Question

An aqueous solution containing $HCl$ and $H_{2}S{O}_4$ required $10$ ml of a $1N$ $NaOH$ solution for neutralization. $20$ ml of the same acid mixture on being treated with an excess of $AgN{O}_3$ gives $0.1435$ g of $AgCl$. The normality of the $HCl$ and the normality of the $H_{2}S{O}_4$ are respectively:

• A. $0.40N$ and $0.05N$
• B. $0.05N$ and $0.35N$
• C. $0.50N$ and $0.25N$
• D. $0.40N$ and $0.5N$

Answer 1

Answer: (B)

$0.05N$ and $0.35N$

$0.1435$ g of $AgCl$ corresponds to $\frac{0.1435}{143.5}=0.001$ moles.

This is equal to the number of moles of $HCl$.
Normality of $HCl$ $=\frac{0.001}{0.20}=0.05N$.
$25$ ml of $0.05N$ $HCl$ corresponds to $25\times 0.05=1.25$ meq.
$10$ ml of $1NNaOH$ corresponds to $10$ meq. out of which $1.25$ meq. are neutralized by $HCl$.
Hence, meq. of sulphuric acid $=10-1.25=8.75$ meq..
Thus, the normality of sulfuric acid is $\frac{8.75}{25}=0.35N$.