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Question: Calculate the concentration of sulfate ions in \(0.01M\) of \({H_2}S{O_4}\) , if \({K_{a1}} = 1 \tim...

Calculate the concentration of sulfate ions in 0.01M0.01M of H2SO4{H_2}S{O_4} , if Ka1=1×10−2{K_{a1}} = 1 \times {10^{ - 2}} and Ka2=1×10−6{K_{a2}} = 1 \times {10^{ - 6}}
A.10−2{10^{ - 2}}
B.0.01×10−80.01 \times {10^{ - 8}}
C.1×10−61 \times {10^{ - 6}}
D.0.01×10−100.01 \times {10^{ - 10}}

Explanation

Solution

We know that Sulphuric acid is represented by the molecular formula H2SO4{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. Ka{K_a} is known as dissociation constant and it tells us the amount of dissociation that has occurred.

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

Complete step-by-step answer: We know that Sulphuric acid is represented by the molecular formula H2SO4{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
H2SO4+H2O→H++HSO4−{H_2}S{O_4} + {H_2}O \to {H^ + } + HS{O_4}^ -
The dissociation constant for this equation is represented as Ka{K_a} . It tells us the amount of dissociation that has occurred. For this equation Ka1=1×10−2{K_{a1}} = 1 \times {10^{ - 2}}
{K_a} = \dfrac{{[{H^ + }][{A^ - }]}}{{[HA]}}
⇒Ka1=[H+][HSO4−][H2SO4]\Rightarrow {K_{a1}} = \dfrac{{[{H^ + }][HS{O_4}^ - ]}}{{[{H_2}S{O_4}]}} …… (1)\left( 1 \right)
Ka1={K_{a1}} = Dissociation constant
[H+][{H^ + }] == The concentration of H+{H^ + } ions in the solution
[HSO4−]=[HS{O_4}^ - ] = The concentration of HSO4−HS{O_4}^ - ions in the solution
Similarly, the second stage of dissociation is represented as
HSO4−⇌H++SO42−HS{O_4}^ - \rightleftharpoons {H^ + } + S{O_4}^{2 - }
For this equation Ka2=1×10−6{K_{a2}} = 1 \times {10^{ - 6}}
Ka2=[H+][SO4−][HSO4−]{K_{a2}} = \dfrac{{[{H^ + }][S{O_4}^ - ]}}{{[HS{O_4}^ - ]}} …… (2)\left( 2 \right)
Ka2{K_{a2}} is the second dissociation constant. To get the concentration of sulfate ions, multiply equation (1)\left( 1 \right) and (2)\left( 2 \right)
⇒Ka1×Ka2=[H+][HSO4−][H2SO4]×[H+][SO4−][HSO4−]\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}^ - ]}}
⇒Ka1×Ka2=[H+]2[SO42−][H2SO4]\Rightarrow {K_{a1}} \times {K_{a2}} = \dfrac{{{{[{H^ + }]}^2}[S{O_4}^{2 - }]}}{{[{H_2}S{O_4}]}}
We are given that concentration of H2SO4{H_2}S{O_4} is 0.01M0.01M . Therefore
⇒[H2SO4]Ka1×Ka2=[SO42−]\Rightarrow [{H_2}S{O_4}]{K_{a1}} \times {K_{a2}} = [S{O_4}^{2 - }]
⇒0.01×10−2×10−6=[SO42−]\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{10^{ - 2}}

Thus the correct option is AA.

Note: We should know the value of the first dissociation constant is much greater than the second dissociation constant that is Ka1>>Ka2{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 HSO4−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