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