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JAMB Chemistry 2015 Paper

Question : 35

Total: 64

What volume of 0.5mldm³ H₂SO₄ will exactly neutralize 20cm³ of 0.1moldm³ NaOH Solution?

Solution:

Concept:

An acid-base titration uses stoichiometry from the balanced neutralization equation to find the unknown volume of acid.

Explanation:

Write the balanced equation:

\mathrm{H}_2\mathrm{SO}_4 + 2\mathrm{NaOH} \rightarrow \mathrm{Na}_2\mathrm{SO}_4 + 2\mathrm{H}_2\mathrm{O}

From the equation,

1

mole of

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

reacts with

2

moles of

\mathrm{NaOH}

Calculate moles of

\mathrm{NaOH}

used:

\text{moles NaOH} = 0.1\,\mathrm{mol\,dm^{-3}} \times 0.020\,\mathrm{dm^3} = 0.002\,\mathrm{mol}

Moles of

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

needed are half the moles of

\mathrm{NaOH}

\text{moles } \mathrm{H}_2\mathrm{SO}_4 = \frac{0.002}{2} = 0.001\,\mathrm{mol}

Find the volume of acid:

V = \frac{\text{moles}}{\text{concentration}} = \frac{0.001\,\mathrm{mol}}{0.5\,\mathrm{mol\,dm^{-3}}} = 0.002\,\mathrm{dm^3}

Convert to

\mathrm{cm^3}

0.002\,\mathrm{dm^3} = 2.0\,\mathrm{cm^3}

Answer:

2.0\,\mathrm{cm^3}

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