Given that the solubility product of radium sulphate `(RaSO_(4))` is `4xx10^(-11)`. Calculate the solubility in (a) pure water (b) 0.10 M `Na_(2)SO_(4)`.

Explanation

To solve the problem of calculating the solubility of radium sulfate (RaSO₄) in pure water and in a sodium sulfate (Na₂SO₄) solution, we will follow these steps:

### Given:
- Solubility product (Ksp) of RaSO₄ = 4 x 10^(-11)

### (a) Solubility in Pure Water

1. **Write the dissociation equation**:
\[ \text{RaSO}_4 (s) \rightleftharpoons \text{Ra}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]

2. **Define the solubility**:
Let the solubility of RaSO₄ in pure water be \( s \) mol/L. Therefore, at equilibrium:
- \([\text{Ra}^{2+}] = s\)
- \([\text{SO}_4^{2-}] = s\)

3. **Write the expression for Ksp**:
\[ Ksp = [\text{Ra}^{2+}][\text{SO}_4^{2-}] = s \cdot s = s^2 \]

4. **Substitute the value of Ksp**:
\[ 4 \times 10^{-11} = s^2 \]

5. **Solve for \( s \)**:
\[ s = \sqrt{4 \times 10^{-11}} = 2 \times 10^{-6} \text{ mol/L} \]

### (b) Solubility in 0.10 M Na₂SO₄

1. **Consider the common ion effect**:
The presence of Na₂SO₄ provides a common ion, \([\text{SO}_4^{2-}] = 0.10 \text{ M}\).

2. **Set up the equilibrium expression**:
Let the solubility of RaSO₄ in this solution be \( s' \). At equilibrium:
- \([\text{Ra}^{2+}] = s'\)
- \([\text{SO}_4^{2-}] = 0.10 + s' \approx 0.10\) (since \( s' \) will be very small compared to 0.10)

3. **Write the Ksp expression**:
\[ Ksp = [\text{Ra}^{2+}][\text{SO}_4^{2-}] = s' \cdot 0.10 \]

4. **Substitute the value of Ksp**:
\[ 4 \times 10^{-11} = s' \cdot 0.10 \]

5. **Solve for \( s' \)**:
\[ s' = \frac{4 \times 10^{-11}}{0.10} = 4 \times 10^{-10} \text{ mol/L} \]

### Final Answers:
- (a) Solubility in pure water: \( 2 \times 10^{-6} \text{ mol/L} \)
- (b) Solubility in 0.10 M Na₂SO₄: \( 4 \times 10^{-10} \text{ mol/L} \)