An aqueous solution contains an unknown concentration `Ba^(2+) ` . When 50 mL of a 1 M solution ` Na_2SO_4` is added , `BaSO_4 ` just begins to precipitate . The final volume is 500 mL .
The solubility product of ` BaSO_4 ` is ` 1xx 10^(-10)` .What is the original concentration of `Ba^(2+) `?

Options

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

Answer

To solve the problem, we need to find the original concentration of \( \text{Ba}^{2+} \) ions in an aqueous solution after adding \( \text{Na}_2\text{SO}_4 \) and causing \( \text{BaSO}_4 \) to precipitate. Here’s a step-by-step breakdown of the solution:

### Step 1: Determine the volume of the original solution
We know that:
- Volume of \( \text{Na}_2\text{SO}_4 \) added = 50 mL
- Final volume of the solution = 500 mL

To find the volume of the original \( \text{Ba}^{2+} \) solution, we subtract the volume of \( \text{Na}_2\text{SO}_4 \) from the final volume:
\[ \text{Volume of original solution} = 500 \, \text{mL} - 50 \, \text{mL} = 450 \, \text{mL} \]

### Step 2: Calculate the concentration of sulfate ions
The concentration of sulfate ions \( \text{SO}_4^{2-} \) from the \( \text{Na}_2\text{SO}_4 \) solution can be calculated using the dilution formula:
\[ C_1V_1 = C_2V_2 \]
Where:
- \( C_1 = 1 \, \text{M} \) (concentration of \( \text{Na}_2\text{SO}_4 \))
- \( V_1 = 50 \, \text{mL} \)
- \( V_2 = 500 \, \text{mL} \) (final volume)

Rearranging the formula to find \( C_2 \):
\[ C_2 = \frac{C_1V_1}{V_2} = \frac{1 \, \text{M} \times 50 \, \text{mL}}{500 \, \text{mL}} = \frac{50}{500} = 0.1 \, \text{M} \]

### Step 3: Set up the expression for the solubility product
The solubility product \( K_{sp} \) for \( \text{BaSO}_4 \) is given as \( 1 \times 10^{-10} \). At the point of precipitation, the ionic product equals the solubility product:
\[ K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] \]

Substituting the known concentration of sulfate ions:
\[ 1 \times 10^{-10} = [\text{Ba}^{2+}] \times 0.1 \]

### Step 4: Solve for the concentration of \( \text{Ba}^{2+} \)
Rearranging the equation to find \( [\text{Ba}^{2+}] \):
\[ [\text{Ba}^{2+}] = \frac{1 \times 10^{-10}}{0.1} = 1 \times 10^{-9} \, \text{M} \]

### Step 5: Calculate the original concentration of \( \text{Ba}^{2+} \)
Now, we need to find the original concentration in the 450 mL solution. We can use the dilution formula again:
\[ C_1V_1 = C_2V_2 \]
Where:
- \( C_1 \) is the original concentration we want to find,
- \( V_1 = 450 \, \text{mL} \),
- \( C_2 = 1 \times 10^{-9} \, \text{M} \),
- \( V_2 = 500 \, \text{mL} \).

Rearranging gives:
\[ C_1 = \frac{C_2V_2}{V_1} = \frac{(1 \times 10^{-9} \, \text{M}) \times (500 \, \text{mL})}{450 \, \text{mL}} = \frac{5 \times 10^{-7}}{450} \approx 1.11 \times 10^{-9} \, \text{M} \]

### Final Answer
The original concentration of \( \text{Ba}^{2+} \) ions is approximately:
\[ \text{Original concentration of } \text{Ba}^{2+} = 1.11 \times 10^{-9} \, \text{M} \]