solubility product of radium sulphate is `4xx10^(-1)`. What will be the solubility of` Ra^(2+)` in` 0.10 M NaSO_(4)` ?

Options

A. `4xx10^(-10)` M
B. `2xx10^(-5)M`
C. `4xx10^(-5)M`
D. `2xx10^(-10)M`

Explanation

To solve the problem, we need to determine the solubility of Ra²⁺ in a 0.10 M Na₂SO₄ solution, given that the solubility product (Ksp) of radium sulfate (RaSO₄) is 4 x 10^(-1).

### Step-by-Step Solution:

1. **Write the Dissociation Equation**:
The dissociation of radium sulfate (RaSO₄) can be represented as:
\[ \text{RaSO}_4 (s) \rightleftharpoons \text{Ra}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]

2. **Define the Solubility Product (Ksp)**:
The solubility product expression for RaSO₄ is given by:
\[ K_{sp} = [\text{Ra}^{2+}][\text{SO}_4^{2-}] \]
Given that \( K_{sp} = 4 \times 10^{-1} \).

3. **Consider the Contribution of SO₄²⁻ from Na₂SO₄**:
In the 0.10 M Na₂SO₄ solution, the concentration of sulfate ions \([\text{SO}_4^{2-}]\) is already 0.10 M due to the dissociation of Na₂SO₄:
\[ \text{Na}_2\text{SO}_4 (s) \rightleftharpoons 2 \text{Na}^+ (aq) + \text{SO}_4^{2-} (aq) \]
Therefore, \([\text{SO}_4^{2-}] = 0.10 \, M\).

4. **Set Up the Equation for Ksp**:
Let the solubility of RaSO₄ in this solution be \(s\). Thus, the concentration of \([\text{Ra}^{2+}]\) will be \(s\) and \([\text{SO}_4^{2-}]\) will be \(0.10 + s\). However, since \(s\) is expected to be very small compared to 0.10 M, we can approximate:
\[ [\text{SO}_4^{2-}] \approx 0.10 \, M \]
Therefore, we can write:
\[ K_{sp} = s \times 0.10 \]

5. **Substitute Ksp Value**:
Now substituting the known values into the Ksp expression:
\[ 4 \times 10^{-1} = s \times 0.10 \]

6. **Solve for s**:
Rearranging the equation to solve for \(s\):
\[ s = \frac{4 \times 10^{-1}}{0.10} = 4 \times 10^{-1} \times 10 = 4 \]

7. **Conclusion**:
The solubility of \(\text{Ra}^{2+}\) in 0.10 M Na₂SO₄ is:
\[ s = 4 \, M \]