From the following chemical reactions determine the relative Bronsted-Lowery acid strengths (strongest to weakest).
$H C l O_{4} (s o l) + C H_{3} C O O H (l) ⟶ C H_{3} C (O H)_{2}^{+} (s o l) + C l O_{4}^{-} (a q)$
$H_{2} S O_{4} (s o l) + C H_{3} C O O H (l) ⇌ C H_{3} C (O H)_{2}^{+} (s o l) + C l O_{4}^{-} (s o l)$

H C l O_{4} > C H_{3} C (O H)_{2}^{+} > H_{2} S O_{4}

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H C l O_{4} > C H_{3} C O O H > H_{2} S O_{4}

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H C l O_{4} > H_{2} S O_{4} > C H_{3} C (O H)_{2}^{+}

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H C l O_{4} > H_{2} S O_{4} > C H_{3} C O O H

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Solution

The correct option is A $H C l O_{4} > C H_{3} C (O H)_{2}^{+} > H_{2} S O_{4}$
The pKa value of ${C H}_{3} C (O {H)}_{2}$ is -8 and the pKa value of $H_{2} {S O}_{4}$ is -3 and ${C H}_{3} C (O {H)}_{2}$ is an inorganic acid. So ${C H}_{3} C (O {H)}_{2}$ is strong acid than $H_{2} {S O}_{4}$ and $H C I O_{4}$ is strong acid than both two acids.
Hence, option $(A)$ is the answer.

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K C l O_{3} + H_{2} S O_{4} \to K H S O_{4} + H C l O_{4} + C l O_{2}

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H_{2} C O_{3}, H_{3} P O_{3}, C H_{3} C O O H, H_{2} S O_{4}, H_{3} P O_{2}, H_{4} P_{2} O_{7}, H C l O_{4}, H_{3} B O_{3}, H N O_{3} .

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Review the equilibrium and choose the correct statement

$H C l O_{4} + H_{2} O ⇌ H_{3} O^{+} + C l O_{4}^{-}$

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