{ - 2}}$
$${K_a} = \dfrac{{[{H^ + }][{A^ - }]}}{{[HA]}}$$
$ \Rightarrow {K_{a1}} = \dfrac{{[{H^ + }][HS{O_4}^ - ]}}{{[{H_2}S{O_4}]}}$ …… $\left( 1 \right)$
${K_{a1}} = $ Dissociation constant
$[{H^ + }]$$ = $ The concentration of ${H^ + }$ ions in the solution
$[HS{O_4}^ - ] = $ The concentration of $HS{O_4}^ - $ ions in the solution
Similarly, the second stage of dissociation is represented as
$HS{O_4}^ - \rightleftharpoons {H^ + } + S{O_4}^{2 - }$
For this equation ${K_{a2}} = 1 \times {10^{ - 6}}$
${K_{a2}} = \dfrac{{[{H^ + }][S{O_4}^ - ]}}{{[HS{O_4}^ - ]}}$ …… $\left( 2 \right)$
${K_{a2}}$ is the second dissociation constant. To get the concentration of sulfate ions, multiply equation $\left( 1 \right)$ and $\left( 2 \right)$
$ \Rightarrow {K_{a1}} \times {K_{a2}} = \dfrac{{[{H^ + }][HS{O_4}^ - ]}}{{[{H_2}S{O_4}]}} \times \dfrac{{[{H^ + }][S{O_4}^ - ]}}{{[HS{O_4}^ - ]}}$
$ \Rightarrow {K_{a1}} \times {K_{a2}} = \dfrac{{{{[{H^ + }]}^2}[S{O_4}^{2 - }]}}{{[{H_2}S{O_4}]}}$
We are given that concentration of ${H_2}S{O_4}$ is $0.01M$ . Therefore
$ \Rightarrow [{H_2}S{O_4}]{K_{a1}} \times {K_{a2}} = [S{O_4}^{2 - }]$
$ \Rightarrow 0.01 \times {10^{ - 2}} \times {10^{ - 6}} = [S{O_4}^{2 - }]$
This gives us the concentration of sulfate ions which is ${10^{ - 2}}$

Thus the correct option is $A$.

Note:We should know the value of the first dissociation constant is much greater than the second dissociation constant that is ${K_{a1}} > > {K_{a2}}$ . This is because sulphuric acid in the first dissociation stage is a neutral molecule, therefore it readily loses a proton. In $HS{O_4}^ - $ , losing proton is much difficult, this is because of the extra electrons that are gained on losing a proton. Therefore first dissociation constant is much greater than the second dissociation constant