What is the molar solubility of PbI₂ in 0.25 M NaI?
K _sp(PbI₂) = 7.9×10^{- 9} (Hint:
Assume "x" is small to make math easier.)

Question

What is the molar solubility of PbI₂ in 0.25 M NaI? K _sp(PbI₂) = 7.9×10^{- 9} (Hint: Assume "x" is small to make math easier.)

Quizplus · Accepted Answer

- Write the equilibrium equation and expression: 
PbIS1U1B12S1U1B0 (s) ⇌ PbS1U1B2S1U1B0 2+ (aq) + 3 I- (aq) 
K S1U1B1spS1U1B0(PbIS1U1B12S1U1B0) = [PbS1U1B2S1U1B0 2+ ][I - ]3 
- Substitute given values: 
K S1U1B1spS1U1B0(PbIS1U1B12S1U1B0) = 7.9×10S1U1P1 - 9S1S1P0 
[I-]=0.25 M 
- Assume "x" is small (compared to 0.25 M) and ignore its contribution to the initial [I-], and assume that [PbIS1U1B12S1U1B0] dissociates by "x". Then, at equilibrium: 
[PbIS1U1B12S1U1B0]= x M 
[PbS1U1B2S1U1B0 2+ ]= x M (Initially 0) 
[I-]= 0.25 M – 3x (since each PbIS1U1B12S1U1B0 molecule dissociates into 3 I-) 
K S1U1B1spS1U1B0(PbIS1U1B12S1U1B0) = [PbS1U1B2S1U1B0 2+ ][I - ]3 
7.9×10S1U1P1 - 9S1S1P0 = x * x * (0.25 M – 3x)3 
- Solve for "x" using approximation (or numerical methods): 
x ≈ 1.3×10S1U1P1 - 7S1S1P0 M 
- Choose the answer choice closest to the calculated value: 
D) [PbIS1U1B12S1U1B0] = 1.3×10S1U1P1 - 7S1S1P0 M (closest to calculated value)

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