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### Transcript of A solution is saturated if it has a solid at equilibrium with its solute. The concentration of...

• Slide 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
• Slide 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
• Slide 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
• Slide 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
• Slide 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
• Slide 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 Chteliers 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 Chteliers 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
• Slide 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
• Slide 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
• Slide 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
• Slide 10
• In todays experiment you will measure the solubility of potassium hydrogen tartrate (KOOC(CHOH) 2 COOH).In todays 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 increas