Question

What would be the volume of water required to dissolve 0.2 g of PbCl$_2$ of molar mass 278 g/mol to prepare a saturated solution of the salt? (K$_{sp}$ of PbCl$_2$ = 3.2 $\times$ 10$^{-8}$)

• A. 1000 ml
• B. 359.7 ml
• C. 278.8 ml
• D. 360.4 ml

Hint:

When calculating solubility, use the K$_{sp}$ expression and relate the concentrations of ions in the saturated solution.

Answer 1

The Correct Option is B

Given:
- Molar mass of PbCl$_2$ = 278 g/mol
- Mass of PbCl$_2$ = 0.2 g
- K$_{sp}$ of PbCl$_2$ = 3.2 $\times$ 10$^{-8}$

Step 1: Calculate the number of moles of PbCl$_2$:
\[ \text{Moles of PbCl}_2 = \frac{\text{Mass of PbCl}_2}{\text{Molar Mass of PbCl}_2} \] \[ \text{Moles of PbCl}_2 = \frac{0.2}{278} = 7.19 \times 10^{-4} \, \text{mol} \]
Step 2: Write the dissociation equation for PbCl$_2$:
\[ \text{PbCl}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2\text{Cl}^- (aq) \]
Step 3: Use the K$_{sp}$ expression:
\[ K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 \] Let the concentration of Pb$^{2+}$ ions be \(s\). Then, the concentration of Cl$^-$ ions will be \(2s\).
Thus, we have: \[ K_{sp} = (s)(2s)^2 = 4s^3 \] Substituting the given \(K_{sp}\) value: \[ 3.2 \times 10^{-8} = 4s^3 \] Solving for \(s\): \[ s^3 = \frac{3.2 \times 10^{-8}}{4} = 8 \times 10^{-9} \] \[ s = \sqrt[3]{8 \times 10^{-9}} = 2 \times 10^{-3} \, \text{mol/L} \]
Step 4: Find the volume of water required:
We know the moles of PbCl$_2$ is \(7.19 \times 10^{-4}\) mol, and the concentration of Pb$^{2+}$ is \(2 \times 10^{-3}\) mol/L.
Using the formula for concentration: \[ \text{Concentration} = \frac{\text{Moles of solute}}{\text{Volume of solution}} \] \[ 2 \times 10^{-3} = \frac{7.19 \times 10^{-4}}{V} \] Solving for \(V\): \[ V = \frac{7.19 \times 10^{-4}}{2 \times 10^{-3}} = 0.3595 \, \text{L} = 359.7 \, \text{mL} \]