Chemistry
Ksp
Solubility Equilibria & Ksp - NOTESa
Definitions
A. Solubility
o = an equilibrium position; e.g., 0.10 g / 100.0 mL H2O at 25oC.
(amount of solute that can dissolve in a solvent (H2O in this class))
B. Solubility Product Constant = Solubility Product (Ksp)
= an equilibrium constant (one value for a given solid at a given temperature)
C. Actual amount that can be dissolved in a solvent depends on temperature. The RATE at
which dissolution can be increased:
(a) increase temperature
(b) agitation
(c) increase surface-to-volume ratio
Applicationa
Tooth enamel = hydroxyapatite; Ca5(PO4)3OH)
Fluoride replaces hydroxyl group forming a stronger substance (fluorapatite; Ca5(PO4)3F)b:
SnF2 + Ca10(OH)2(PO4)6 + H2O  CaF2 + Sn2OHPO4 + Sn3F3PO4+ Ca(SnF3)2
When CaF2(s) is first added to water, it dissolves into the ions.
CaF2(s)  Ca2+(aq) + 2 F–(aq)
However, as more Ca2+(aq) and F–(aq) are formed, the chances increase that the two ions will
collide and re-form the solid CaF2. Eventually, the solution will become saturated when a
dynamic equilibrium is reached:
𝐻2 𝑂
CaF2(s) ⇔
Document1 (2/9/2016 11:18:00 AM)
Ca2+(aq) + 2 F–(aq)
Ksp = [Ca2+] [F–]2
Chemistry
Solubility - Introduction
p. 2
1. Will a Precipitate Form? c
Just as with dynamic equilibrium (Keq), we use the reaction quotient (Q) to determine the final
concentrations. If the Ksp is reached or exceeded, no more of the solid will dissolve.
A 275-mL sample of 6.0 x 10–3 M strontium nitrate, Sr(NO3)2, is mixed with 825-mL of 0.040 M
potassium chromate, K2CrO4. Will SrCrO4 precipitate form? (Ksp = 3.6 x 10–5)
1. First determine the concentrations because Ksp = [Sr2+] [CrO42–]
0.275 𝐿
a. nSr2+ = (
b. nCrO4– = (
0.825 𝐿
𝑥
𝑥
6.0 𝑥 10−3 𝑚𝑜𝑙
1𝐿
0.040 𝑥 𝑚𝑜𝑙
1𝐿
) / (0.275 + 0.825)𝐿 = 1.5 x 10–3 M mol Sr2+
) / (0.275 + 0.825)𝐿
= 3.0 x 10–2 M CrO42–
2. Is Ksp > Q (no precipitate)? Ksp = Q (just enough for form solid)? K=Q (saturated)?
a. Q = (1.5 x 10–3) (3.0 x 10–2) = 4.5 x 10–5
Ksp = 3.6 x 10–5
b. Ksp < Q. A SrCrO4 precipitate will form. (Just barely.)
Equilibrium curve for silver chloride.
AgCl(s) is in contact with Ag+ and Cl– ions in an aqueous
solution. The reaction quotient (Q) of the ion
concentrations is equal to the curve (Ksp) when
equilibrium exists. If Keq < Q, AgCl(s) tends to
precipitate until equilibrium is reached. If Keq > Q,
additional AgCl(s) will continue to dissolve.
Chemistry
Solubility - Introduction
p. 3
2. Determining the Equilibrium Concentrations of Ions
Just as with dynamic equilibrium (Keq), we use the reaction quotient (Q) to determine the final
concentrations in an equation.
Ca3(PO4)2 is a water-insoluble mineral used in commercial fertilizers. (Ksp = 1 x 10–33). What is
the concentration of Ca2+ in equilibrium with solid Ca3(PO4)2 if [PO4)3–] = 5 x 10–5 M?
1. Write the balanced chemical equation and the Ksp expression.
𝐻2 𝑂
a. 1 Ca3(PO4)2(s) ⇔
3 Ca2+(aq) + 2 PO43–(aq)
b. Ksp = 1 x 10–33 = [Ca2+]3 [PO43–]2
2. Rearrange to solve for the unknown [Ca2+]:
b. [Ca2+]3 =
1 𝑥 10−33
[𝑃𝑂43− ]2
1 𝑥 10−33
= (5 𝑥 10−5 )2 = 4 x 10–25
3
c. [Ca2+] = √4 𝑥 10−25
d. [Ca2+] = 7 x 10–9 M
Ca3(PO43–)2 is a solid so not included in Ksp
Chemistry
Solubility - Introduction
p. 4
3. Converting Between Solubility and Kspa
A. Converting Solubility to Kspa
Copper(I) bromide solubility: 2.0 x 10–4 mol/L. Calculate Ksp.
