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.

1
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 CnxAmy
• 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 Bi32S23
• 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)
• What is the Ksp for this reaction?
• Ksp Ag2CrO42

3
SOLUBILITY PRODUCT CONSTANT

• 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 Ca2F2 2.7x1011 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 2.7x1011 Ca2F2 (x)(2x)2 4x3
• x3 2.7x1011 / 4 6.75x1012
• F 2x 3.8x104 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 1.9x104 M.

6
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 Châteliers principle.
• CaF2(s) Ca2(aq) 2F(aq)
• Ksp Ca2F2 2.7x1011 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
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 2.7x1011 Ca2F2 Ca2(12)
• Ca2 2.7x1011 M
• Compared to the previous example, did the
  concentration of Ca2(aq) increase or decrease?
• It decreased from 1.9x104 M to 2.7x1011 M.
• Did the concentration of F(aq) increase or
  decrease?
• It increased from 3.8x104 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 todays 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 KOOC(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 todays 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. 2004. Chemistry, 4th ed.
  Upper Saddle River, NJ Prentice Hall.
• Petrucci, R.H. 1985. General Chemistry Principles
  and Modern Applications, 4th ed. New York, NY
  Macmillan Publishing Company.
• Traverso M. 2006. Titration using Phenolphthalein
  as an Indicator. Available www.chemistry.wustl.ed
  u/.../AcidBase/phph.htm accessed 14 September
  2006.