$H_{2} S O_{4} ⟶ H^{+} + H S O_{4}^{-}$ (100% ionisation)

$H S O_{4}^{-} ⟶ H^{+} + S O_{4}^{2 -}$ (10% ionisation)

The given reaction takes place when sulphuric acid dissolves in water.
If we start with $0.2 M$ aqueous solution of $H_{2} S O_{4}$ then $[S O_{4}^{2 -}]$ in the above case will be:

0.1 M

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0.01 M

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0.2 M

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0.02 M

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Solution

The correct option is D $0.02 M$
$S t e p I : H_{2} S O_{4} ⟶ H^{+} + H S O_{4}^{-} - - - - - 100$ %

$S t e p I I : H S O_{4}^{-} ⟶ H^{+} + S O_{4}^{2 -} - - - - - 10$ %

$H^{+} + H_{2} O ⟶ H_{3} O^{+}$

$S O_{4}^{2 -} i s f r o m s t e p I I$

$S t e p I i s 100 %$ $t h u s [H^{+}] = 0.2 M$

and $[H S O_{4}^{-}] = 0.2 M$

$S t e p I I i s 10$ % $t h u s [H^{+}] = 0.2 \times 0.1 = 0.02 M$

and $[S O_{4}^{2 -}] = 0.02 M$

Total $[H_{3} O^{+}] = 0.2 + 0.02 = 0.22 M$

$[S O_{4}^{2 -}] = 0.02 M$

and $[H S O_{4}^{-}] = 0.20 - 0.02 = 0.18 M$ .

Hence, the correct option is $(D)$ .

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Similar questions

H_{2} S O_{4} ⟶ H^{+} + H S O_{4}^{-}

100 %

H_{2} S O_{4}^{-} ⟶ H^{+} + S O_{4}^{2 -}

10 %

0.2 M

H_{2} S O_{4}

[H S O_{4}^{-}]

H S O_{4}^{⊝}

S O_{4}^{2 -}

\begin{matrix} List-I & List-II (I) & 100 ml of 0.2 M A l C l_{3} solution & (P) & Total concentration of cation(s)=0.12M + 400 ml of 0.1 M H C l solution (I I) & 50 ml of 0.4 M K C l + 50 ml H_{2} O & (Q) & [S O_{4}^{2 -}] = 2.5 M (I I I) & 30 ml of 0.2 M K_{2} S O_{4} + 70 ml H_{2} O & (R) & [C l^{-} =] 2.5 M (I V) & 200 ml 24.5 % (w/v) H_{2} S O_{4} & (S) & [S O_{4}^{2 -} =] 0.06 M (T) & [C l^{-} =] 0.2 M (U) & Total concentration of cation(s) = 0.2 M \end{matrix}

N a H S O_{4}

100.64^{\circ} C

[K_{b} of H_{2} O = 0.512 K k g m o l^{- 1}]

H S O_{4}^{-} ⇌ H^{+} + S O_{4}^{2 -}

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