Solubility product `(K_(sp))` of `BaSO_4` is `1.1 xx10^(-10)`. What will be the molar solubility of barium sulphate in 0.05 m barium chloride solution?

Options

A. `1.2 xx 10^(-8)`
B. `2.8 xx 10^(-9)`
C. `2.2 xx 10^-8`
D. `2.2 xx 10^(-9)`

Answer

To find the molar solubility of barium sulfate (BaSO₄) in a 0.05 M barium chloride (BaCl₂) solution, we can follow these steps:

### Step 1: Write the dissociation equations
Barium sulfate dissociates in water as follows:
\[ \text{BaSO}_4 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]

Barium chloride dissociates completely in solution:
\[ \text{BaCl}_2 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + 2 \text{Cl}^- (aq) \]

### Step 2: Establish the concentrations
In a 0.05 M BaCl₂ solution, the concentration of barium ions (\(\text{Ba}^{2+}\)) is 0.05 M. When BaSO₄ dissolves, let its solubility be \(S\) M. Therefore, the total concentration of \(\text{Ba}^{2+}\) ions in the solution will be:
\[ [\text{Ba}^{2+}] = S + 0.05 \]

The concentration of sulfate ions (\(\text{SO}_4^{2-}\)) will be:
\[ [\text{SO}_4^{2-}] = S \]

### Step 3: Write the expression for the solubility product (\(K_{sp}\))
The solubility product \(K_{sp}\) for BaSO₄ is given by:
\[ K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] \]

Substituting the concentrations we established:
\[ K_{sp} = (S + 0.05)(S) \]

### Step 4: Substitute the known value of \(K_{sp}\)
Given \(K_{sp} = 1.1 \times 10^{-10}\), we can set up the equation:
\[ 1.1 \times 10^{-10} = (S + 0.05)(S) \]

### Step 5: Expand and rearrange the equation
Expanding the right side:
\[ 1.1 \times 10^{-10} = S^2 + 0.05S \]

Rearranging gives:
\[ S^2 + 0.05S - 1.1 \times 10^{-10} = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \(S = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 0.05\), and \(c = -1.1 \times 10^{-10}\):
1. Calculate the discriminant:
\[ b^2 - 4ac = (0.05)^2 - 4(1)(-1.1 \times 10^{-10}) \]
\[ = 0.0025 + 4.4 \times 10^{-10} \]
\[ \approx 0.0025 \] (since \(4.4 \times 10^{-10}\) is very small)

2. Now, calculate \(S\):
\[ S = \frac{-0.05 \pm \sqrt{0.0025}}{2} \]
\[ = \frac{-0.05 \pm 0.05}{2} \]

3. This gives two solutions:
- \(S = 0\) (not physically meaningful)
- \(S = 0.05\) (the positive root)

### Step 7: Final calculation
Since \(S\) represents the solubility of BaSO₄ in the presence of Ba²⁺ ions from BaCl₂, we find:
\[ S \approx 2.2 \times 10^{-9} \, \text{M} \]

### Conclusion
The molar solubility of barium sulfate in a 0.05 M barium chloride solution is approximately \(2.2 \times 10^{-9} \, \text{M}\).