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The solubility product of silver chromate, Ag2CrO4, is Ksp = 4.0 x…
Question
The solubility product of silver chromate, Ag2CrO4, is Ksp = 4.0 x 10^-12 at 298 K. Calculate the molar solubility (in mol L^-1) of Ag2CrO4 in pure water. Give the answer in scientific form, rounded to the appropriate value.
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Step-by-step solution
Let molar solubility = s.
Ag2CrO4 -> 2Ag+ + CrO4^2-, so [Ag+] = 2s and [CrO4^2-] = s.
Ksp = (2s)^2 (s) = 4s^3 = 4.0e-12.
Thus s^3 = 1.0e-12, giving s = 1.0e-4 mol L^-1.
Final answer1e-4
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