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 the reaction quotient (Q) for the dissociation of lead(II) chloride (PbCl₂) in a mixture of lead(II) nitrate (Pb(NO₃)₂) and sodium chloride (NaCl). We will compare this value with the solubility product constant (Ksp) to see if precipitation occurs.

### Step-by-Step Solution:

**Step 1: Calculate the moles of Pb²⁺ ions from Pb(NO₃)₂.**

Given:
- Volume of Pb(NO₃)₂ solution = 300 mL = 0.300 L
- Concentration of Pb(NO₃)₂ = 0.134 M

Moles of Pb²⁺ from Pb(NO₃)₂:
\[ \text{Moles of Pb}^{2+} = \text{Volume} \times \text{Concentration} = 0.300 \, \text{L} \times 0.134 \, \text{mol/L} = 0.0402 \, \text{mol} \]

**Step 2: Calculate the moles of Cl⁻ ions from NaCl.**

Given:
- Volume of NaCl solution = 100 mL = 0.100 L
- Concentration of NaCl = 0.4 M

Moles of Cl⁻ from NaCl:
\[ \text{Moles of Cl}^- = \text{Volume} \times \text{Concentration} = 0.100 \, \text{L} \times 0.4 \, \text{mol/L} = 0.0400 \, \text{mol} \]

**Step 3: Calculate the total volume of the mixture.**

Total volume = Volume of Pb(NO₃)₂ + Volume of NaCl
\[ \text{Total Volume} = 300 \, \text{mL} + 100 \, \text{mL} = 400 \, \text{mL} = 0.400 \, \text{L} \]

**Step 4: Calculate the concentrations of Pb²⁺ and Cl⁻ in the mixture.**

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

Concentration of Cl⁻:
\[ [\text{Cl}^-] = \frac{\text{Moles of Cl}^-}{\text{Total Volume}} = \frac{0.0400 \, \text{mol}}{0.400 \, \text{L}} = 0.1000 \, \text{M} \]

**Step 5: Calculate the reaction quotient Q.**

The reaction quotient Q for the dissociation of PbCl₂ is given by:
\[ Q = [\text{Pb}^{2+}][\text{Cl}^-]^2 \]
Substituting the concentrations:
\[ Q = (0.1005)(0.1000)^2 = (0.1005)(0.0100) = 0.001005 = 1.005 \times 10^{-3} \]

**Step 6: Compare Q with Ksp.**

Given:
\[ K_{sp} = 1.6 \times 10^{-5} \]

Since \( Q = 1.005 \times 10^{-3} \) is greater than \( K_{sp} = 1.6 \times 10^{-5} \), this indicates that the solution is supersaturated and precipitation of PbCl₂ will occur.

### Final Answer:
The reaction quotient \( Q \) is greater than \( K_{sp} \), indicating that PbCl₂ will precipitate.

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