Question

A sample of $0.636\text{g}$ of pure sodium carbonate is titrated with $HCl$ solution. A volume of $60\text{mL}$ is required to reach the end point. Calculate the molarity of the acid.

• A. $0.1\text{M}$
• B. $0.2\text{M}$
• C. $0.4\text{M}$
• D. $0.8\text{M}$

Answer 1

The correct option is B $0.2\text{M}$

n-factor of $N{a}_{2}C{O}_3=2$
n-factor for $HCl=1$
At the equivalence point,
Equivalents of $N{a}_{2}C{O}_3$ = equivalents of $HCl$
$\text{n-factor}\times \frac{\text{weight of}N{a}_{2}C{O}_3}{\text{molar mass}}=\text{molarity}\times \text{volume in L}$
$2\times \frac{0.636}{106}=M\times 0.06$
On solving, the molarity of $HCl$ $=0.2\text{M}$