25 ml of a solution of $N{a}_{2}C{O}_3$having a specific gravity of 1.25 gram ml required 32.9 ml of a solution of HCl containing 109.5 gram of the acid per litre for complete neutralization. Calculate the volume of ${\text{0.84 N H}}_{\text{2}}{\text{SO}}_{\text{4}}$that will be completely neutralized by 125 gram of $N{a}_{2}C{O}_3$solution.

Since $N{a}_{2}C{O}_3$ is completely neutralized by $HCl$

Also, given molarity of $HCl$= $\frac{109.5}{36.5}=3$

Therefore, milliequivalents of $N{a}_{2}C{O}_3$= milliequivalents of $HCl$

$$\begin{gathered} N\times 25 = 32.9\times 3 \\ N = 3.948 \end{gathered}$$

Now, a fresh $N{a}_{2}C{O}_3$solution reacts with $H_{2}S{O}_4$

Volume of 125 gm of  $N{a}_{2}C{O}_3$= $\frac{125}{1.25} = 100$ml

Again we can write,

Milliequivalents of  $N{a}_{2}C{O}_3$= Milliequivalents of $H_{2}S{O}_4$

$$\begin{gathered} 100\times 3.948 = 0.84\times V \\ V =470 ml \end{gathered}$$