`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 determine whether ZnS and/or CuS will precipitate when H₂S is added to a solution containing 0.1 moles of Zn²⁺ and 0.1 moles of Cu²⁺ until the concentration of sulfide ions (S²⁻) reaches 8.1 × 10⁻¹⁰ moles. We will compare the ionic product (IP) of both salts with their respective solubility product constants (Ksp).

### Step-by-step Solution:

1. **Identify the Dissociation Reactions:**
- For CuS:
\[ \text{CuS (s)} \rightleftharpoons \text{Cu}^{2+} + \text{S}^{2-} \]
- For ZnS:
\[ \text{ZnS (s)} \rightleftharpoons \text{Zn}^{2+} + \text{S}^{2-} \]

2. **Write the Expression for Ksp:**
- For CuS:
\[ K_{sp} = [\text{Cu}^{2+}][\text{S}^{2-}] \]
- For ZnS:
\[ K_{sp} = [\text{Zn}^{2+}][\text{S}^{2-}] \]

3. **Calculate the Ionic Product (IP) for CuS:**
- Given:
- \([\text{Cu}^{2+}] = 0.1 \, \text{mol/dm}^3\)
- \([\text{S}^{2-}] = 8.1 \times 10^{-10} \, \text{mol/dm}^3\)
- Calculate IP for CuS:
\[ IP_{CuS} = [\text{Cu}^{2+}][\text{S}^{2-}] = (0.1)(8.1 \times 10^{-10}) = 8.1 \times 10^{-11} \]

4. **Compare IP with Ksp for CuS:**
- Given \( K_{sp} \) for CuS = \( 8 \times 10^{-36} \)
- Since \( IP_{CuS} = 8.1 \times 10^{-11} \) is greater than \( K_{sp} = 8 \times 10^{-36} \), CuS will precipitate.

5. **Calculate the Ionic Product (IP) for ZnS:**
- Given:
- \([\text{Zn}^{2+}] = 0.1 \, \text{mol/dm}^3\)
- \([\text{S}^{2-}] = 8.1 \times 10^{-10} \, \text{mol/dm}^3\)
- Calculate IP for ZnS:
\[ IP_{ZnS} = [\text{Zn}^{2+}][\text{S}^{2-}] = (0.1)(8.1 \times 10^{-10}) = 8.1 \times 10^{-11} \]

6. **Compare IP with Ksp for ZnS:**
- Given \( K_{sp} \) for ZnS = \( 3 \times 10^{-22} \)
- Since \( IP_{ZnS} = 8.1 \times 10^{-11} \) is greater than \( K_{sp} = 3 \times 10^{-22} \), ZnS will also precipitate.

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

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