Question

$0.1$ milli mole of $Cd{SO}_4$ are present in $10mL$ acid solution of $0.08N$ $HCl$. Now $H_{2}S$ is passed to precipitate all the ${Cd}^{2+}$ ions. The $pH$ of solution after filtering off precipitate, boiling of $H_{2}S$ and making the solution $100mL$ by adding $H_{2}O$ will be :

• A. 2
• B. 4
• C. 8
• D. None of these

Answer 1

Answer: (A)

2

$CdS{O}_4+HCl+H_{2}S\to CdS+H_{2}S{O}_4$

Initially 0.1 0.8 0 0

Finally 0 0.8 0.1 0.1

$\therefore$ Milli mole of $H^{+}$ left in solutio from $HCl$ from $H_2{SO}_4$ respectively $=0.8+0.1\times 2=1.0$
Total volume$=100mL$

$\therefore$ $\left[H^{+}\right]=\left(1/100\right)={10}^{-2}M$
$\therefore$ $pH=2$