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Ksp of PbCl2 = 1.6 xx 10^(-5) On mixing ...

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

A

`Q gt Ksp`

B

`Q lt Ksp`

C

`Q = Ksp`

D

Relation does not exit

Text Solution

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To solve the problem, we need to determine whether the precipitation of lead(II) chloride (PbCl₂) will occur when mixing the given solutions of lead(II) nitrate [Pb(NO₃)₂] and sodium chloride (NaCl). We will calculate the concentrations of the ions in the mixed solution and then compare the reaction quotient (Q) to the solubility product constant (Ksp) of PbCl₂. ### Step-by-Step Solution: 1. **Calculate the total volume of the mixed solution:** - Volume of Pb(NO₃)₂ solution = 300 mL - Volume of NaCl solution = 100 mL - Total volume (V_total) = 300 mL + 100 mL = 400 mL 2. **Calculate the moles of Pb²⁺ and Cl⁻ ions in the mixed solution:** - For Pb(NO₃)₂: - Molarity (M) = 0.134 M - Volume (V) = 300 mL = 0.300 L - Moles of Pb²⁺ = M × V = 0.134 mol/L × 0.300 L = 0.0402 moles - For NaCl: - Molarity (M) = 0.4 M - Volume (V) = 100 mL = 0.100 L - Moles of Cl⁻ = M × V = 0.4 mol/L × 0.100 L = 0.040 moles 3. **Calculate the concentrations of Pb²⁺ and Cl⁻ in the mixed solution:** - Concentration of Pb²⁺ (C_Pb²⁺) = Moles of Pb²⁺ / Total Volume - C_Pb²⁺ = 0.0402 moles / 0.400 L = 0.1005 M - Concentration of Cl⁻ (C_Cl⁻) = Moles of Cl⁻ / Total Volume - C_Cl⁻ = 0.040 moles / 0.400 L = 0.1 M 4. **Calculate the reaction quotient (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 expression for Q is: \[ Q = [\text{Pb}^{2+}][\text{Cl}^-]^2 \] - Substitute the concentrations: \[ Q = (0.1005)(0.1)^2 = (0.1005)(0.01) = 0.001005 \] 5. **Compare Q with Ksp:** - Given Ksp of PbCl₂ = 1.6 × 10⁻⁵ - Q = 0.001005 = 1.005 × 10⁻³ 6. **Determine if precipitation occurs:** - Since Q (1.005 × 10⁻³) > Ksp (1.6 × 10⁻⁵), precipitation of PbCl₂ will occur. ### Conclusion: The correct conclusion is that the reaction quotient Q is greater than the solubility product Ksp, indicating that lead(II) chloride will precipitate out of the solution.

To solve the problem, we need to determine whether the precipitation of lead(II) chloride (PbCl₂) will occur when mixing the given solutions of lead(II) nitrate [Pb(NO₃)₂] and sodium chloride (NaCl). We will calculate the concentrations of the ions in the mixed solution and then compare the reaction quotient (Q) to the solubility product constant (Ksp) of PbCl₂. ### Step-by-Step Solution: 1. **Calculate the total volume of the mixed solution:** - Volume of Pb(NO₃)₂ solution = 300 mL - Volume of NaCl solution = 100 mL - Total volume (V_total) = 300 mL + 100 mL = 400 mL ...
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