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WAEC Chemistry 2018 Paper

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Question : 27

Total: 50

Consider the following reaction equation:

2\mathrm{HCl} + \mathrm{Ca}(\mathrm{OH})_{2} \to \mathrm{CaCl}_{2} + \mathrm{H}_{2}\mathrm{O}

. What is the volume of 0.1

\text{mol dm}^{-3}

HCl that would completely neutralize 25

\text{cm}^{3}

\text{mol dm}^{-3}

_{2}

?

150 $\text{cm}^{3}$
75 $\text{cm}^{3}$
30 $\text{cm}^{3}$
25 $\text{cm}^{3}$

Solution:

Concept:

Neutralization proceeds according to the mole ratio in the balanced equation.

Explanation:

The balanced equation is:

2\mathrm{HCl} + \mathrm{Ca}(\mathrm{OH})_{2} \to \mathrm{CaCl}_{2} + 2\mathrm{H}_{2}\mathrm{O}

So,

2

moles of

\mathrm{HCl}

react with

1

mole of

\mathrm{Ca}(\mathrm{OH})_{2}

.

Use the neutralization formula:

\frac{C_A V_A}{C_B V_B} = \frac{n_A}{n_B}

Here,

C_A = 0.1\ \text{mol dm}^{-3}

,

V_B = 25\ \text{cm}^{3}

,

C_B = 0.3\ \text{mol dm}^{-3}

,

n_A = 2

,

n_B = 1

.

Substitute the values:

\frac{0.1 \times V_A}{0.3 \times 25} = \frac{2}{1}

V_A = \frac{0.3 \times 25 \times 2}{0.1 \times 1}

V_A = 150\ \text{cm}^{3}

Answer:

A.

150\ \text{cm}^{3}

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