Silver chromate is sparingly soluble in aqueous solutions. The Ksp 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?
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
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
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