silver chromate is sparingly soluble in aqueous solutions. The ksp of silver chromate is 1.12×10^-12?

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

1 Answer

  • 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.

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