What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, `K_(sp)=6.3xx10^(-18))`.

Explanation

To find the maximum concentration of equimolar solutions of ferrous sulfate (FeSO₄) and sodium sulfide (Na₂S) such that there is no precipitation of iron sulfide (FeS) when mixed in equal volumes, we can follow these steps:

### Step-by-Step Solution

1. **Define the Molarity of Solutions**:
Let the molarity of both the ferrous sulfate (FeSO₄) and sodium sulfide (Na₂S) solutions be \( Y \) mol/L.

2. **Volume of Solutions**:
Assume we take equal volumes \( V \) of both solutions. Therefore, the total volume after mixing will be \( 2V \).

3. **Calculate Concentrations After Mixing**:
After mixing, the concentration of Fe²⁺ ions from FeSO₄ will be:
\[ [Fe^{2+}] = \frac{Y \cdot V}{2V} = \frac{Y}{2} \]
Similarly, the concentration of S²⁻ ions from Na₂S will also be:
\[ [S^{2-}] = \frac{Y \cdot V}{2V} = \frac{Y}{2} \]

4. **Use the Solubility Product (Ksp)**:
The solubility product constant \( K_{sp} \) for iron sulfide (FeS) is given as \( 6.3 \times 10^{-18} \). The precipitation of FeS occurs when the product of the concentrations of Fe²⁺ and S²⁻ ions exceeds \( K_{sp} \). Therefore, we set up the equation:
\[ K_{sp} = [Fe^{2+}][S^{2-}] \]
Substituting the concentrations we found:
\[ K_{sp} = \left(\frac{Y}{2}\right)\left(\frac{Y}{2}\right) = \frac{Y^2}{4} \]

5. **Set Up the Equation**:
To avoid precipitation, we need:
\[ \frac{Y^2}{4} \leq K_{sp} \]
Substituting the value of \( K_{sp} \):
\[ \frac{Y^2}{4} = 6.3 \times 10^{-18} \]

6. **Solve for Y**:
Rearranging the equation gives:
\[ Y^2 = 4 \times 6.3 \times 10^{-18} \]
\[ Y^2 = 25.2 \times 10^{-18} \]
\[ Y = \sqrt{25.2 \times 10^{-18}} = 5.02 \times 10^{-9} \text{ mol/L} \]

### Final Answer:
The maximum concentration of equimolar solutions of ferrous sulfate and sodium sulfide, so that when mixed in equal volumes there is no precipitation of iron sulfide, is:
\[ Y = 5.02 \times 10^{-9} \text{ mol/L} \]