1. 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.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:For a saturated solution of ionic solid in water: C x A y(s) = xC n+ (aq) + yA m– (aq) Solubility Product Constant = K sp = [C n+ ] x [A m– ] ySolubility Product Constant = K sp = [C n+ ] x [A m– ] y WhereWhere C x A y(s) is a slightly soluble ionic solid.C x A y(s) is a slightly soluble ionic solid. [C n+ ] and [A m– ] are the equilibrium concentrations of ions in moles per liter.[C n+ ] and [A m– ] are the equilibrium concentrations of ions in moles per liter. x and y are the stoichiometric coefficients from the balance reaction.x and y are the stoichiometric coefficients from the balance reaction. The solubility product constant (K sp ) is the equilibrium constant for the dissolution of a slightly soluble ionic compound at a specified temperature.The solubility product constant (K sp ) is the equilibrium constant for the dissolution of a slightly soluble ionic compound at a specified temperature. By convention the “concentration” of the solid, C x A y, is NOT used to calculate K sp. (That is, the activity of a pure solid is 1.)By convention the “concentration” of the solid, C x A y, is NOT used to calculate K sp. (That is, the activity of a pure solid is 1.) CH 104: DETERMINATION OF A SOLUBILITY PRODUCT CONSTANT

2. What is the reaction for a saturated solution of Bi 2 S 3(s) in water?What is the reaction for a saturated solution of Bi 2 S 3(s) in water? Bi 2 S 3(s) = 2Bi 3+ (aq) + 3S 2 – (aq) What is the K sp for this reaction?What is the K sp for this reaction? K sp = [Bi 3+ ] 2 [S 2– ] 3 Notice the “concentration” of Bi 2 S 3(s) is NOT used to calculate K sp.Notice the “concentration” of Bi 2 S 3(s) is NOT used to calculate K sp. What is the reaction for a saturated solution of Ag 2 CrO 4(s) in water?What is the reaction for a saturated solution of Ag 2 CrO 4(s) in water? Ag 2 CrO 4(s) = 2Ag + (aq) + CrO 4 2 – (aq) What is the K sp for this reaction?What is the K sp for this reaction? K sp = [Ag + ] 2 [CrO 4 2 – ] SOLUBILITY PRODUCT CONSTANT

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

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

5. Step #3: Use the K sp equation to solve for [Ca 2+ ] and [F – ].Step #3: Use the K sp equation to solve for [Ca 2+ ] and [F – ]. K sp = 2.7x10 –11 = [Ca 2+ ][F – ] 2 = (x)(2x) 2 = 4x 3 x 3 = 2.7x10 –11 / 4 = 6.7 5 x10 –12 [F – ] = 2x = 3.8x10 –4 M Step #4: Solve for the solubility of CaF 2(s).Step #4: Solve for the solubility of CaF 2(s). One mole of Ca 2+ (aq) is produced for every mole of CaF 2(s) that dissolves; therefore, the solubility of CaF 2(s) = [Ca 2+ ] = 1.9x10 –4 M. CALCULATING SOLUBILITY FROM K sp

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

7. The common ion F – (aq) is added to a saturated solution of CaF 2(s) in distilled water.What is the concentration of Ca 2+ (aq) in equilibrium with 1.0 M F – (aq) and CaF 2(s) at 25° C?The common ion F – (aq) is added to a saturated solution of CaF 2(s) in distilled water. What is the concentration of Ca 2+ (aq) in equilibrium with 1.0 M F – (aq) and CaF 2(s) at 25° C? K sp = 2.7x10 –11 = [Ca 2+ ][F – ] 2 = [Ca 2+ ](1 2 ) [Ca 2+ ] = 2.7x10 –11 M Compared to the previous example, did the concentration of Ca 2+ (aq) increase or decrease?Compared to the previous example, did the concentration of Ca 2+ (aq) increase or decrease? It decreased from 1.9x10 –4 M to 2.7x10 –11 M.It decreased from 1.9x10 –4 M to 2.7x10 –11 M. Did the concentration of F – (aq) increase or decrease?Did the concentration of F – (aq) increase or decrease? It increased from 3.8x10 –4 M to 1.0 M.It increased from 3.8x10 –4 M to 1.0 M. Does this agree with the common ion effect?Does this agree with the common ion effect? Yes. The concentration of Ca 2+ (aq) decreased. The concentration of F – (aq) increased. And the solubility of CaF 2(s) decreased.Yes. The concentration of Ca 2+ (aq) decreased. The concentration of F – (aq) increased. And the solubility of CaF 2(s) decreased. THE COMMON ION EFFECT

