A solution was prepared by dissolving 476 mg of silver chromate (Ag2CrO4, MW = 331.73 g mol-1 ) in 633 mL of water a) moles of Ag2CrO4 b) millimoles of Ag2CrO4 c) molarity of Ag2CrO4 d) molarity of Ag+ e) molarity of CrO4 2- f) parts per million of Ag2CrO4 g) % (weight/volume) Ag2CrO4 (in units of g/mL)

Silver chromate, Ag2CrO4 , a sparingly soluble compound, has  a solubility product, Ksp , of 1.2 x 10-12

What is the molar  solubility of Ag2CrO4 in pure water?

Explanation:

Use the ICE approach:       Write the Ksp equilibrium:

     Ag2CrO4(s) = 2 Ag+(aq) + CrO4  2- (aq)            Ksp = 1.2 x 10 -12

I :           1         0         0

C :      (“0”)    +2x      +x

E :          1        2x        x

Note: above, we let          x = [CrO4   2- ] and so [Ag+] = 2x )

Ksp =   [Ag+] 2 [CrO4   2-]/1 = (2x)2(x) = 4x3 = 1.2 x 10-12

Or,        x3  = 3.0 x 10 -13        Take cube root of both sides:

        => x = 6.7 x 10 -5 M = [ CrO4   2- ] => [ Ag+ ] =2x = 1.3 x 10  -4 M

the answer for sodium is 0.186

the answer for rubidium is 0.251

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