Ksp of `PbCl_2 = 1.6 × 10^(-5) On mixing `300 mL, 0.134M Pb(NO_3)_2(aq.) + 100 mL, 0.4M NaCl(aq.)`

Options

A. Not enough data provided
B. `Q=k_("sp")`
C. `Qltk_("sp")`
D. `Qgtk_("sp")`

Answer

To solve the problem, we need to determine the solubility product quotient (Qsp) of lead(II) chloride (PbCl2) when mixing the given solutions and compare it with the solubility product constant (Ksp) to see if precipitation occurs.

### Step-by-Step Solution:

1. **Write the Reaction:**
The reaction between lead(II) nitrate and sodium chloride can be written as:
\[ \text{Pb(NO}_3\text{)}_2 (aq) + 2 \text{NaCl} (aq) \rightarrow \text{PbCl}_2 (s) + 2 \text{NaNO}_3 (aq) \]

2. **Calculate the Millimoles of Reactants:**
- For Pb(NO3)2:
\[ \text{Millimoles of Pb(NO}_3\text{)}_2 = \text{Molarity} \times \text{Volume} = 0.134 \, \text{mol/L} \times 300 \, \text{mL} = 40.2 \, \text{mmol} \]
- For NaCl:
\[ \text{Millimoles of NaCl} = 0.4 \, \text{mol/L} \times 100 \, \text{mL} = 40 \, \text{mmol} \]

3. **Identify the Limiting Reagent:**
The stoichiometry of the reaction shows that 1 mole of Pb(NO3)2 reacts with 2 moles of NaCl. Therefore, the required amount of NaCl for 40.2 mmol of Pb(NO3)2 would be:
\[ \text{Required NaCl} = 2 \times 40.2 \, \text{mmol} = 80.4 \, \text{mmol} \]
Since we only have 40 mmol of NaCl, NaCl is the limiting reagent.

4. **Calculate the Amount of PbCl2 Formed:**
Using the limiting reagent (NaCl):
\[ \text{Millimoles of PbCl}_2 = \frac{40 \, \text{mmol NaCl}}{2} = 20 \, \text{mmol PbCl}_2 \]

5. **Calculate the Total Volume of the Mixture:**
The total volume after mixing is:
\[ \text{Total Volume} = 300 \, \text{mL} + 100 \, \text{mL} = 400 \, \text{mL} \]

6. **Calculate the Concentration of PbCl2:**
\[ \text{Concentration of PbCl}_2 = \frac{\text{Millimoles of PbCl}_2}{\text{Total Volume (in L)}} = \frac{20 \, \text{mmol}}{0.400 \, \text{L}} = 0.050 \, \text{M} \]

7. **Calculate the Concentration of Ions:**
- For Pb²⁺:
\[ [\text{Pb}^{2+}] = 0.050 \, \text{M} \]
- For Cl⁻:
\[ [\text{Cl}^-] = 2 \times [\text{PbCl}_2] = 2 \times 0.050 \, \text{M} = 0.100 \, \text{M} \]

8. **Calculate Qsp:**
\[ Qsp = [\text{Pb}^{2+}] \times [\text{Cl}^-]^2 = (0.050) \times (0.100)^2 = 0.050 \times 0.0100 = 0.0005 = 5.0 \times 10^{-4} \]

9. **Compare Qsp with Ksp:**
Given that \( Ksp \) for PbCl2 is \( 1.6 \times 10^{-5} \):
\[ Qsp = 5.0 \times 10^{-4} > Ksp = 1.6 \times 10^{-5} \]
Since \( Qsp > Ksp \), precipitation of PbCl2 will occur.

### Final Answer:
The correct option is **D: Q > Ksp**.