Q7 / 160 · AP EAMCET 2018 (23 Apr Shift 2) · Chemistry

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

  • A
    10
  • B
    5
  • Correct answer
    C
    20
  • D
    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 $=20 \mathrm{~mL}$

Hence, the correct answer is option: 'C'.

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