However, the compounds that we have referred to as “insoluble” in the past are realistically “slightly soluble.” Everything is at least slightly soluble in water.
Note: Every reaction considered here is assumed to be at 25oC.
Solubility Constant = Ksp = really just another version of Kc. It is the equilibrium constant for the solubility equilibrium of a slightly soluble (or nearly insoluble) ionic compound.
Example:
AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)
Kc = [Ag+][Cl-] = Ksp
Remember not to include the [AgCl] as part of the Ksp because it is a solid, so just like before, it is ignored.
It's important to be able to recognize which ions an ionic compound will break up into. Once the equilibrium is written, it's easy to find the Ksp. Here's a reminder of the polyatomic ions:
Examples:
CaSO4 (s) ⇌ Ca2+ (aq) + SO42- (aq)
Ksp = [Ca2+][SO42-]
AgC2H3O2 (s) ⇌ Ag+ (aq) + C2H3O2- (aq)
Ksp = [Ag+][C2H3O2-]
CaF2 (s) ⇌ Ca2+ (aq) + 2F- (aq)
Ksp = [Ca2+][F-]2
Ca3(PO4)2 (s) ⇌ 3Ca2+ (aq) + 2PO43- (aq)
Ksp = [Ca2+]3[PO43-]2
Zn(OH)2 (s) ⇌ Zn2+ (aq) + 2OH- (aq)
Ksp = [Zn2+][OH-]2
Solubility is generally expressed in grams per liter: g/L. Molar solubility is expressed in mols per liter: mols/L. Note that this is the same thing as molarity.
Example:
Solubility of AgCl = 1.9 x 10-3 g/L
What is the Ksp?
The values that are needed to find the Ksp are expressed in Molarity, which is mols per liter, and we have grams per liter. So the first step is to convert from grams to mols using the periodic table.
For reference:
[AgCl] = (1.9 x 10-3 g / 1 L) x (1 mol AgCl / 143 g AgCl) = 1.33 x 10-5 mol/L AgCl
AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)
Ksp = x2
Since we know that 1.33 x 10-5 M of AgCl will dissociate into 1.33 x 10-5 M of Ag+ and 1.33 x 10-5 M of Cl-, we know that x = 1.33 x 10-5 mol/L.
Ksp = (1.33 x 10-5)2 = 1.77 x 10-10
In some problems, we won't be given the solubility directly when we are asked to find the Ksp.
Example:
One liter of a solution saturated with TcF is evaporated, leaving 5.32 x 10-5g of TcF. What is the Ksp?
TcF (s) ⇌ Tc+ (aq) + F- (aq)
Ksp = [Tc+][F-]
We have been given the information that for one liter of solution, we have 5.32 x 10-5g of TcF, which is solubility in g/L. So the first step is to convert from grams to mols:
( 5.32 x 10-5 g / 1L) x (1 mol TcF / 117 g TcF) = 4.55 x 10-7 M TcF
And we don't need to build a table to see that 4.55 x 10-7 M of TcF will dissociate into 4.55 x 10-7 M of Tc+ and 4.55 x 10-7 M of F-. Also, that Ksp = (4.55 x 10-7)2 = 2.07 x 10-13
Everything up to this point has been pretty simple because everything has been 1:1. Remember that when ions have coefficients associated with them, these coefficients show up not only in the calculation for Ksp as exponents, but also in the table we'll build to solve for x.
Example:
One liter of a solution saturated with W2S3 is evaporated, leaving 9.35 x 10-4g of W2S3. What is the Ksp?
The first step is to convert from grams to mols:
(9.35 x 10-4 g / L) x (1 mol W2S3 / 464 g W2S3) = 2.02 x 10-6 M W2S3
W2S3 (s) ⇌ 2W3+ (aq) + 3S2- (aq)
Ksp = [W3+]2[S2-]3
Ksp = (2x)2(3x)3
We know 2.02 x 10-6 M W2S3 dissociates, so we know x = 2.02 x 10-6
Ksp = (2 x 2.02 x 10-6)2(3 x 2.02 x 10-6)3 = 3.63 x 10-27
Other problems may give the Ksp and ask for the solubility.
Example:
Find the molar solubility and the solubility of Pb3(AsO4)2. The Ksp for Pb3(AsO4)2 = 4 x 10-36
Pb3(AsO4)2 (s) ⇌ 3Pb2+ (aq) + 2AsO43- (aq)
Ksp = [Pb2+]3[AsO43-]2 = (3x)3(2x)2 = (27x3)(4x2) = 108x5 = 4 x 10-36
(divide by 108 and take the fifth root)
x = 3.26 x 10-8 M = molar solubility
To find solubility, use the periodic table to convert this value to grams per liter:
(3.26 x 10-8 mols / L) x (899 g Pb3(AsO4)2/ mol Pb3(AsO4)2) =
2.93 x 10-5 g / L Pb3(AsO4)2
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