1. The solid initially comes in contact with water, there’s only CuBr(s) and H2O.
[Cu2+]0 = [Br–]0 = 0
2. The process of dissolving involves separating Cu+ and Br– ions.
𝐻2 𝑂
CuBr(s) ⇔
Cu2+(aq) + Br–(aq)
Ksp = [Cu2+] [Br–]
3. Ksp can be obtained from the measured solubility. This means 2.0 x 10–4 mol of solid/L
dissolves:
From the balanced chemical equation:
CuBr(s)  Cu+(aq) + Br–(aq)
2.0 x 10–4 mol/L CuBr(s)  2.0 x 10–4 mol/L Cu+(aq) + 2.0 x 10–4 mol/L Br–(aq)
4. Equilibrium concentrations:
R:
CuBr(s)
I:
–
C:
–
E:
–
Cu+(aq)
0
+ 2.0 x 10–4
2.0 x 10–4
Br–(aq)
0
+ 2.0 x 10–4
2.0 x 10–4
5. Ksp = [Cu2+] [Br–] = [2.0 x 10–4] [2.0 x 10–4] = 4.0 x 10–8
(continued)
Chemistry
Solubility - Introduction
p. 5
B. Converting Ksp to Solubilitya
Calculate the solubility of copper(II) iodate, Cu(IO3)2. Ksp = 1.4 x 10–7 at 25oC.
1. System initially contains H2O and solid Bi2S3: [Cu2+]0 = [IO3–]0 = 0
2. When the solid dissolves:
𝐻2 𝑂
Cu(IO3)2(s) ⇔
Cu2+(aq) + 2 IO3–(aq)
Ksp = [Cu2+] [IO3–]2
3. In order to calculate the solubility, we need to know the equilibrium concentrations of the
ions. Using ‘x’ for these:
x mol/L Cu(IO3)2(s)  (x mol Cu2+(aq)) + (2x mole/L IO3–(aq))
4. Equilibrium concentrations:
R:
Cu(IO3)2(s)
I:
–
C:
–
E:
–
5. Ksp = [Cu2+] [IO3–]2 = [x] [2x]
IO3–
Cu2+
0
+x
x
0
+ 2x
2x
1.4 x 10–7 = (x)(2x)2 = 4x3
3.5 x 10–8 = x3
3
x = √3.5 𝑥 10−8
x = 3.3 x 10–3 M
6. Thus, the solubility of Cu(IO3)2 in water at 25oC is 3.3 x 10–3 mol/L.
Chemistry
Solubility - Introduction
p. 6
4. Common Ion Effectd
So far, we have considered only one solute in a solution. However, we now consider when
another solute – when it has one of the same ions as the first solute - is added to the solution.
The solubility of the solute changes. The basis for this is Le Châtelier's principle because the
common ion can be removed (e.g., precipitation) or added.
An attempt is made to dissolve lead (II) chloride in a solution of 0.100 M sodium chloride. What
is the equilibrium concentration of Pb2+?
A. What is the lead (II) concentration in a saturated solution of lead(II) chloride?
(Ksp = 1.7 x 10–5)
Pb2+(aq) + 2 Cl–(aq)
1. PbCl2(s)
Ksp = [Pb2+] [Cl–]2
2. We know Ksp but we don’t know the concentrations of the ions. Set s = [Pb2+]
1.7 x 10–5 = (s) (2s)2
(2s from the balanced chemical equation)
1.7 x 10–5 = 4s3
s = 1.62 x 10–2 M
 [Pb2+] = 1.62 x 10–2 M
B. What will happen to the solubility of PbCl2 if additional Cl– ions are added? (0.100 M NaCl)
1. We can assume that the original concentration of Cl– ions is negligible compared with
that from the NaCl because Ksp is very small. (5% rule)
Ksp
= [Pb2+] [Cl–]2
1.7 x 10–5 = [Pb2+] [0.100]2
[Pb2+] =
1.7 𝑥 10−5
0.0100
[Pb2+] = 1.7 x 10–3 M
C.
[Pb2+] in Original Solution
0.0162 M
[Pb2+] in 0.100 M NaCl Solution
0.0017 M
Thus, adding NaCl decreased the solubility of PbCl2 by a factor of ~10.
The solubility would be decreased even more if the NaCl solution was more
concentrated.
Chemistry
(a)
(b)
(c)
(d)
Solubility - Introduction
Zumdahl, 9th ed. 2014.
http://www.ncbi.nlm.nih.gov/pubmed/281374
Materton, 6th ed. 2009.
http://www.chemguide.co.uk/physical/ksp/commonion.html
p. 7