Question

When sulphuric acid dissolves in water, the following reactions take place:

$H_{2}S{O}_4⟶H^{+}+HS{O}_4^{-}$ ($100\%$ ionisation)

$H_{2}S{O}_4^{-}⟶H^{+}+S{O}_4^{2-}$ ($10\%$ ionisation)

We start with $0.2M$ aqueous solution of $H_{2}S{O}_4$.

The species that has the maximum concentration is :

• A. $H_{2}S{O}_4$
• B. $H_3{O}^{+}$
• C. $S{O}_4^{2-}$
• D. $HS{O}_4^{-}$

Answer 1

Answer: (B)

$H_3{O}^{+}$

$StepI:H_{2}S{O}_4⟶H^{+}+HS{O}_4^{-}$ $100$%

$StepII:HS{O}_4^{-}⟶H^{+}+S{O}_4^{2-}$ $10$%

$H^{+}+H_{2}O⟶H_3{O}^{+}$

$S{O}_4^{2-}isfromstepII$.

$StepI:100$% and hence, $\left[H^{+}\right]=0.2M$

.
and $\left[HS{O}_4^{-}\right]=0.2M$.

$StepII:10$% and hence, $\left[H^{+}\right]=0.2\times 0.1=0.02M$

and $\left[S{O}_4^{2-}\right]=0.02M$.

Total $\left[H_3{O}^{+}\right]=0.2+0.02=0.22M$,

$\left[S{O}_4^{2-}\right]=0.02M$;
$\left[HS{O}_4^{-}\right]=0.20-0.02=0.18M$.

$\therefore$ The species that has maximum concentration is $\left[H_3{O}^{+}\right]=0.22M$

Hence, the correct option is $\left(B\right)$