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. `4.1 xx 10^(-5)M`
B. `5.1 xx 10^(-5)M`
C. `8.1 xx 10^(-8)M`
D. `8.1 xx 10^(-7)M`

Answer

To solve the problem, we need to determine the concentration of `Ba^(2+)` ions at which a precipitate of `BaCO3` will start to form when solid `Ba(NO3)2` is dissolved in a `1.0 x 10^(-4) M Na2CO3` solution. We will use the solubility product constant (`Ksp`) of `BaCO3` to find this concentration.

### Step-by-Step Solution:

1. **Understand the Dissociation of Barium Carbonate**:
The dissociation of `BaCO3` in water can be represented as:
\[ BaCO3 (s) \rightleftharpoons Ba^{2+} (aq) + CO3^{2-} (aq) \]

2. **Write the Expression for Ksp**:
The solubility product constant (`Ksp`) for `BaCO3` is given by:
\[ Ksp = [Ba^{2+}][CO3^{2-}] \]
where `[Ba^{2+}]` is the concentration of barium ions and `[CO3^{2-}]` is the concentration of carbonate ions.

3. **Identify the Concentration of Carbonate Ions**:
The concentration of `Na2CO3` is given as `1.0 x 10^(-4) M`. Since `Na2CO3` dissociates completely in solution:
\[ Na2CO3 \rightarrow 2Na^+ + CO3^{2-} \]
The concentration of `CO3^{2-}` ions will also be `1.0 x 10^(-4) M`.

4. **Set Up the Ksp Equation**:
Substituting the known concentration of `CO3^{2-}` into the Ksp expression:
\[ Ksp = [Ba^{2+}][CO3^{2-}] = [Ba^{2+}](1.0 \times 10^{-4}) \]
Given that \( Ksp \) for `BaCO3` is \( 5.1 \times 10^{-9} \), we can write:
\[ 5.1 \times 10^{-9} = [Ba^{2+}](1.0 \times 10^{-4}) \]

5. **Solve for [Ba^{2+}]**:
Rearranging the equation to find the concentration of `Ba^{2+}`:
\[ [Ba^{2+}] = \frac{5.1 \times 10^{-9}}{1.0 \times 10^{-4}} = 5.1 \times 10^{-5} M \]

6. **Conclusion**:
The concentration of `Ba^{2+}` at which a precipitate of `BaCO3` will begin to form is \( 5.1 \times 10^{-5} M \).

### Final Answer:
The concentration of `Ba^{2+}` at which a precipitate begins to form is \( 5.1 \times 10^{-5} M \).

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