What is the solubility in (Mol/L) of silver chromate in...
a) 1.3 M potassium chromate
b) 1.3 M silver nitrate
c) pure water
??? thanks a lot for whoever helps me out
Ag2CrO4 (s) <==> 2 Ag+ (aq) + CrO4-- (aq)
Ksp = [Ag+]^2 [CrO4--]
If x moles of Ag2CrO4 dissolve in a liter of water (which defines molar solubility), then there are 2x moles of Ag+ and x moles of CrO4-- in solution.
Doing c) first:
Substitute what we know into the Ksp equation:
1.12x10^-12 = [2x]^2 [x] = 4x^3;
x = ((1.12x10^-12)/4)^(1/3) = 6.54x10^-5 M is the solubility in pure water.
a) Since initially [CrO4--] = 1.3 M, then the Ksp equation becomes
1.12x10^-12 = [2x]^2 [1.3 + x].
We can simplify this equation by remembering that since Ksp is small, x will be small too. Therefore, 1.3 >> x and hence 1.3 + x is very close to 1.3. The equation becomes
1.12x10^-12 = [2x]^2 [1.3] = 5.2x^2;
x = ((1.12x10^-12)/5.2)^(1/2) = 4.64x10^-7 M is the solubility in 1.3 M K2CrO4.
b) In the same way as a) above, since initially [Ag+] = 1.3 M, the Ksp equation becomes:
1.12x10^-12 = [1.3 + 2x]^2 [x].
Making the same approximation as above, 1.3 + 2x is very close to 1.3, and the equation becomes
1.12x10^-12 = [1.3]^2 [x] = 1.69x;
x = (1.12x10^-12)/1.69 = 6.63x10^-13 M is the solubility in 1.3 M AgNO3.