The correct order of increasing $[H_{3} O^{+}]$ in the following aqueous solutions is:

0.01 M H_{2} S < 0.01 M H_{2} {S O}_{4} < 0.01 M N a C l < 0.01 M N a {N O}_{2}

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0.01 M N a C l < 0.01 M N a {N O}_{2} < 0.01 M H_{2} S < 0.01 M H_{2} {S O}_{4}

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0.01 M N a {N O}_{2} < 0.01 M N a C l < 0.01 M H_{2} S < 0.01 M H_{2} {S O}_{4}

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0.01 M H_{2} S < 0.01 M N a {N O}_{2} < 0.01 M N a C l < 0.01 M H_{2} {S O}_{4}

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Solution

The correct option is D $0.01 M N a {N O}_{2} < 0.01 M N a C l < 0.01 M H_{2} S < 0.01 M H_{2} {S O}_{4}$
$H_{2} {S O}_{4}$ is strong acid having $p H < 7$ .

$N a {N O}_{2}$ on hydrolysis gives alkaline solution of $p H > 7. N a C l$ is neutral solution with $p H = 7$ and $H_{2} S$ is weak acid with $p H < 7$ .

So order is $0.01 M N a {N O}_{2} < 0.01 M N a C l < 0.01 M H_{2} S < 0.01 M H_{2} {S O}_{4}$
Option C is correct.

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