Question

30 mL of a $\frac{N}{10}$ $HCl$ is required to neutralize 50 mL of sodium carbonate solution. What volume of water (in mL) must be added to a 30 mL $N{a}_{2}C{O}_3$ solution so that the final solution has a concentration of $\frac{N}{50}$?

• A. 20 mL
• B. 30 mL
• C. 50 mL
• D. 60 mL

Answer 1

The correct option is D 60 mL

Since $HCl$ neutralizes $N{a}_{2}C{O}_3$ solution,
Milliequivalent of $HCl$ = Milliequivalent of $N{a}_{2}C{O}_3$
$(N_1{V}_1{)}_{HCl}$ = $(N_2{V}_2{)}_{N{a}_{2}C{O}_3}$
$N_1$,$V_1$ are normality and volume of $HCl$solution respectively. $N_2$,$V_2$ are normality and volume of $N{a}_{2}C{O}_3$ solution respectively.
$\frac{1}{10}\times 30$ = $N_2\times 50$
$N_2$ = $\frac{3}{50}N$
Now to this solution of $N{a}_{2}C{O}_3$, 30 mL of water is added. The final concentration becomes $\frac{N}{50}$.
Again applying $N\times V$ = constant
$\frac{3}{50}=\frac{1}{50}(30+V)$
V= 60 mL