8. Common ions decrease the solubility of ionic solids.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”.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.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. THE SALT EFFECT

9. The salt effect is not as striking as the common ion effect.The salt effect is not as striking as the common ion effect. The presence of the common ion CrO 4 2 – (aq), from K 2 CrO 4, decreases the solubility of Ag 2 CrO 4 by a factor of 35.The presence of the common ion CrO 4 2 – (aq), from K 2 CrO 4, decreases the solubility of Ag 2 CrO 4 by a factor of 35. In contrast, the presence of the uncommon ions K + (aq) and NO 3 – (aq), from KNO 3, increase the solubility of Ag 2 CrO 4 by a factor of only 0.25.In contrast, the presence of the uncommon ions K + (aq) and NO 3 – (aq), from KNO 3, increase the solubility of Ag 2 CrO 4 by a factor of only 0.25. THE SALT EFFECT

10. In today’s experiment you will measure the solubility of potassium hydrogen tartrate (KOOC(CHOH) 2 COOH).In today’s experiment you will measure the solubility of potassium hydrogen tartrate (KOOC(CHOH) 2 COOH). KOOC(CHOH) 2 COOH(s) = K + (aq) + – OOC(CHOH) 2 COOH (aq) What is the K sp for this reaction?What is the K sp for this reaction? K sp = [K + ][ – OOC(CHOH) 2 COOH] You will make a saturated solution of KOOC(CHOH) 2 COOH in 0.10 M NaCl and in 0.10 M KNO 3.You will make a saturated solution of KOOC(CHOH) 2 COOH in 0.10 M NaCl and in 0.10 M KNO 3. Is the NaCl a source of a common ion or uncommon ions?Is the NaCl a source of a common ion or uncommon ions? Na + and Cl – are uncommon ions.Na + and Cl – are uncommon ions. This NaCl should increase or decrease the solubility of KOOC(CHOH) 2 COOH?This NaCl should increase or decrease the solubility of KOOC(CHOH) 2 COOH? Increase.Increase. Is the KNO 3 a source of a common ion or uncommon ions?Is the KNO 3 a source of a common ion or uncommon ions? K + is a common ion. NO 3 – is an uncommon ion.K + is a common ion. NO 3 – is an uncommon ion. This KNO 3 should increase or decrease the solubility of KOOC(CHOH) 2 COOH?This KNO 3 should increase or decrease the solubility of KOOC(CHOH) 2 COOH? Decrease. The common ion effect is usually greater than the uncommon ion effect.Decrease. The common ion effect is usually greater than the uncommon ion effect. CALCULATING K sp FROM SOLUBILITY

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

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

13. Give at least 1 safety concern for the following procedure.Give at least 1 safety concern for the following procedure. Using KOOC(CHOH) 2 COOH, NaCl, KNO 3, NaOH, and phenolphthalein.Using KOOC(CHOH) 2 COOH, NaCl, KNO 3, 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.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.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.Ask your instructor if you ever have any questions about safety.SAFETY

14. SOURCES McMurry, J., R.C. Fay. 2004. Chemistry, 4th ed. Upper Saddle River, NJ: Prentice Hall.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.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.edu/.../AcidBase/phph.htm [accessed 14 September 2006].Traverso M. 2006. Titration using Phenolphthalein as an Indicator. Available: www.chemistry.wustl.edu/.../AcidBase/phph.htm [accessed 14 September 2006]. www.chemistry.wustl.edu/.../AcidBase/phph.htm