Question

The ratio of the molar amounts of $H_{2}S$ needed to precipitate the metal ions from 20 ml each of 1M $Cd(N{O}_3{)}_2$ and $0.5MCuS{O}_4$ is:

• A. 1 : 1
• B. 2 : 1
• C. 1 : 2
• D. indefinite

Answer 1

Answer: (B)

2 : 1

$nf$ for $H_{2}S=2$ as $2H^{+}$ are produced

$nf$ for $Ca(N{O}_3{)}_2=2$ as $C{a}^2+2e^{-}⟶Cd$

$nf$ for $CuS{O}_4=2$ as $C_4^{2+}+2e^{-}⟶Cu$

$\Rightarrow n_1\times nf\left(H_{2}S\right)=M_1{V}_1\times nf\left(Cd\right(N{O}_3{)}_2)$

$\Rightarrow n_1\times 2=1\times 20\times 2$

$\Rightarrow n_1=20mmol$

$\Rightarrow n_2\times nf\left(H_{2}S\right)=M_2{V}_2\times nf\left(CuS{O}_4\right)$

$\Rightarrow n_2\times 2=0.5\times 20\times 2$

$\Rightarrow n_2=10mmol$

Thus, $\frac{n_1}{n_2}=\frac{20}{10}=2:1$ .