A solution of `Na_(2)CO_(3)` is added drop by drop to litre of a solution containing `10^(-4)` mole of `Ba^(2+)` and `10^(-5)` mole of Ag, if `K_(SP)` for `BaCO_(3)` is `8.1xx10^(-9)` and `K_(SP)` for `Ag_(2)CO_(3)is 6.9xx10^(-12)`, then which is not true ?

Options

A. No precIpitate of `BaCO_(3)` will appear until `[CO_(3)^(2-)]` reaches `8.1xx10^(-5)` mol per litre
B. A precipitate of `Ag_(2)CO_(3)` will appear when `[CO_(3)^(2-)]` reaches `6.9xx10^(-5)` mol `"litre"^(-1)`
C. No precipitate of `Ag_(2)CO_(3)` will apppear when `[CO_(3)^(2-)]` reaches `6.9x10^(-6)`mole per litre
D. `BaCO_(3)` will be precipitated first

Explanation

To solve the problem, we need to determine the conditions under which barium carbonate (BaCO₃) and silver carbonate (Ag₂CO₃) will precipitate when sodium carbonate (Na₂CO₃) is added to a solution containing barium ions (Ba²⁺) and silver ions (Ag⁺).

### Step-by-Step Solution:

1. **Identify the given data:**
- Concentration of Ba²⁺ = \(10^{-4}\) moles/L
- Concentration of Ag⁺ = \(10^{-5}\) moles/L
- \(K_{sp}\) for BaCO₃ = \(8.1 \times 10^{-9}\)
- \(K_{sp}\) for Ag₂CO₃ = \(6.9 \times 10^{-12}\)

2. **Calculate the concentration of carbonate ions required for BaCO₃ precipitation:**
- The solubility product expression for BaCO₃ is:
\[ K_{sp} = [Ba^{2+}][CO_3^{2-}] \]
- Rearranging gives:
\[ [CO_3^{2-}] = \frac{K_{sp}}{[Ba^{2+}]} \]
- Substituting the values:
\[ [CO_3^{2-}] = \frac{8.1 \times 10^{-9}}{10^{-4}} = 8.1 \times 10^{-5} \text{ moles/L} \]

3. **Calculate the concentration of carbonate ions required for Ag₂CO₃ precipitation:**
- The solubility product expression for Ag₂CO₃ is:
\[ K_{sp} = [Ag^{+}]^2[CO_3^{2-}] \]
- Rearranging gives:
\[ [CO_3^{2-}] = \frac{K_{sp}}{[Ag^{+}]^2} \]
- Substituting the values:
\[ [CO_3^{2-}] = \frac{6.9 \times 10^{-12}}{(10^{-5})^2} = \frac{6.9 \times 10^{-12}}{10^{-10}} = 6.9 \times 10^{-2} \text{ moles/L} \]

4. **Determine which precipitate forms first:**
- The concentration of carbonate ions required for BaCO₃ precipitation is \(8.1 \times 10^{-5}\) moles/L.
- The concentration of carbonate ions required for Ag₂CO₃ precipitation is \(6.9 \times 10^{-2}\) moles/L.
- Since \(8.1 \times 10^{-5} < 6.9 \times 10^{-2}\), BaCO₃ will precipitate first.

5. **Analyze the statements:**
- **Statement A:** No precipitate of BaCO₃ will appear until the carbonate ion concentration reaches \(8.1 \times 10^{-5}\) moles/L. (True)
- **Statement B:** The precipitate of Ag₂CO₃ will appear when the carbonate ion concentration reaches \(6.9 \times 10^{-5}\) moles/L. (False, it should be \(6.9 \times 10^{-2}\) moles/L)
- **Statement C:** No precipitate of Ag₂CO₃ will appear when the carbonate ion concentration reaches \(6.9 \times 10^{-5}\) moles/L. (True)
- **Statement D:** BaCO₃ will precipitate first. (True)

### Conclusion:
The statement that is **not true** is **Statement B**.