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JAMB Chemistry 2024 Paper

Question : 22

Total: 138

^{-3}

_2

_4

Solution:

Concept:

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

is a strong diprotic acid, so one mole releases two moles of

\mathrm{H}^+

ions.

Explanation:

Given concentration of

\mathrm{H}_2\mathrm{SO}_4 = 0.001\ \mathrm{mol\ dm}^{-3}

When it dissociates completely:

\mathrm{H}_2\mathrm{SO}_4 \rightarrow 2\mathrm{H}^+ + \mathrm{SO}_4^{2-}

Thus, the concentration of

\mathrm{H}^+

ions is:

[\mathrm{H}^+] = 2 \times 0.001 = 0.002\ \mathrm{mol\ dm}^{-3}

Write this as

0.002 = 2 \times 10^{-3}

The pH formula is:

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

Substitute the hydrogen ion concentration:

\mathrm{pH} = -\log_{10}(2 \times 10^{-3})

Using logarithm rules:

\mathrm{pH} = -(\log_{10}2 + \log_{10}10^{-3})

\mathrm{pH} = -(0.3010 - 3)

\mathrm{pH} = -(-2.699) = 2.699 \approx 2.7

Answer:

\mathrm{pH} = 2.7

Option A.

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