solubility product of radium sulphate is `4xx10^(-9)`. 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 find the solubility of \( \text{Ra}^{2+} \) in a \( 0.10 \, \text{M} \) \( \text{NaSO}_4 \) solution, we will use the solubility product constant (\( K_{sp} \)) of radium sulfate (\( \text{RaSO}_4 \)), which is given as \( 4 \times 10^{-9} \).

### Step-by-Step Solution:

1. **Write the Dissociation Equation for Radium Sulfate:**
\[ \text{RaSO}_4 (s) \rightleftharpoons \text{Ra}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]

2. **Express the Solubility Product Constant:**
The solubility product (\( K_{sp} \)) is given by the equation:
\[ K_{sp} = [\text{Ra}^{2+}][\text{SO}_4^{2-}] \]
where \( [\text{Ra}^{2+}] \) is the molarity of radium ions and \( [\text{SO}_4^{2-}] \) is the molarity of sulfate ions.

3. **Determine the Concentration of Sulfate Ions:**
Since \( \text{NaSO}_4 \) dissociates completely in solution:
\[ \text{NaSO}_4 \rightarrow 2 \text{Na}^+ + \text{SO}_4^{2-} \]
The concentration of \( \text{SO}_4^{2-} \) from \( 0.10 \, \text{M} \) \( \text{NaSO}_4 \) is:
\[ [\text{SO}_4^{2-}] = 0.10 \, \text{M} \]

4. **Substitute into the \( K_{sp} \) Expression:**
Now, substituting the known values into the \( K_{sp} \) expression:
\[ 4 \times 10^{-9} = [\text{Ra}^{2+}][0.10] \]

5. **Solve for the Concentration of Radium Ions:**
Rearranging the equation to find \( [\text{Ra}^{2+}] \):
\[ [\text{Ra}^{2+}] = \frac{4 \times 10^{-9}}{0.10} \]
\[ [\text{Ra}^{2+}] = 4 \times 10^{-8} \, \text{M} \]

6. **Final Result:**
The solubility of \( \text{Ra}^{2+} \) in \( 0.10 \, \text{M} \) \( \text{NaSO}_4 \) is:
\[ [\text{Ra}^{2+}] = 4 \times 10^{-8} \, \text{M} \]