Question

$5mL$ of $8NHN{O}_3,4.8mL$ of $5NHCl$ and a certain volume of $17M{H}_{2}S{O}_4$ are mixed together and made upto $2L$. $30mL$ of this acid mixture exactly neutralizes $42.9mL$ of ${Na}_{2}C{O}_3$ solution containing $1g$ of ${Na}_{2}C{O}_3$. $10H_{2}O$ in $100mL$ of water. The mass (in $g$) of sulphate ions present in the solution is (only integer part) :

Answer 1

The milliequivalents of acids in mixture can be calculated as:
Meq. of $HN{O}_3=5\times 8=40$
Meq. of $HCl$ $=4.8\times 5=24$

Meq. of $H_{2}S{O}_4=V\times 17\times 2=34V$ (Let $V$ mL of $H_{2}S{O}_4$ be used)
$\therefore$ Total meq. of acids in mixture $=40+24+34V=64+34V$
The mixture reacts with ${Na}_{2}C{O}_3$.
$\because N_{{Na}_{2}C{O}_3}=\frac{1}{\frac{286}{2}}\times \frac{1000}{100}$ as given.
Meq. of acid in $30$ mL solution $=$ Meq. of ${Na}_{2}C{O}_3$ used for it
$=42.9\times \frac{1\times 1000}{\frac{286}{2}\times 100}=3$
$\therefore$ Meq. of acid in $2$ litres solution$=\frac{3\times 2000}{30}=200$
Therefore, $64+34V=200$
$34V=200-64=136$
Further, Meq. of $H_{2}S{O}_4$= Meq. of $S{O}_4^{2-}$
$34V=136$
Meq. of $S{O}_4^{2-}=$ $136$
$\frac{w}{\frac{96}{2}}\times 1000=136$
Mass of $S{O}_4^{2-}=6.528$ g
So, the answer is $6$.