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MHTCET 16 May 2024 Shift 1 Paper

Question : 58 of 150

Marks: +1, -0

Calculate the pH of 0.01 M sulphuric acid.

Solution:

Concept:

Sulphuric acid is a strong diprotic acid, so it releases two hydronium ions per molecule.

Explanation:

For

0.01 \, \mathrm{M}

\mathrm{H}_2\mathrm{SO}_4

, complete dissociation gives:

\left[\mathrm{H}_3\mathrm{O}^+\right] = 2 \times 0.01 = 0.02 \, \mathrm{M} = 2 \times 10^{-2} \, \mathrm{M}

Now calculate pH using

\mathrm{pH} = -\log_{10}\left[\mathrm{H}_3\mathrm{O}^+\right]

\mathrm{pH} = -\log_{10}\left(2 \times 10^{-2}\right)

\mathrm{pH} = -\log_{10} 2 + 2 = 2 - 0.3010 = 1.699

Answer:

\mathrm{pH} = 1.699

, so option A is correct.

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