Q40 / 180 · AIIMS 2008 · Chemistry

The correct order of increasing \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\)in the following aqueous solutions is

  • A
    \(0.01 \mathrm{M} \mathrm{H}_2 \mathrm{~S} < 0.01 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4 < 0.01 \mathrm{M} \mathrm{NaCl} < 0.01 \mathrm{M} \mathrm{NaNO}_2\)
  • B
    \(0.01 \mathrm{M} \mathrm{NaCl} < 0.01 \mathrm{M} \mathrm{NaNO}_2 < 0.01 \mathrm{M} \mathrm{H}_2 \mathrm{~S} < 0.01 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4\)
  • Correct answer
    C
    \(0.01 \mathrm{M} \mathrm{NaNO}_2 < 0.01 \mathrm{M} \mathrm{NaCl} < 0.01 \mathrm{M} \mathrm{H}_2 \mathrm{~S} < 0.01 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4\)
  • D
    \(0.01 \mathrm{M} \mathrm{H}_2 \mathrm{~S} < 0.01 \mathrm{M} \mathrm{NaNO}_2 < 0.01 \mathrm{M} \mathrm{NaCl} < 0.01 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4\).

Solution

$\mathrm{H}_2 \mathrm{SO}_4$ is strong acid having $\mathrm{pH} < 7 . \mathrm{NaNO}_2$ on hydrolysis gives alkaline solution of $\mathrm{pH}>7$. $\mathrm{NaCl}$ is neutral and $\mathrm{H}_2 \mathrm{~S}$ is weak acid.

Hence, the correct answer is option: 'C'.

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