At `25^(@)`C, `K_(sp)` for `PbBr_(2)` is equal to `8xx10^(-5)`. If the salt is `80%` dissociated, What is the solubility of `PbBr_(2)` in mol//litre?

Options

A. `[(10^(-4))/(1.6xx1.6)]^(1//2)`
B. `[(10^(-4))/(1.6xx1.6)]^(1//3)`
C. `[(10^(-4))/(0.8xx0.8)]^(1//3)`
D. `[(10^(-5))/(1.6xx1.6)]^(1//2)`

Answer

To solve the problem, we need to find the solubility of PbBr₂ when it is 80% dissociated. Let's break down the solution step by step.

### Step 1: Define the solubility
Let the solubility of PbBr₂ be represented as \( S \) (in mol/L). This means that if \( S \) moles of PbBr₂ dissolve in 1 L of water, it will dissociate into its ions.

### Step 2: Write the dissociation equation
The dissociation of PbBr₂ in water can be represented as:
\[ \text{PbBr}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2 \text{Br}^- (aq) \]
From this equation, we can see that 1 mole of PbBr₂ produces 1 mole of Pb²⁺ and 2 moles of Br⁻.

### Step 3: Calculate the concentrations of ions
If PbBr₂ is 80% dissociated, then:
- The concentration of Pb²⁺ ions will be \( 0.8S \).
- The concentration of Br⁻ ions will be \( 2 \times 0.8S = 1.6S \).

### Step 4: Write the expression for Ksp
The solubility product constant \( K_{sp} \) for PbBr₂ is given by:
\[ K_{sp} = [\text{Pb}^{2+}][\text{Br}^-]^2 \]
Substituting the concentrations we found:
\[ K_{sp} = (0.8S)(1.6S)^2 \]

### Step 5: Substitute the value of Ksp
We know that \( K_{sp} = 8 \times 10^{-5} \). Therefore, we can set up the equation:
\[ 8 \times 10^{-5} = (0.8S)(1.6S)^2 \]

### Step 6: Simplify the equation
Now, simplify the right-hand side:
\[ (1.6S)^2 = 2.56S^2 \]
Thus, we have:
\[ 8 \times 10^{-5} = 0.8S \cdot 2.56S^2 \]
\[ 8 \times 10^{-5} = 2.048S^3 \]

### Step 7: Solve for S
Now, we can solve for \( S \):
\[ S^3 = \frac{8 \times 10^{-5}}{2.048} \]
Calculating the right side:
\[ S^3 = 3.90625 \times 10^{-5} \]
Taking the cube root:
\[ S = \sqrt[3]{3.90625 \times 10^{-5}} \approx 0.0342 \text{ mol/L} \]

### Final Answer
The solubility of PbBr₂ at 25°C, when 80% dissociated, is approximately:
\[ \boxed{0.0342 \text{ mol/L}} \]