$30.0 \mathrm{~mL}$ of the given $\mathrm{HCl}$ solution requires $20.0 \mathrm{~mL}$ of $0.1 \mathrm{M}$…
$30.0 \mathrm{~mL}$ of the given $\mathrm{HCl}$ solution requires $20.0 \mathrm{~mL}$ of $0.1 \mathrm{M}$ sodium carbonate solution for complete neutralisation. What is the volume of this $\mathrm{HCl}$ solution required to neutralise $30.0 \mathrm{~mL}$ of $0.2 \mathrm{M} \mathrm{NaOH}$ solution?
25 mL
50 mL
90 mL
45 mL
Solution
Given,
Volume of $\mathrm{HCl}$ solution $\left(V_1\right)=30 \mathrm{~mL}$
Volume of sodium carbonate solution $\left(\mathrm{Na}_2 \mathrm{CO}_3\right)$
$\left(V_2\right)=20 \mathrm{~mL}$
Concentration of $\mathrm{Na}_2 \mathrm{CO}_3$ solution $\left(M_2\right)=0.1 \mathrm{M}$ Volume of $\mathrm{NaOH}$ solution $\left(V_3\right)=30.0 \mathrm{~mL}$
Concentration of $\mathrm{NaOH}$ solution $\left(M_3\right)=0.2 \mathrm{M}$
$M_1 V_1=M_2 V_2$
where, $M_1$ is concentration of $\mathrm{HCl}$ solution.
$M_1=\frac{M_2 V_2}{V_1}=\frac{0.1 \times 20}{30}=0.066 \mathrm{M}$
Also, $\mathrm{HCl}$ solution required to neutralise $30 \mathrm{~mL}$ of $0.2 \mathrm{NaOH}$ is calculated as follows :
$M_1 V_f=M_3 V_3$ (where, $V_f$ is the volume of $\mathrm{HCl}$ solution required)
$V_f=\frac{M_3 V_3}{M_1}=\frac{0.2 \times 30}{0.066}=90 \mathrm{~mL}$
Hence, option (c) is the correct answer.