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WAEC Chemistry 2020 Paper

Question : 37

Total: 49

20\,\mathrm{cm}^3

20\,\mathrm{cm}^3

0.01\,\mathrm{mol}\,\mathrm{dm}^3

(\mathrm{VII})

acid, what is the concentration of the solution?

Solution:

Concept:

The concentration of sodium hydroxide is found using the mole ratio from the balanced neutralization equation.

Explanation:

Tetraoxosulphate(VI) acid is

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

and sodium hydroxide is

\mathrm{NaOH}

Balanced equation:

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

Mole ratio of acid to base is

1:2

Moles of acid used:

\displaystyle n_a = 0.01 \times \frac{20}{1000} = 0.0002\,\mathrm{mol}

Moles of sodium hydroxide required:

n_b = 2 \times 0.0002 = 0.0004\,\mathrm{mol}

Concentration of sodium hydroxide:

\displaystyle C_b = \frac{0.0004}{20/1000} = 0.020\,\mathrm{mol}\,\mathrm{dm}^{-3}

Answer:

0.020\,\mathrm{mol}\,\mathrm{dm}^{-3}

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