The volume of $0.1 \mathrm{M} \mathrm{HCl}$ required in $\mathrm{mL}$ to neutralise $20 \mathrm{~mL}$ of a…
The volume of $0.1 \mathrm{M} \mathrm{HCl}$ required in $\mathrm{mL}$ to neutralise $20 \mathrm{~mL}$ of a solution containing $0.106 \mathrm{~g}$ of $\mathrm{Na}_2 \mathrm{CO}_3$ is
10
5
20
40
Solution
Given,
Molarity $(\mathrm{HCl})\left[M_1\right]=0.1 \mathrm{M}$
Volume of $\mathrm{Na}_2 \mathrm{CO}_3$ used $=20 \mathrm{~mL}$
Mass of $\mathrm{Na}_2 \mathrm{CO}_3$ used $=0.106 \mathrm{~g}$
thus, molarity $\left(M_2\right)$ of $\mathrm{Na}_2 \mathrm{CO}_3=\frac{0.106}{106} \times \frac{1000}{20}$
$
\left(M_2\right) \mathrm{Na}_2 \mathrm{CO}_3=0.05 \mathrm{M}
$
$\because$ Equation for neutralisation is
$
M_1 V_1 Z_1=M_2 V_2 Z_2
$
Where $Z_1$ and $Z_2$ are number of moles of $\mathrm{H}^{+}$and ions, given by $\mathrm{HCl}$ and $\mathrm{Na}_2 \mathrm{CO}_3$ in water
$
\begin{aligned}
& Z_1=1(\text { for } \mathrm{HCl}) ; Z_2=2\left(\text { for } \mathrm{Na}_2 \mathrm{CO}_3\right) \\
& V_1=\frac{M_2 \times V_2 \times Z_2}{M_1 \times Z_1}=\frac{0.05 \times 20 \times 2}{0.1 \times 1}=20 \mathrm{~mL}
\end{aligned}
$
Hence, volume of $\mathrm{HCl}$ required $=20 \mathrm{~mL}$