Question

50 mL of 0.060 M K₂CrO₄ is mixed with 50 mL of 0.080 M AgNO₃. Calculate the following:

a. The solubility of Ag₂CrO₄ (K_sp = 1.9 X 10^-12) in the solution in moles per liter.

b. The concentrations of the following ions Ag⁺, CrO₄^-2, K⁺, and NO₃^-.

Answer 1

50 mL of 0.060 M K2CrO4 is mixed with 50 mL of 0.080 M AgNO3. Calculate the following:
After mixing, [K2CrO4] = total number of moles / total volume
                                              = 50*0.06 / (50+50)
                                              = 0.03 M

So, [K+] = 2*[K2CrO4] = 2*0.03M = 0.06 M
[[CrO4 2- ] = [K2CrO4] = 0.03 M

After mixing, [AgNO3] = total number of moles / total volume
                                              = 50*0.08 / (50+50)
                                              = 0.04 M
[Ag+] = [AgNO3] = 0.04 M
[NO3-] = [AgNO3] = 0.04 M

a)
                                                Ag2CrO4 ---> 2Ag+ + CrO42-
initial:                                                                 0.04         0.03
at eqilibrium:                                               0.04+s        0.03+s
Qsp = [Ag+]^2 [CrO42-] = (0.04)^2 *0.03 = 4.8*10^-5
This is greater than Ksp. So Ag2CrO4 will not be soluble.
so solubility = 0

b)
As calculated above:
, [K+] = 2*[K2CrO4] = 2*0.03M = 0.06 M
[CrO4 2- ] = [K2CrO4] = 0.03 M
[Ag+] = [AgNO3] = 0.04 M
[NO3-] = [AgNO3] = 0.04 M