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The solubility of Ag2Cro4 at 85°c is 8.0 ×10.5 m. Find the solubility product of Ag2Cro4. |
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Answer» Ag2Cro4 → 2 Ag+ + cro4-2 using law at mass action K = \(\frac {[Ag^+]^2[Cro_4^{-2}}{Ag_2Cro_4}\) = K [Ag2Cro4] = [Ag+]2 [Crp4-2] = Ksp = [Ag+]2 [Cro4-2] (Ksp = K [Ag2Cro4]) Let say solubility of Ag2Cro4 = S M = 8.0 x 10-5m ∴ Ksp = [2s]2 [s] Ksp = 4s3 Ksp = 4 x (8.0 x 10-5)3 Ksp = 2048 x 10-15 Ksp = 2.048 x 10-12 Hence the solubility product at Ag2Cro4 will be 2.048 x 10-12 |
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