The correct order of increasing `[H_(3)O^(o+)]` in the following aqueous solution is

Options

A. `0.01M H_(2)S lt 0.01M H_(2)SO_(4) lt 0.01 M NaC1 lt 0.01M NaNO_(3)`
B. `0.01M NaC1 = 0.01 M NaNO_(3) lt 0.01 M H_(2)S lt 0.01M H_(2)SO_(4)`
C. `0.01M H_(2)S lt 0.01M NaNO_(3) = 0.01 M NaC1 lt 0.01M H_(2)SO_(4)`
D. `0.01M H_(2)S lt 0.01M NaNO_(3) lt 0.01M NaC1 lt 0.01M H_(2)SO_(4)`

Answer

B

Explanation

Higher `[H_(3)O^(o+)]`, lower is the `pH`.
`H_(2)SO_(4)-= S_(A) rArr` Highest `[H_(3)O^(o+)]`
`H_(2)S -= W_(A) rArr` Lower `[H_(3)O^(o+)]`
`NaC1` and `NaNO_(3) -=` Salt of `S_(A)//S_(B) (pH = 7)`
`rArr` Lowest `[H_(3)O^(o+)] = 10^(-7)m`
Hence, the order is (b).