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`        If the solubility product of pbBr2 in water is s mol/L, then what is the solubility product considering 80% ionisation?`
10 months ago

## Answers : (1)

Arun
22540 Points
```							PbBr2 -----> Pb++ + 2Br-Ksp = 8 x 10-5 Solubility  ξ = [Pb++]  + 2[Br-]2 = ξ x 2ξ2 = 4ξ3 ξ = (Ksp/4)1/3 = (s/4)1/3 mol/ L for 100 % ionisationFor 80% = s/4 x (80/100) = s/5 mol/LSolubility = s/5 * 367 (molwt of PbBr2) = 73.4 s g/L
```
10 months ago
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