Question

Silver chromate is sparingly soluble in aqueous solutions. The K_sp of Ag₂CrO₄ is 1.12× 10^–12. What is the solubility (in mol/L) of silver chromate.

a) in 1.00 M potassium chormate aqueous solution?

b) in 1.00 M silver nitrate aqueous solution?

c) in pure water?

Answer 1

a)

K2CrO4 ---> 2K+ and CrO4-2

Common ion, CrO4-2 = 1.1 M

If

Ag2CrO4 <---< 2Ag+ and CrO4-2

Ksp = [Ag+]²[CrO4-2]

Since Ksp = 1.12*10^-12 and [CrO4-2] = 1.1; solve for Ag+

[Ag+] = sqrt(Ksp/([CrO4-2]))

[Ag+] = sqrt((1.12*10^-12 )/(1.1)) = 1.009*10^-6

[Ag+] = 1.01*10^-6

But we will have 2 moles of Ag+ per mol of Ag2CrO4 , divide by 2

[Ag2CrO4] = 1.01*10^-6 / 2 = 5*10^-7 M

[Ag2CrO4] =  5*10^-7 M

b)

If AgNO3 --> Ag+ and NO3-

[Ag+] = 1.1

Ksp = 1.12*10^-12

Ksp expresion:

Ksp = [Ag+]2[CrO4-2]

1.12*10^-12 = (1.1^2)([CrO4-2])

Solve for [CrO4-2]

[CrO4-2] = (1.12*10^-12) / (1.1^2) = 9.25*10^-13

Since 1 mol of CrO4-2 is present for each mol of Ag2CrO4, expect the same molarity

[Ag2CrO4] = 9.25*10^-13 mol per liter

c)

If in pure water:

Ag2CrO4 <-> 2Ag+ and CrO4-2

Ksp = 1.12*10^-12

Ksp = [Ag+]2[CrO4-2]

Assume [CrO4-2] = s and [Ag+] = 2s

Then Ksp:

Ksp = (2s)^2 (s) = 4s^3

1.12*10^-12 = 4 s^3

1.12/4 * 10^-12 = s^3

S = (2.8*10^-13)^(1/3)

s = 6.54*10^-5

[Ag2CrO4] = 6.54*10^-5 M

Answer 2

a)

Ag2CrO4 (s)   -------------> 2 Ag+ (aq) + CrO42- (aq)

                                            2 S                  1.00

Ksp = [2S]^2[1.00]

1.12× 10^–12 = (2S)^2 x 1

S = 5.29 x 10^-7

solubility = 5.29 x 10^-7 M

b)

Ag2CrO4 (s)   -------------> 2 Ag+ (aq) + CrO42- (aq)

                                            1.00                 S

Ksp = [1.00]^2[S]

1.12× 10^–12 = 1^2 x S

S = 1.12× 10^–12

solubility = 1.12× 10^–12 M

c)

Ag2CrO4 (s)   -------------> 2 Ag+ (aq) + CrO42- (aq)

                                            2S                 S

Ksp = [2S]^2[S]

1.12 x 10^-12 = 4S^3

S = 6.54 x 10^-5 M

solubility = 6.54 x 10^-5 M