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)

If $0.2M$ aqueous solution of $H_{2}S{O}_4$ then $\left[H_3{O}^{+}\right]$ in aqueous solution is:

• A. $0.28$ M
  
• B. $0.22$ M
• C. $0.02$ M
• D. $0.10$ M

Answer 1

The correct option is B $0.22$ M

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

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

$H^{+}+H_{2}O⟶H_3{O}^{+}$

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

$StepI:100$% 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 $\left[H_3{O}^{+}\right]=0.2+0.02=0.22M$,

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

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