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WAEC Chemistry 2014 Paper

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Question : 27

Total: 45

Consider the reaction represented by the following equation: Na₂CO_3(aq) + 2HCl_(aq)

⟶

2NaCl_(aq) + H₂O_(l) + CO_2(g) . What volume of 0.02 mol dm^-3 Na₂CO_3(aq) would be required to completely neutralize 40 cm³ of 0.10 mol dm^-3 HCl_(aq)?

200 cm³
100 cm³
40 cm³
20 cm³

Solution:

Concept:

Use the stoichiometric molar ratio from the balanced equation and the relation

n = cV

.

Explanation:

First, find the moles of HCl present.

Volume of HCl:

40\ \mathrm{cm^3} = \frac{40}{1000}\ \mathrm{dm^3} = 0.040\ \mathrm{dm^3}

.

Moles of HCl:

0.10\ \mathrm{mol\,dm^{-3}} \times 0.040\ \mathrm{dm^3} = 0.004\ \mathrm{mol}

.

From the equation:

1\ \mathrm{mol\ Na_2CO_3}

reacts with

2\ \mathrm{mol\ HCl}

.

Therefore, moles of

\mathrm{Na_2CO_3}

required:

\frac{0.004}{2} = 0.002\ \mathrm{mol}

.

Let the required volume be

V

(in

\mathrm{dm^3}

).

Using

c = \frac{n}{V}

, we get

V = \frac{n}{c}

.

V = \frac{0.002\ \mathrm{mol}}{0.02\ \mathrm{mol\,dm^{-3}}} = 0.10\ \mathrm{dm^3}

.

Converting to

\mathrm{cm^3}

:

0.10 \times 1000 = 100\ \mathrm{cm^3}

.

Answer:

The required volume is

100\ \mathrm{cm^3}

, which is option B.

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