Question

The ratio of the molar amounts of $\mathrm{H}_{2}\mathrm{S}$ needed to precipitate the metal ions from $20 \mathrm{mL}$ each of $1 ~\mathrm{M}~ \mathrm{Cd}\left(\mathrm{NO}_{3}\right){2}$ and $0.5 ~\mathrm{M ~CuSO}_{4}$ is

(A) $1:1$

(B) $2:1$

(C) $1:2$

(D) Indefinite

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Answer

To solve the problem, we need to determine how many moles of $\mathrm{H}_2\mathrm{S}$ are required to precipitate metal ions from solutions of $\mathrm{Cd(NO}_3\mathrm{)_2}$ and $\mathrm{CuSO}_4$.

  1. Chemical reaction for precipitation of Cadmium as Cadmium Sulfide: $$ \mathrm{Cd}^{2+} + \mathrm{S}^{2-} \rightarrow \mathrm{CdS} $$ Each mole of $\mathrm{Cd}^{2+}$ requires 1 mole of $\mathrm{S}^{2-}$ to form $\mathrm{CdS}$.

  2. Calculating moles of $\mathrm{Cd}^{2+}$ in the given solution:

    • Molarity of $\mathrm{Cd(NO}_3\mathrm{)_2}$ is $1 \mathrm{M}$.

    • Volume of solution = $20 \mathrm{mL}$ = $0.020 \mathrm{L}$.

    • Moles of $\mathrm{Cd}^{2+}$ = Molarity × Volume = $1 \times 0.020 = 0.020 \mathrm{moles}$.

  3. Chemical reaction for precipitation of Copper as Copper Sulfide: $$ \mathrm{Cu}^{2+} + \mathrm{S}^{2-} \rightarrow \mathrm{CuS} $$ Each mole of $\mathrm{Cu}^{2+}$ requires 1 mole of $\mathrm{S}^{2-}$ to form $\mathrm{CuS}$.

  4. Calculating moles of $\mathrm{Cu}^{2+}$ in the given solution:

    • Molarity of $\mathrm{CuSO}_4$ is $0.5 \mathrm{M}$.

    • Volume of solution remains $20 \mathrm{mL}$ = $0.020 \mathrm{L}$.

    • Moles of $\mathrm{Cu}^{2+}$ = Molarity × Volume = $0.5 \times 0.020 = 0.010 \mathrm{moles}$.

  5. Ratio of $\mathrm{H}_2\mathrm{S}$ required:

    • Moles of $\mathrm{H}_2\mathrm{S}$ needed for $\mathrm{CdS}$: $0.020 \mathrm{moles}$ (as each mole of $\mathrm{Cd}^{2+}$ requires an equivalent mole of $\mathrm{S}^{2-}$).

    • Moles of $\mathrm{H}_2\mathrm{S}$ needed for $\mathrm{CuS}$: $0.010 \mathrm{moles}$ (as each mole of $\mathrm{Cu}^{2+}$ requires an equivalent mole of $\mathrm{S}^{2-}$).

    Thus, the ratio is given by: $$ \text{Ratio} = \frac{0.020}{0.010} = 2:1 $$

Answer: The correct option is (B) $2:1$. This means that twice as much $\mathrm{H}_2\mathrm{S}$ is required for precipitating $\mathrm{Cd}^{2+}$ compared to $\mathrm{Cu}^{2+}$.


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