Question

$25$ ml of a solution containing $HCl$ and $H_{2}S{O}_4$ required $10$ ml of a $1NNaOH$ solution for neutralization. $20$ ml of the same acid mixture on being treated with an excess of $AgN{O}_3$ gives $0.1435g$ 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

The correct option is B $0.05N$ and $0.35N$

Let normality of HCl is $N_1$ and $H_{2}S{O}_4$ is $N_2$

$\therefore$ Mili eq. of HCl + Mili eq. of $H_{2}S{O}_4$ = Mili eq. of NaOH

$25\times N_1+25\times N_2=10\times 1$

$N_1+N_2=0.4$ .....(i)

BY POAC

Moles of $C{l}^{-}$ = moles of AgCl

$\frac{20\times N_1}{1000}=\frac{0.1435}{143.5}={10}^{-3}$

$N_1$ = 0.05N

Putting value in equation (i) we get

$N_2$ = 0.35N

Hence, option B is correct.