Ionic EquilibriumMedium
Question
The solubility of PbF2 in water at 25oC is ~ 10-3 M. What is its solubility in 0.05 M NaF solution ?
Options
A.1.6 × 10-6 M
B.1.2 × 10-6 M
C.1.2 × 10-5 M
D.1.6 × 10-4 M
Solution
Solubility of PbF2 ≈ 103 M
∴ Ksp = 4S3 = 4 × 10-9
In 0.05 M NaF we have 0.05 M of F- ion contributed by NaF. If the solubility of PbF2
In theis solution is S M, then Total [F-] = [2s + 0.05]M
∴ S[2S = 0.05]2 = 4 × 10-9
Assuming 2S << 0.05,
S × 25 × 10-4 = 4 × 10-9
∴ S = 0.16 × 10-5 M ⇒ 1.6 × 10-6M
We observe that our approximation that 2S << 0.05 is justified.
∴ Ksp = 4S3 = 4 × 10-9
In 0.05 M NaF we have 0.05 M of F- ion contributed by NaF. If the solubility of PbF2
In theis solution is S M, then Total [F-] = [2s + 0.05]M
∴ S[2S = 0.05]2 = 4 × 10-9
Assuming 2S << 0.05,
S × 25 × 10-4 = 4 × 10-9
∴ S = 0.16 × 10-5 M ⇒ 1.6 × 10-6M
We observe that our approximation that 2S << 0.05 is justified.
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