An aqueous solution contains an unknown concentratioon of `Ba^(2+)`, When 50mL of 1M solutioon of `Na_(2)SO_4` is added `BaSO_4` just begin to precipitate. The final volume is 500mL. The solubility product of `BaSO_4` is `1 xx 10^(-10)`. What is the original concentration of `Ba^(2+)?`

Options

A. ` 1.0 xx 10^(-10)M`
B. `5 xx 10^(-9)M`
C. `2 xx 10^(-9)M`
D. `1.1 xx 10^(-9)M`

Answer

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

### Step 1: Determine the concentration of sulfate ions after mixing
We have a 50 mL solution of 1 M \( \text{Na}_2\text{SO}_4 \). The number of moles of sulfate ions (\( \text{SO}_4^{2-} \)) in this solution can be calculated as follows:

\[ \text{Moles of } \text{SO}_4^{2-} = \text{Molarity} \times \text{Volume} = 1 \, \text{mol/L} \times 0.050 \, \text{L} = 0.050 \, \text{mol} \]

Since \( \text{Na}_2\text{SO}_4 \) dissociates completely, the concentration of sulfate ions in the final 500 mL solution can be calculated using the dilution formula:

\[ C_1V_1 = C_2V_2 \]

Where:
- \( C_1 = 1 \, \text{M} \) (initial concentration of \( \text{Na}_2\text{SO}_4 \))
- \( V_1 = 50 \, \text{mL} = 0.050 \, \text{L} \)
- \( V_2 = 500 \, \text{mL} = 0.500 \, \text{L} \)

Using the formula:

\[ 1 \times 0.050 = C_2 \times 0.500 \]

Solving for \( C_2 \):

\[ C_2 = \frac{1 \times 0.050}{0.500} = 0.10 \, \text{M} \]

### Step 2: Use the solubility product to find the concentration of \( \text{Ba}^{2+} \)
The solubility product \( K_{sp} \) for \( \text{BaSO}_4 \) is given as \( 1 \times 10^{-10} \). The \( K_{sp} \) expression for \( \text{BaSO}_4 \) is:

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

Substituting the known concentration of sulfate ions:

\[ 1 \times 10^{-10} = [\text{Ba}^{2+}][0.10] \]

Let \( [\text{Ba}^{2+}] = x \):

\[ 1 \times 10^{-10} = x \times 0.10 \]

Solving for \( x \):

\[ x = \frac{1 \times 10^{-10}}{0.10} = 1 \times 10^{-9} \, \text{M} \]

### Step 3: Relate this concentration to the original concentration of \( \text{Ba}^{2+} \)
The final volume of the solution is 500 mL, and we need to find the original concentration of \( \text{Ba}^{2+} \) in the 450 mL of the original solution. Let \( M_1 \) be the original concentration of \( \text{Ba}^{2+} \):

Using the dilution equation again:

\[ M_1 \times 0.450 = 1 \times 10^{-9} \times 0.500 \]

Solving for \( M_1 \):

\[ M_1 = \frac{1 \times 10^{-9} \times 0.500}{0.450} = \frac{5 \times 10^{-10}}{0.450} \approx 1.11 \times 10^{-9} \, \text{M} \]

Thus, the original concentration of \( \text{Ba}^{2+} \) is approximately \( 1.11 \times 10^{-9} \, \text{M} \).

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