1. SOLUBILITY PRODUCT CONSTANT CxAy(s) = xCn+(aq) + yAm–(aq)
CH 104: DETERMINATION OF A
SOLUBILITY PRODUCT CONSTANT
A solution is saturated if it has a solid at equilibrium with its solute. The concentration of solute in a saturated solution is the largest that is normally possible.
For a saturated solution of ionic solid in water:
CxAy(s) = xCn+(aq) + yAm–(aq)
Solubility Product Constant = Ksp = [Cn+]x[Am–]y
Where
CxAy(s) is a slightly soluble ionic solid.
[Cn+] and [Am–] are the equilibrium concentrations of ions in moles per liter.
x and y are the stoichiometric coefficients from the balance reaction.
The solubility product constant (Ksp) is the equilibrium constant for the dissolution of a slightly soluble ionic compound at a specified temperature.
By convention the “concentration” of the solid, CxAy, is NOT used to calculate Ksp. (That is, the activity of a pure solid is 1.)

2. SOLUBILITY PRODUCT CONSTANT
What is the reaction for a saturated solution of Bi2S3(s) in water?
Bi2S3(s) = 2Bi3+(aq) + 3S2–(aq)
What is the Ksp for this reaction?
Ksp = [Bi3+]2[S2–]3
Notice the “concentration” of Bi2S3(s) is NOT used to calculate Ksp.
What is the reaction for a saturated solution of Ag2CrO4(s) in water?
Ag2CrO4(s) = 2Ag+(aq) + CrO42–(aq)
Ksp = [Ag+]2[CrO42–]

3. SOLUBILITY PRODUCT CONSTANT CaF2(s) = Ca2+(aq) + 2F–(aq)
Calcium fluoride (CaF2) is slightly soluble in water. In a saturated solution the CaF2(s) is dissolving at the same rate that Ca2+(aq) and F–(aq) crystallize. That is, the solid and solute are at equilibrium.
CaF2(s) = Ca2+(aq) + 2F–(aq)

4. CALCULATING SOLUBILITY FROM Ksp
A saturated solution is made by adding excess CaF2(s) to distilled water. What is the solubility of this CaF2(s) at 25° C?
Step #1: Write the balanced reaction and Ksp equation.
CaF2(s) = Ca2+(aq) + 2F–(aq)
Ksp = [Ca2+][F–]2 = 2.7x10–11 at 25° C
Step #2: The initial concentrations of Ca2+(aq) and F–(aq) are 0. The equilibrium concentrations of Ca2+(aq) and F–(aq) are given algebraic variables based on the stoichiometric coefficients from the balance reaction. Write these equilibrium concentrations of Ca2+(aq) and F–(aq).
[Ca2+] = x
[F–] = 2x

5. CALCULATING SOLUBILITY FROM Ksp
Step #3: Use the Ksp equation to solve for [Ca2+] and [F–].
Ksp = 2.7x10–11 = [Ca2+][F–]2 = (x)(2x)2 = 4x3
x3 = 2.7x10–11 / 4 = 6.75x10–12
[F–] = 2x = 3.8x10–4 M
Step #4: Solve for the solubility of CaF2(s).
One mole of Ca2+(aq) is produced for every mole of CaF2(s) that dissolves; therefore,
the solubility of CaF2(s) = [Ca2+] = 1.9x10–4 M.

6. CaF2(s) = Ca2+(aq) + 2F–(aq) Ksp = [Ca2+][F–]2 = 2.7x10–11 at 25° C
THE COMMON ION EFFECT
In the previous example the pure solid (CaF2(s)) was the only source of its dissolved ions (Ca2+(aq) and F–(aq)).
However, if the common ion F–(aq) is added it will react with Ca2+(aq) to decrease the solubility of CaF2(s). The new concentration of Ca2+(aq) is less than in the original equilibrium. And the new concentration of F–(aq) is greater than in the original equilibrium. This is called Le Châtelier’s principle.
CaF2(s) = Ca2+(aq) + 2F–(aq)
Ksp = [Ca2+][F–]2 = 2.7x10–11 at 25° C
Similarly, if the common ion Ca2+(aq) is added it will react with F–(aq) to the solubility of CaF2(s). The new concentration of F–(aq) is than in the original equilibrium. And the new concentration of Ca2+(aq) is than in the original equilibrium.
decrease
less
greater

