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Question
The solubility of Ag2CrO4 at 25°C is 8.0 × 10-5 moles litre-1. Calculate its solubility product.
Solution
Let the solubility of Ag2CrO4 in s mol L-1
`underset(S)Ag2CrO4 ⇌ underset(2s)2Ag^+ + underset(2s) CrO_4^-`
`K_(sp (Ag_(2)CrO4))=[Ag+]^2[CrO_4^(2-)] `
=[2s]2 [s]
= 4s3
= 4 × (8.0 × 10-5)3
Ksp = 2.048 × 10-12
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