Question

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

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

The given reaction takes place when sulphuric acid dissolves in water.

If we start with $0.2M$ aqueous solution of $H_{2}S{O}_4$ then maximum concentration is of __________ species.

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

Answer 1

The correct option is B $H_3{O}^{+}$

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

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

$H^{+}+H_{2}O⟶H_3{O}^{+}$

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

$StepIis100\%$ $thus\left[H^{+}\right]=0.2M$

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

$StepIIis10$% $thus\left[H^{+}\right]=0.2\times 0.1=0.02M$

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 $\left(B\right)$.