Solid `Ba(NO_(3))` is gradually dissolved in a `1.0xx10^(-4) M Na_(2)CO_(3)` solution. At what concentrations of `Ba^(2+)`, will a precipitate begin to form?
(`K_(SP)` for `BaCO_(3)=5.1xx10^(-9)`)

Options

A. `4.1xx10^(-5) M`
B. `5.1xx10^(-5) M`
C. `8.1xx10^(-8) M`
D. `8.1xx10^(-7) M`

Explanation

To find the concentration of \( \text{Ba}^{2+} \) at which a precipitate of \( \text{BaCO}_3 \) begins to form, we will use the solubility product constant \( K_{sp} \) for \( \text{BaCO}_3 \) and the concentration of carbonate ions from the sodium carbonate solution.

### Step-by-Step Solution:

1. **Identify the Dissociation of Sodium Carbonate:**
\[ \text{Na}_2\text{CO}_3 \rightarrow 2\text{Na}^+ + \text{CO}_3^{2-} \]
Given the concentration of \( \text{Na}_2\text{CO}_3 \) is \( 1.0 \times 10^{-4} \, M \), the concentration of \( \text{CO}_3^{2-} \) will also be \( 1.0 \times 10^{-4} \, M \).

2. **Write the Dissociation Equation for Barium Carbonate:**
\[ \text{BaCO}_3 (s) \rightleftharpoons \text{Ba}^{2+} + \text{CO}_3^{2-} \]

3. **Write the Expression for the Solubility Product \( K_{sp} \):**
\[ K_{sp} = [\text{Ba}^{2+}][\text{CO}_3^{2-}] \]
Given \( K_{sp} \) for \( \text{BaCO}_3 \) is \( 5.1 \times 10^{-9} \).

4. **Substitute the Known Concentration of \( \text{CO}_3^{2-} \):**
\[ 5.1 \times 10^{-9} = [\text{Ba}^{2+}](1.0 \times 10^{-4}) \]

5. **Solve for \( [\text{Ba}^{2+}] \):**
\[ [\text{Ba}^{2+}] = \frac{5.1 \times 10^{-9}}{1.0 \times 10^{-4}} = 5.1 \times 10^{-5} \, M \]

### Conclusion:
A precipitate of \( \text{BaCO}_3 \) will begin to form when the concentration of \( \text{Ba}^{2+} \) reaches \( 5.1 \times 10^{-5} \, M \).