The `K_(sp)` for the following dissociation is `1.6xx10^(-5)`
`PbCl_(2) (s) to Pb^(2+)(aq)+ 2Cl^(-)(aq)` Which of the following choices is correct for a mixture of 300 mL 0.134 M `Pb(NO_(3))_(2)` and 100 mL 0.4 M NaCl?

Options

A. `Q gt K_(sp)`
B. `Q = K_(sp)`
C. Not enough data provided
D. `Q lt K_(sp)`

Explanation

To solve the problem, we need to determine whether a precipitate of lead(II) chloride (PbCl₂) will form when mixing the given solutions. We will follow these steps:

### Step 1: Calculate the moles of Pb²⁺ and Cl⁻ ions in the mixture.

1. **Calculate moles of Pb²⁺ from Pb(NO₃)₂:**
- Volume = 300 mL = 0.300 L
- Molarity = 0.134 M
- Moles of Pb²⁺ = Molarity × Volume = 0.134 mol/L × 0.300 L = 0.0402 moles

2. **Calculate moles of Cl⁻ from NaCl:**
- Volume = 100 mL = 0.100 L
- Molarity = 0.4 M
- Moles of Cl⁻ = Molarity × Volume = 0.4 mol/L × 0.100 L = 0.040 moles

### Step 2: Calculate the total volume of the mixture.

- Total Volume = Volume of Pb(NO₃)₂ + Volume of NaCl = 300 mL + 100 mL = 400 mL = 0.400 L

### Step 3: Calculate the concentrations of Pb²⁺ and Cl⁻ in the mixture.

1. **Concentration of Pb²⁺:**
\[ \text{Concentration of Pb}^{2+} = \frac{\text{Moles of Pb}^{2+}}{\text{Total Volume}} = \frac{0.0402 \text{ moles}}{0.400 \text{ L}} = 0.1005 \text{ M} \]

2. **Concentration of Cl⁻:**
\[ \text{Concentration of Cl}^- = \frac{\text{Moles of Cl}^-}{\text{Total Volume}} = \frac{0.040 \text{ moles}}{0.400 \text{ L}} = 0.100 \text{ M} \]

### Step 4: Calculate the ionic product (Q) for PbCl₂.

The dissociation of PbCl₂ can be represented as:
\[ \text{PbCl}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2 \text{Cl}^- (aq) \]

The ionic product (Q) is given by:
\[ Q = [\text{Pb}^{2+}][\text{Cl}^-]^2 \]
Substituting the concentrations:
\[ Q = (0.1005)(0.100)^2 = (0.1005)(0.01) = 0.001005 = 1.005 \times 10^{-3} \]

### Step 5: Compare Q with Ksp.

Given \( K_{sp} \) for PbCl₂ is \( 1.6 \times 10^{-5} \).

- Since \( Q = 1.005 \times 10^{-3} \) and \( K_{sp} = 1.6 \times 10^{-5} \):
\[ Q > K_{sp} \]

### Step 6: Conclusion.

Since \( Q \) is greater than \( K_{sp} \), a precipitate of PbCl₂ will form.

### Final Answer:
**A precipitate of PbCl₂ will form.**

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