The Ksp for Ag2CrO4 is 9x10-12 M3. What is the molar solubility of Ag2CrO4 in pure water?
Asked by Topperlearning User
| 4th Jun, 2014,
01:23: PM
Expert Answer:
Let x be the molar solubility of Ag2CrO4, then
Ag2CrO4 = 2 Ag+ + CrO42-
2 x x
(2 x)2 (x) = Ksp
x = {(9x10-12)/4}(1/3)
= 1.3x10-4 M
Hence, the molar solubility of Ag2CrO4 is 1.3x10-4 M.
Answered by
| 4th Jun, 2014,
03:23: PM
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