A saturated solution of sparingly soluble lead chloride on analysis was found to contain 11.84 g/ litre of the salt at room temperature. Calculate the solubility product constant at room temperature. (At. wt . : Pb = 207, Cl = 35.5 )

Explanation

To calculate the solubility product constant (Ksp) for lead chloride (PbCl2) based on the given solubility, we will follow these steps:

### Step 1: Calculate the molar mass of PbCl2
To find the molar mass of lead chloride (PbCl2), we need to sum the atomic weights of lead and chlorine.

- Atomic weight of Pb = 207 g/mol
- Atomic weight of Cl = 35.5 g/mol

Since there are two chlorine atoms in PbCl2, the calculation will be:

\[ \text{Molar mass of PbCl2} = \text{Atomic weight of Pb} + 2 \times \text{Atomic weight of Cl} \]
\[ = 207 + 2 \times 35.5 = 207 + 71 = 278 \text{ g/mol} \]

### Step 2: Convert the solubility from grams per liter to moles per liter
Given that the solubility of PbCl2 is 11.84 g/L, we can convert this to moles per liter using the molar mass calculated in Step 1.

\[ \text{Solubility (in mol/L)} = \frac{\text{mass (g/L)}}{\text{molar mass (g/mol)}} \]
\[ = \frac{11.84 \text{ g/L}}{278 \text{ g/mol}} \approx 0.0426 \text{ mol/L} \]

### Step 3: Determine the concentrations of ions in solution
When PbCl2 dissolves in water, it dissociates into its ions:

\[ \text{PbCl2} \rightleftharpoons \text{Pb}^{2+} + 2 \text{Cl}^- \]

From the dissociation, we can see that for every 1 mole of PbCl2 that dissolves, we get 1 mole of Pb²⁺ and 2 moles of Cl⁻. Therefore, the concentrations of the ions will be:

- \([\text{Pb}^{2+}] = 0.0426 \text{ mol/L}\)
- \([\text{Cl}^-] = 2 \times 0.0426 \text{ mol/L} = 0.0852 \text{ mol/L}\)

### Step 4: Calculate the solubility product constant (Ksp)
The solubility product constant (Ksp) is given by the expression:

\[ Ksp = [\text{Pb}^{2+}][\text{Cl}^-]^2 \]

Substituting the concentrations we found in Step 3:

\[ Ksp = (0.0426)(0.0852)^2 \]
\[ = (0.0426)(0.00725) \approx 0.000308 \]
\[ = 3.08 \times 10^{-4} \]

### Final Answer
Thus, the solubility product constant (Ksp) for lead chloride at room temperature is approximately:

\[ Ksp \approx 3.09 \times 10^{-4} \]

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