7. Ksp = 2.7x10–11 = [Ca2+][F–]2 = [Ca2+](12)
THE COMMON ION EFFECT
The common ion F–(aq) is added to a saturated solution of CaF2(s) in distilled water. What is the concentration of Ca2+(aq) in equilibrium with 1.0 M F–(aq) and CaF2(s) at 25° C?
Ksp = 2.7x10–11 = [Ca2+][F–]2 = [Ca2+](12)
[Ca2+] = 2.7x10–11 M
Compared to the previous example, did the concentration of Ca2+(aq) increase or decrease?
It decreased from 1.9x10–4 M to 2.7x10–11 M.
Did the concentration of F–(aq) increase or decrease?
It increased from 3.8x10–4 M to 1.0 M.
Does this agree with the common ion effect?
Yes. The concentration of Ca2+(aq) decreased. The concentration of F–(aq) increased. And the solubility of CaF2(s) decreased.

8. THE SALT EFFECT
Common ions decrease the solubility of ionic solids.
In contrast, the presence of “uncommon” ions tends to increase solubility of ionic solids. This is called the “salt effect”, the “uncommon ion effect”, or the “diverse ion effect”.
Soluble uncommon ions increase the interionic attractions of a solution. As a result, these uncommon ions decrease the effective concentrations (or activities) of other solutes and increase the solubility of ionic solids.

9. THE SALT EFFECT
The salt effect is not as striking as the common ion effect.
The presence of the common ion CrO42–(aq), from K2CrO4, decreases the solubility of Ag2CrO4 by a factor of 35.
In contrast, the presence of the uncommon ions K+(aq) and NO3–(aq), from KNO3, increase the solubility of Ag2CrO4 by a factor of only 0.25.

10. CALCULATING Ksp FROM SOLUBILITY
In today’s experiment you will measure the solubility of potassium hydrogen tartrate (KOOC(CHOH)2COOH).
KOOC(CHOH)2COOH(s) = K+(aq) + –OOC(CHOH)2COOH(aq)
What is the Ksp for this reaction?
Ksp = [K+][–OOC(CHOH)2COOH]
You will make a saturated solution of KOOC(CHOH)2COOH in 0.10 M NaCl and in 0.10 M KNO3.
Is the NaCl a source of a common ion or uncommon ions?
Na+ and Cl– are uncommon ions.
This NaCl should increase or decrease the solubility of KOOC(CHOH)2COOH?
Increase.
Is the KNO3 a source of a common ion or uncommon ions?
K+ is a common ion. NO3– is an uncommon ion.
This KNO3 should increase or decrease the solubility of KOOC(CHOH)2COOH?
Decrease. The common ion effect is usually greater than the uncommon ion effect.

11. CALCULATING Ksp FROM SOLUBILITY
In today’s experiment you will measure the concentration of –OOC(CHOH)2COOH(aq) by titration with standardized sodium hydroxide (NaOH) to a phenolphthalein endpoint.
Potassium hydrogen tartrate is a monoprotic acid; that is, only 1 hydrogen will be neutralized by titration with NaOH.
–OOC(CHOH)2COOH(aq) + OH–(aq) →
–OOC(CHOH)2COO–(aq) + H2O(l)

12. TITRATION USING PHENOLPHTHALEIN AS AN INDICATOR
Stop adding base when the indicator just begins to turn a faint but stable pink. This is the endpoint.

13. SAFETY
Give at least 1 safety concern for the following procedure.
Using KOOC(CHOH)2COOH, NaCl, KNO3, NaOH, and phenolphthalein.
These are irritants. Wear your goggles at all times. Immediately clean all spills. If you do get either of these in your eye, immediately flush with water.
Your laboratory manual has an extensive list of safety procedures. Read and understand this section.
Ask your instructor if you ever have any questions about safety.

14. SOURCES
McMurry, J., R.C. Fay Chemistry, 4th ed. Upper Saddle River, NJ: Prentice Hall.
Petrucci, R.H General Chemistry Principles and Modern Applications, 4th ed. New York, NY: Macmillan Publishing Company.
Traverso M Titration using Phenolphthalein as an Indicator. Available: [accessed 14 September 2006].