Given $\underset{strongacid}{H_{2} {SO}_{4}} {+ H}_{2} O \to H_{2} O^{+} {+ HSO}_{4}^{-}$
$\underset{weak acid}{{HSO}_{4}^{-}} {+ H}_{2} O ⇌ H_{3} O^{+} {+ SO}_{4}^{2-} {, K}_{a} {= 1.1 × 10}^{-2}$
Thus, at equilibrium, $[H_{3} O^{+}], [{HSO}_{4}^{-}] and [{SO}_{4}^{2-}] {in 0.50 M H}_{2} {SO}_{4}$ respectively, are

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$[H_{3} O^{+}]$ $[{HSO}_{4}^{-}]$ $[{SO}_{4}^{2-}]$

0.51M 0.49M 0.01M

$0.50 M 0.50 M 0.50 M$

$0.49 M 0.49 M 0.01 M$

$0.51 M 0.51 M 0.51 M$

answer is A.

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Detailed Solution

$[H_{2} {SO}_{4}] = 0.50M [H_{3} O^{+}] = 0.5 M [{HSO}_{4}^{-}] = 0.50M$

${HSO}_{4}^{-} \Leftrightarrow H_{2} O^{+} {+ SO}_{4}^{2-}$

$0.5 0.5M 0$

$[H_{3} O^{+}] = (0.5 + x) M [{SO}_{4}^{2-}] = xm$

$H_{2} {SO}_{4}^{-} = (0.5 - x) {M K}_{a} = \frac{[H_{3} O^{+}] [{SO}_{4}^{2-}]}{[{HSO}_{4}^{-}]} x = 0.01$

$[H_{3} O^{+}] = 0.51 M [{SO}_{4}^{2-}] = 0.01M [{HSO}_{4}^{-}] = 0.49M$

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