What is the molarity of {H}₂ {SO}₄ solution having {pH}=4 ?

Correct Answer is : 5×10−5 N5 × 10^{-5} {~N}5×10−5 N

What is the molarity of {H}₂ {SO}₄ solution having {pH}=4 ?

Correct Answer is : 5×10−5 N5 × 10^{-5} {~N}5×10−5 N

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MHTCET 18 Apr 2026 Shift 1 Paper

Question : 62 of 150

Marks: +1, -0

\mathrm{H}_{2} \mathrm{SO}_{4}

solution having

\mathrm{pH}=4

$4 \times 10^{-1} \mathrm{~N}$
$4 \times 10^{-2} \mathrm{~N}$
$5 \times 10^{-5} \mathrm{~N}$
$1 \times 10^{-4} \mathrm{~N}$

Validate

Solution:

Concept:

The pH of a solution gives the concentration of hydrogen ions. For a strong dibasic acid like sulphuric acid, the acid concentration is half of the hydrogen ion concentration.

Explanation:

Given that the pH of the

\mathrm{H}_{2}\mathrm{SO}_{4}

solution is

4

The hydrogen ion concentration is calculated from the pH as:

[\mathrm{H}^{+}] = 10^{-\mathrm{pH}} = 10^{-4}\ \mathrm{M}

Sulphuric acid dissociates completely in water as:

\mathrm{H}_{2}\mathrm{SO}_{4} \rightarrow 2\mathrm{H}^{+} + \mathrm{SO}_{4}^{2-}

Thus, one mole of

\mathrm{H}_{2}\mathrm{SO}_{4}

produces two moles of

\mathrm{H}^{+}

ions.

If the molarity of the acid is

M

, then:

[\mathrm{H}^{+}] = 2M

Substituting the known value of

[\mathrm{H}^{+}]

2M = 10^{-4}

Solving for

M

M = \frac{10^{-4}}{2} = 5 \times 10^{-5}\ \mathrm{M}

This value matches the one given in the options.

Answer:

The molarity of the

\mathrm{H}_{2}\mathrm{SO}_{4}

solution is

5 \times 10^{-5}\ \mathrm{M}

, which corresponds to option C.

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