A solution containing 0.1 M `Zn^(2+)` and 0.01 M `Cu^(2+)` is saturated with `H_2S`. The `S^(2-)` concentration is `8.1 xx 10^(-31)` M. Will ZnS or CuS precipitate ? `K_(sp)(ZnS)=3.0xx10^(-23)` and `K_(sp)(CuS)=8.0xx10^(-34)`.

Explanation

To determine whether ZnS or CuS will precipitate from a solution containing 0.1 M Zn²⁺ and 0.01 M Cu²⁺, saturated with H₂S, we need to calculate the ionic products (IP) for both salts and compare them with their respective solubility products (Ksp).

### Step 1: Write the expressions for Ksp
The solubility product (Ksp) expressions for ZnS and CuS are:
- For ZnS: \( K_{sp}(ZnS) = [Zn^{2+}][S^{2-}] \)
- For CuS: \( K_{sp}(CuS) = [Cu^{2+}][S^{2-}] \)

### Step 2: Calculate the ionic product for ZnS
Given:
- \( [Zn^{2+}] = 0.1 \, M \)
- \( [S^{2-}] = 8.1 \times 10^{-31} \, M \)

Now, calculate the ionic product (IP) for ZnS:
\[ IP_{ZnS} = [Zn^{2+}][S^{2-}] = (0.1)(8.1 \times 10^{-31}) = 8.1 \times 10^{-32} \]

### Step 3: Compare IP for ZnS with Ksp for ZnS
Given:
- \( K_{sp}(ZnS) = 3.0 \times 10^{-23} \)

Now compare:
\[ IP_{ZnS} = 8.1 \times 10^{-32} < K_{sp}(ZnS) = 3.0 \times 10^{-23} \]
Since the ionic product is less than the Ksp, ZnS will **not** precipitate.

### Step 4: Calculate the ionic product for CuS
Given:
- \( [Cu^{2+}] = 0.01 \, M \)
- \( [S^{2-}] = 8.1 \times 10^{-31} \, M \)

Now, calculate the ionic product (IP) for CuS:
\[ IP_{CuS} = [Cu^{2+}][S^{2-}] = (0.01)(8.1 \times 10^{-31}) = 8.1 \times 10^{-33} \]

### Step 5: Compare IP for CuS with Ksp for CuS
Given:
- \( K_{sp}(CuS) = 8.0 \times 10^{-34} \)

Now compare:
\[ IP_{CuS} = 8.1 \times 10^{-33} > K_{sp}(CuS) = 8.0 \times 10^{-34} \]
Since the ionic product is greater than the Ksp, CuS will **precipitate**.

### Conclusion
- **ZnS will not precipitate.**
- **CuS will precipitate.**