An aqueous solution contains an unknown concentration of `Ba^(2+)`. When 50 mL of a 1 M solution of `Na_(2)SO_(4)` is added, `BaSO_(4 )`just begins to precipitate. The final volume is 500 mL. The solubility product of `BaSO_(4)` is `1xx10^(–10)`. What is the original concentration of `Ba^(2+)`?

Options

A. `1.1xx10^(-9)M`
B. `1.1xx10^(-10)M`
C. `5.1xx10^(-9)M`
D. `2xx10^(-9)M`

Explanation

To find the original concentration of \( \text{Ba}^{2+} \) in the solution, we can follow these steps:

### Step 1: Understand the Reaction and Given Data
We know that \( \text{BaSO}_4 \) precipitates when \( \text{Ba}^{2+} \) ions react with \( \text{SO}_4^{2-} \) ions. The solubility product (\( K_{sp} \)) of \( \text{BaSO}_4 \) is given as \( 1 \times 10^{-10} \).

### Step 2: Calculate the Concentration of \( \text{SO}_4^{2-} \)
When 50 mL of a 1 M solution of \( \text{Na}_2\text{SO}_4 \) is added to the solution, the total volume becomes 500 mL. The concentration of \( \text{SO}_4^{2-} \) can be calculated as follows:

\[ \text{Concentration of } \text{SO}_4^{2-} = \frac{\text{moles of } \text{SO}_4^{2-}}{\text{total volume in L}} = \frac{(1 \, \text{mol/L}) \times (0.050 \, \text{L})}{0.500 \, \text{L}} = 0.1 \, \text{M} \]

### Step 3: Set Up the Expression for \( K_{sp} \)
The \( K_{sp} \) expression for \( \text{BaSO}_4 \) is given by:

\[ K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] \]

Substituting the known values:

\[ 1 \times 10^{-10} = [\text{Ba}^{2+}][0.1] \]

### Step 4: Solve for \( [\text{Ba}^{2+}] \)
Rearranging the equation to solve for \( [\text{Ba}^{2+}] \):

\[ [\text{Ba}^{2+}] = \frac{1 \times 10^{-10}}{0.1} = 1 \times 10^{-9} \, \text{M} \]

### Step 5: Calculate the Initial Concentration of \( \text{Ba}^{2+} \)
Using the dilution formula \( M_1V_1 = M_2V_2 \):

Let \( C \) be the original concentration of \( \text{Ba}^{2+} \) and \( V_1 \) be the initial volume of the solution before adding \( \text{Na}_2\text{SO}_4 \) (which is 450 mL):

\[ C \times 0.450 \, \text{L} = (1 \times 10^{-9} \, \text{M}) \times (0.500 \, \text{L}) \]

Solving for \( C \):

\[ C = \frac{(1 \times 10^{-9}) \times 0.500}{0.450} = \frac{5 \times 10^{-10}}{4.5} = 1.11 \times 10^{-9} \, \text{M} \]

### Final Answer
The original concentration of \( \text{Ba}^{2+} \) is approximately \( 1.11 \times 10^{-9} \, \text{M} \).

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