The correct order of increasing $[H_{3} O^{+}]$ in the following aqueous solution 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 $0.01 M H_{2} S < 0.01 M N a N O_{2} < 0.01 M N a C l < 0.01 M H_{2} S O_{4}$
Since, $H_{2} S O_{4}$ is stronger acid than $H C l H N O_{2} H_{2} S$ , $[H_{3} O^{+}]$ will be highest for it. It will be followed by $N a C l$ as it is a salt of strong acid & strong base. $H_{2} S$ will have the least $[H_{3} O^{+}]$ concentration as it is a weaker acid.
Hence, the required order is:
$0.01 M H_{2} S < 0.01 M N a N O_{2} < 0.01 M N a C l < 0.01 M H_{2} S O_{4}$

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