`H_(2)S` is passed into one `dm^(3)` of a solution containing 0.1 mole of `Zn^(2+)` and 0.1 mole of the `Cu^(2+)` till the sulphide ion concentration reaches `8.1 xx 10^(-10)` moles. Which one of the following statements is true? `[K_(sp)` of ZnS and CuS are `3 xx 10^(-22)` and `8 xx 10^(-36)` respectively]

Options

A. only ZnS precipitates
B. Both Cus and ZnS precipitate
C. Only CuS precipitates
D. No precipitation occurs

Answer

To solve the problem, we need to analyze the precipitation of zinc sulfide (ZnS) and copper sulfide (CuS) when hydrogen sulfide (H₂S) is added to a solution containing 0.1 moles of Zn²⁺ and 0.1 moles of Cu²⁺ until the sulfide ion concentration reaches \(8.1 \times 10^{-10}\) moles.

### Step-by-Step Solution:

1. **Identify the Relevant Equations:**
- The solubility product constant (Ksp) expressions for ZnS and CuS are:
\[ K_{sp} \text{ for ZnS} = [Zn^{2+}][S^{2-}] \]
\[ K_{sp} \text{ for CuS} = [Cu^{2+}][S^{2-}] \]

2. **Determine the Concentrations:**
- Since the volume of the solution is 1 dm³, the concentrations of Zn²⁺ and Cu²⁺ are both 0.1 M.
- The concentration of sulfide ions (S²⁻) is given as \(8.1 \times 10^{-10}\) M.

3. **Calculate the Ionic Product for CuS:**
- The ionic product (Q) for CuS can be calculated as:
\[ Q_{CuS} = [Cu^{2+}][S^{2-}] = (0.1)(8.1 \times 10^{-10}) = 8.1 \times 10^{-11} \]
- Compare this with the Ksp of CuS:
\[ K_{sp} \text{ for CuS} = 8 \times 10^{-36} \]
- Since \(Q_{CuS} > K_{sp}\), CuS will precipitate.

4. **Calculate the Ionic Product for ZnS:**
- The ionic product (Q) for ZnS can be calculated similarly:
\[ Q_{ZnS} = [Zn^{2+}][S^{2-}] = (0.1)(8.1 \times 10^{-10}) = 8.1 \times 10^{-11} \]
- Compare this with the Ksp of ZnS:
\[ K_{sp} \text{ for ZnS} = 3 \times 10^{-22} \]
- Since \(Q_{ZnS} > K_{sp}\), ZnS will also precipitate.

5. **Conclusion:**
- Both CuS and ZnS will precipitate when H₂S is added to the solution.

### Final Answer:
Both CuS and ZnS precipitate.