When solid lead iodide is added to water, the equilibrium concentration of `I^(-)` becomes `2.6 xx 10^(-3)M` . What is the `K_(sp)` for `PbI_(2)` ?

Options

A. `2.2xx10^(-9)`
B. `8.8xx10^(-9)`
C. `1.8xx10^(-8)`
D. `3.5xx10^(-8)`

Answer

To find the solubility product constant (Ksp) for lead iodide (PbI2), we can follow these steps:

### Step 1: Write the Dissolution Equation
The dissolution of lead iodide in water can be represented by the following equilibrium equation:

\[ \text{PbI}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2 \text{I}^- (aq) \]

### Step 2: Determine Ion Concentrations at Equilibrium
From the equilibrium concentration of iodide ions \([I^-]\), which is given as \(2.6 \times 10^{-3} \, M\), we can find the concentration of lead ions \([Pb^{2+}]\).

According to the stoichiometry of the dissolution equation:
- For every 1 mole of PbI2 that dissolves, 1 mole of Pb²⁺ and 2 moles of I⁻ are produced.
- Therefore, the concentration of lead ions \([Pb^{2+}]\) will be half of the concentration of iodide ions.

Calculating \([Pb^{2+}]\):
\[ [Pb^{2+}] = \frac{1}{2} [I^-] = \frac{1}{2} \times 2.6 \times 10^{-3} \, M = 1.3 \times 10^{-3} \, M \]

### Step 3: Write the Expression for Ksp
The solubility product constant \(K_{sp}\) is given by the expression:
\[ K_{sp} = [Pb^{2+}][I^-]^2 \]

### Step 4: Substitute the Concentrations into the Ksp Expression
Now, we can substitute the values of \([Pb^{2+}]\) and \([I^-]\) into the Ksp expression:
\[ K_{sp} = (1.3 \times 10^{-3})(2.6 \times 10^{-3})^2 \]

### Step 5: Calculate Ksp
Calculating the square of the iodide ion concentration:
\[ (2.6 \times 10^{-3})^2 = 6.76 \times 10^{-6} \]

Now, substituting back into the Ksp expression:
\[ K_{sp} = (1.3 \times 10^{-3}) \times (6.76 \times 10^{-6}) = 8.788 \times 10^{-9} \approx 8.8 \times 10^{-9} \]

### Final Answer
Thus, the value of \(K_{sp}\) for lead iodide (PbI2) is approximately:
\[ K_{sp} \approx 8.8 \times 10^{-9} \]

---