A solution is 0.10 M `Ba(NO_(3))_(2)` and 0.10 M `Sr(NO_(3))_(2.)` If solid `Na_(2)CrO_(4)` is added to the solution, what is `[Ba^(2+)]` when `SrCrO_(4)` begins to precipitate?
`[K_(sp)(BaCrO_(4))=1.2xx10^(-10),K_(sp)(SrCrO_(4))=3.5xx10^(-5)]`

Options

A. `7.4xx10^(-7)`
B. `2.0xx10^(-7)`
C. `6.1xx10^(-7)`
D. `3.4xx10^(-7)`

Answer

To solve the problem, we need to determine the concentration of barium ions \([Ba^{2+}]\) when strontium chromate \((SrCrO_4)\) begins to precipitate. We are given the solubility product constants \((K_{sp})\) for both barium chromate \((BaCrO_4)\) and strontium chromate \((SrCrO_4)\).

### Step-by-Step Solution:

1. **Identify the Dissociation of Compounds:**
- Barium nitrate \((Ba(NO_3)_2)\) dissociates into \(Ba^{2+}\) and \(2NO_3^{-}\).
- Strontium nitrate \((Sr(NO_3)_2)\) dissociates into \(Sr^{2+}\) and \(2NO_3^{-}\).
- Sodium chromate \((Na_2CrO_4)\) dissociates into \(2Na^{+}\) and \(CrO_4^{2-}\).

Given concentrations:
\[ [Ba^{2+}] = 0.10 \, M, \quad [Sr^{2+}] = 0.10 \, M \]

2. **Determine the Precipitation Condition for \(SrCrO_4\):**
The precipitation of \(SrCrO_4\) occurs when the product of the concentrations of \(Sr^{2+}\) and \(CrO_4^{2-}\) exceeds the \(K_{sp}\):
\[ K_{sp}(SrCrO_4) = [Sr^{2+}][CrO_4^{2-}] = 3.5 \times 10^{-5} \]

Since \([Sr^{2+}] = 0.10 \, M\), we can find \([CrO_4^{2-}]\) at the point of precipitation:
\[ [CrO_4^{2-}] = \frac{K_{sp}}{[Sr^{2+}]} = \frac{3.5 \times 10^{-5}}{0.10} = 3.5 \times 10^{-4} \, M \]

3. **Determine the Concentration of \(Ba^{2+}\) at the Precipitation Point:**
Now, we will use the \(K_{sp}\) for \(BaCrO_4\) to find \([Ba^{2+}]\) when \([CrO_4^{2-}] = 3.5 \times 10^{-4} \, M\):
\[ K_{sp}(BaCrO_4) = [Ba^{2+}][CrO_4^{2-}] = 1.2 \times 10^{-10} \]

Rearranging gives:
\[ [Ba^{2+}] = \frac{K_{sp}}{[CrO_4^{2-}]} = \frac{1.2 \times 10^{-10}}{3.5 \times 10^{-4}} \]

Calculating this gives:
\[ [Ba^{2+}] = \frac{1.2}{3.5} \times 10^{-10 + 4} = \frac{1.2}{3.5} \times 10^{-6} \approx 0.0343 \times 10^{-6} = 3.43 \times 10^{-7} \, M \]

### Final Result:
\[ [Ba^{2+}] \approx 3.4 \times 10^{-7} \, M \]