What is expression for Ksp for `PbCl_(2)` ?

Options

A. `[Pb^(2+)][2Cl^(-)]^(2)`
B. `[Pb^(+)]//[2Cl^(-)]^(2)`
C. `[Pb^(2+)][Cl^(-)]^(2)`
D. `[Pb^(2+)]//[Cl^(-)]^(2)`

Answer

To derive the expression for the solubility product constant (Ksp) of lead(II) chloride (PbCl₂), we will follow these steps:

### Step 1: Write the dissociation equation for PbCl₂
Lead(II) chloride dissociates in water according to the following equation:
\[ \text{PbCl}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2 \text{Cl}^{-} (aq) \]

### Step 2: Identify the ions produced
From the dissociation equation, we see that one formula unit of PbCl₂ produces:
- 1 ion of Pb²⁺
- 2 ions of Cl⁻

### Step 3: Define the solubility product (Ksp)
The solubility product (Ksp) is defined as the product of the molar concentrations of the ions in a saturated solution, each raised to the power of their coefficients in the balanced equation.

### Step 4: Write the Ksp expression
Based on the dissociation, the Ksp expression for PbCl₂ can be written as:
\[ K_{sp} = [\text{Pb}^{2+}][\text{Cl}^{-}]^2 \]

### Step 5: Substitute the coefficients
In this case, since there is 1 Pb²⁺ ion and 2 Cl⁻ ions produced, we can express Ksp as:
\[ K_{sp} = [\text{Pb}^{2+}][\text{Cl}^{-}]^2 \]

### Final Expression
Thus, the final expression for the solubility product constant (Ksp) of lead(II) chloride is:
\[ K_{sp} = [\text{Pb}^{2+}][\text{Cl}^{-}]^2 \]

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