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Grade: 12
        
If the solubility product of pbBr2 in water is s mol/L, then what is the solubility product considering 80% ionisation?
11 days ago

Answers : (1)

Arun
16138 Points
							

PbBr-----> Pb++ + 2Br-

Ksp = 8 x 10-5 

Solubility  ξ = [Pb++]  + 2[Br-]2 = ξ x 2ξ= 4ξ

ξ = (Ksp/4)1/3 = (s/4)1/3 mol/ L for 100 % ionisation

For 80% = s/4 x (80/100) = s/5 mol/L

Solubility = s/5 * 367 (molwt of PbBr2) = 73.4 s g/L

11 days ago
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