The `K_(sp)` value of `CaCO_(3)` and `CaC_(2)O_(4)` in water are `4.7 xx 10^(-9)` and `1.3 xx 10^(-9)`, respectively, at `25^(@)C`. If a mixture of two is washed with `H_(2)O`, what is `Ca^(2+)` ion concentration in water?

Options

A. `7.7 xx 10^(-5)`
B. `5.831 xx 10^(-5)`
C. `6.856 xx 10^(-5)`
D. `3.606 xx 10^(-5)`

Explanation

To solve the problem, we need to determine the concentration of \( \text{Ca}^{2+} \) ions in water after washing a mixture of \( \text{CaCO}_3 \) and \( \text{CaC}_2\text{O}_4 \) with water, given their solubility product constants (\( K_{sp} \)).

### Step-by-Step Solution:

1. **Identify the Dissociation Reactions:**
- For \( \text{CaCO}_3 \):
\[ \text{CaCO}_3 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + \text{CO}_3^{2-} (aq) \]
- For \( \text{CaC}_2\text{O}_4 \):
\[ \text{CaC}_2\text{O}_4 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + \text{C}_2\text{O}_4^{2-} (aq) \]

2. **Define Solubility:**
- Let the solubility of \( \text{CaCO}_3 \) be \( x \) (mol/L).
- Let the solubility of \( \text{CaC}_2\text{O}_4 \) be \( y \) (mol/L).

3. **Write the Expressions for \( K_{sp} \):**
- For \( \text{CaCO}_3 \):
\[ K_{sp} = [\text{Ca}^{2+}][\text{CO}_3^{2-}] = x \cdot x = x^2 \]
Given \( K_{sp} = 4.7 \times 10^{-9} \):
\[ x^2 = 4.7 \times 10^{-9} \implies x = \sqrt{4.7 \times 10^{-9}} \approx 6.86 \times 10^{-5} \text{ mol/L} \]

- For \( \text{CaC}_2\text{O}_4 \):
\[ K_{sp} = [\text{Ca}^{2+}][\text{C}_2\text{O}_4^{2-}] = y \cdot y = y^2 \]
Given \( K_{sp} = 1.3 \times 10^{-9} \):
\[ y^2 = 1.3 \times 10^{-9} \implies y = \sqrt{1.3 \times 10^{-9}} \approx 3.61 \times 10^{-5} \text{ mol/L} \]

4. **Consider the Common Ion Effect:**
- The \( \text{Ca}^{2+} \) concentration from both salts will affect each other due to the common ion effect.
- The total concentration of \( \text{Ca}^{2+} \) ions in solution will be \( [\text{Ca}^{2+}] = x + y \).

5. **Set Up the Equations:**
- For \( \text{CaCO}_3 \):
\[ K_{sp} = (x + y) \cdot x = 4.7 \times 10^{-9} \]
- For \( \text{CaC}_2\text{O}_4 \):
\[ K_{sp} = (y + x) \cdot y = 1.3 \times 10^{-9} \]

6. **Solve the Equations:**
- From the first equation:
\[ 4.7 \times 10^{-9} = (x + y)x \]
- From the second equation:
\[ 1.3 \times 10^{-9} = (y + x)y \]

7. **Substitute and Solve:**
- Divide the first equation by the second:
\[ \frac{4.7 \times 10^{-9}}{1.3 \times 10^{-9}} = \frac{(x + y)x}{(y + x)y} \]
- This simplifies to:
\[ \frac{4.7}{1.3} = \frac{x}{y} \]
- From this, we can express \( x \) in terms of \( y \):
\[ x = \frac{4.7}{1.3}y \]

8. **Substitute Back to Find \( y \):**
- Substitute \( x \) back into one of the \( K_{sp} \) equations to find \( y \).

9. **Calculate \( \text{Ca}^{2+} \) Concentration:**
- Finally, calculate the total \( \text{Ca}^{2+} \) concentration:
\[ [\text{Ca}^{2+}] = x + y \]

### Final Answer:
After performing the calculations, the concentration of \( \text{Ca}^{2+} \) ions in water is approximately \( 7.7 \times 10^{-5} \) mol/L.