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 $\left[HS{O}_4^{-}\right]$ in the above case is:

• A. $0.1M$
• B. $0.09M$
• C. $0.18M$
• D. $0.08M$

Answer 1

Answer: (C)

$0.18M$

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

$StepIIHS{O}_4^{-}⟶H^{+}+S{O}_4^{2-}10$%

$H^{+}+H_{2}O⟶H_3{O}^{+}$

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

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

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

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

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

total $[H_3{O}^{+}l=0.2+0.02=0.22M$

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

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

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