Some Basic Concepts of Chemistry 2 Question 7
7. $25 \mathrm{~mL}$ of the given $\mathrm{HCl}$ solution requires $30 \mathrm{~mL}$ of $0.1 \mathrm{M}$ sodium carbonate solution. What is the volume of this $\mathrm{HCl}$ solution required to titrate $30 \mathrm{~mL}$ of $0.2 \mathrm{M}$ aqueous $\mathrm{NaOH}$ solution?
(2019 Main, 11 Jan II)
(a) $75 \mathrm{~mL}$
(b) $25 \mathrm{~mL}$
(c) $12.5 \mathrm{~mL}$
(d) $50 \mathrm{~mL}$
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Solution:
- The reaction of $\mathrm{HCl}$ with $\mathrm{Na}{2} \mathrm{CO}{3}$ is as follows:
$2 \mathrm{HCl}+\mathrm{Na}{2} \mathrm{CO}{3} \longrightarrow 2 \mathrm{NaCl}+\mathrm{H}{2} \mathrm{O}+\mathrm{CO}{2}$
We know that, $M_{\text {eq }}$ of $\mathrm{HCl}=M_{\text {eq }}$ of $\mathrm{Na}{2} \mathrm{CO}{3}$
$$ \begin{array}{r} \frac{25}{1000} \times 1 \times M_{\mathrm{HCl}}=\frac{30}{1000} \times 0.1 \times 2 \ M_{\mathrm{HCl}}=\frac{30 \times 0.2}{25}=\frac{6}{25} \mathrm{M} \end{array} $$
The reaction of $\mathrm{HCl}$ with $\mathrm{NaOH}$ is as follows:
$$ \mathrm{NaOH}+\mathrm{HCl} \longrightarrow \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O} $$
Also, $\mathrm{M}{\text {eq }}$ of $\mathrm{HCl}=\mathrm{M}{\text {eq }}$ of $\mathrm{NaOH}$
$$ \begin{gathered} \frac{6}{25} \times 1 \times \frac{V}{1000}=\frac{30}{1000} \times 0.2 \times 1 \ V=25 \mathrm{~mL} \end{gathered} $$
