Question

When $5$ mL of $8N$ nitric acid, $4.8$ mL of $5N$ hydrochloric acid and a certain volume of $17M$ sulphuric acid are mixed together and made upto $2$ L, $30$ mL of this acid mixture exactly neutralizes $42.9$ mL of sodium carbonate solution containing $1$ g of $N{a}_{2}C{O}_3.10H_{2}O$ in $100$ mL of water. Calculate the amount (in grams) of the sulphate ions in the solution.

• A. $3.25$
• B. $13$
• C. $8.5$
• D. $6.5$

Answer 1

The correct option is D $6.5$

For $H_{2}S{O}_4$, $n$ factor is $2$.

Let there be $n$ milliequivalents of $H_{2}S{O}_4$.

Milliequivalents of all acids in $2$ L $=(5\times 8)+(48.5\times 5)+n$ $=64+n$

Normality $=\frac{(6n+n)}{2000}N$

Normality of $N{a}_{2}C{O}_3.10H_{2}O$ is $\frac{1}{\frac{286}{2}}\times \frac{1}{100}\times 1000=0.07N$

$\therefore N_1{V}_1=N_2{V}_2$ $⟹\frac{64+n}{2000}\times 30=0.07\times 42.9$

$⟹n=136.2$ milliequivalents of $H_{2}S{O}_4$

Moles of $H_{2}S{O}_4=\frac{136.2}{2}\times {10}^{-3}$ mol
Mass of $S{O}_4^{2-}=\frac{136.2}{2}\times {10}^{-3}\times 96$ $=6.5376$ g