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 $0.01 M N a C l < 0.01 M N a N O_{2} < 0.01 M H_{2} S < 0.01 M H_{2} S O_{4}$
Among $0.01 M$ $H_{2} S$ , $0.01 M$ $H_{2} {S O}_{4}$ , $0.01 M$ $N a C l$ and $0.01 M$ ${N a N O}_{2}$ , $0.01$ $H_{2} {S O}_{4}$ has highest $[H_{3} O^{+}]$ and $0.01 M$ $N a C l$ has lowest $[H_{3} O^{+}]$ .
Because $H_{2} {S O}_{4}$ is very strong acid and it is almost fully ionized to give $0.02 N$ $[H_{3} O^{+}]$ , but $N a C l$ does not hydrolyses to give $[H_{3} O^{+}]$ . Since it is forms strong acid $(H C l)$ and strong base $(N a O H)$ . Again $[H_{3} O^{+}]$ will higher for $H_{2} S$ than that of ${N a N O}_{2}$ .
$∴$ correct order will be : $(B)$
$0.01 M N a C l < 0.01 M {N a N O}_{2} < 0.01 M H_{2} S < 0.01 M H_{2} {S O}_{4}$

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