Slide 1 / 39 Slide 2 / 39 AP Chemistry Equilibrium Part C : Solubility Equilibrium 2014-10-29 www.njctl.org Slide 3 / 39 Slide 4 / 39 Table of Contents click on the topic to go to that section Molar Solubility Calculating Ksp Molar Solubility Le-Chatlier and Solubility Equilibrium Return to Table of Contents Slide 5 / 39 Slide 6 / 39 Solubility Equilibrium Solubility Equilibrium Ionic compounds dissociate into their ions to different degrees when placed in water and reach equilibrium with the non-dissociated solid phase when the solution is saturated. A saturated solution of CaCO 3 (s) Answer Ca 2+ Ca 2+ CO 32- CO 32- Many shells are made of relatively insoluble calcium carbonate, so CaCO 3 (s) the shells are not at huge risk of dissolving in the ocean. Class Question: Calcium carbonate is a relatively insoluble ionic salt. Would the picture look different for a soluble ionic salt such as Na 2 CO 3 ? Which solution would be the better electrolyte?
Slide 7 / 39 Slide 8 / 39 Molar Solubility Solubility Equilibrium The molar solubility of an ionic salt is the molar equivalent (mol/L) of The degree to which an ionic compound dissociates in water can be the solid that has dissociated into its ions. determined by measuring it's "K sp " or solubility product equilibrium constant. The molar solubility can be determined either by: 1. Measuring the concentration of ions in solution directly CaCO 3 (s) --> Ca 2+ (aq) + CO 32- (aq) Ksp @ 25 C = 5.0 x 10 -9 Answer Answer or by ... MgCO 3 (s) --> Mg 2+ (aq) + CO 32- (aq) Ksp @ 25 C = 6.8 x 10 -6 2. Using the equilibrium constant to first calculate the concentration of ions and thereby use this to find the molar solubility. In both cases above, the equilibrium lies far to the left, meaning relatively few aqueous ions would be present in solution. Class Question: Using the K sp values on the prior slide, which carbonate (MgCO 3 or CaCO 3 ) would have the higher molar Class Question: Which saturated solution above would have the solubility and why? higher conductivity and why? Slide 9 / 39 Slide 10 / 39 Molar Solubility Molar Solubility Calculating the molar solubility from ion concentrations simply Calculating the molar solubility from an equilibrium constant requires involves using stoichiometrical ratios. writing and using an equilibrium expression. Example: What is the molar solubility of a saturated aqueous Example: What is the molar solubility of a saturated silver carbonate solution in which the [Ag + ] = 2.4 x 10 -8 ? solution of PbI 2 ? (Ksp @25 C = 1.39 x 10 -8 ) Answer Ag 2 CO 3 (s) --> 2Ag + (aq) + CO 32- (aq) PbI 2 (s) --> Pb 2+ (aq) + 2I - (aq) For every 2 silver ions in solution, 1 Ag 2 CO 3 would have been Ksp = 1.39 x 10 -8 = [Pb 2+ ][I - ] 2 required to dissociate. Since neither ion concentration is known, we will substitute "x" So... 2.4 x 10 -8 M Ag + x 1 M Ag 2 CO 3 = 1.2 x 10 -8 M for the [Pb 2+ ] and "2x" for the [I - ]. 2 M Ag+ 1.39 x 10 -8 = (x)(2x) 2 = 4x 3 "x" = [Pb 2+ ] = 1.51 x 10 -3 M Class Question: What would one need to know to find the number of grams of silver carbonate that were dissolved? Since 1 Pb 2+ required 1 PbI 2 , the molar solubility of the PbI 2 (s) = Then, explain how this would be calculated. 1.51 x 10 -3 M. Slide 11 / 39 Slide 11 (Answer) / 39 1 When 30 grams of NaCl are dissolved into 100 mL of 1 When 30 grams of NaCl are dissolved into 100 mL of distilled water, all of the solid NaCl has dissolved. The distilled water, all of the solid NaCl has dissolved. The solution must be saturated and the Ksp for the NaCl must solution must be saturated and the Ksp for the NaCl must be very high. be very high. True True Answer False False B or D [This object is a pull tab]
Slide 12 / 39 Slide 12 (Answer) / 39 2 In two separate beakers with identical volumes of water, 2 In two separate beakers with identical volumes of water, 10 grams of CuCO 3 was added to beaker 1 and 10 grams 10 grams of CuCO 3 was added to beaker 1 and 10 grams of PbCO 3 was added to beaker 2. After stirring, the of PbCO 3 was added to beaker 2. After stirring, the contents of each beaker were poured through filter paper contents of each beaker were poured through filter paper and dried. The mass of the the solid retained in the filter and dried. The mass of the the solid retained in the filter paper from beaker 1 was 9.995 grams while that from paper from beaker 1 was 9.995 grams while that from beaker 2 was 9.992 grams. It can be deduced that beaker 2 was 9.992 grams. It can be deduced that PbCO 3 has the smaller K sp . PbCO 3 has the smaller K sp . Answer False True True False False [This object is a pull tab] Slide 13 / 39 Slide 13 (Answer) / 39 3 Which of the following ionic salts would have the highest 3 Which of the following ionic salts would have the highest molar solubility? molar solubility? A NiCO 3 (s) Ksp = 6.61 x 10 -9 A NiCO 3 (s) Ksp = 6.61 x 10 -9 B MnCO 3 (s) Ksp = 1.82 x 10 -11 B MnCO 3 (s) Ksp = 1.82 x 10 -11 Answer D C ZnCO 3 (s) Ksp = 1.45 x 10 -11 C ZnCO 3 (s) Ksp = 1.45 x 10 -11 D Ag 2 CrO 4 (s) Ksp = 9.00 x 10 -12 D Ag 2 CrO 4 (s) Ksp = 9.00 x 10 -12 E All have the same molar solubility E All have the same molar solubility [This object is a pull tab] Slide 14 / 39 Slide 14 (Answer) / 39 4 A student places 4.000 grams of PbI 2 (s) into distilled 4 A student places 4.000 grams of PbI 2 (s) into distilled water to produce a 200.0 mL solution. After stirring, the water to produce a 200.0 mL solution. After stirring, the student filters the solution and dries the solid. What would student filters the solution and dries the solid. What would be expected mass of the solid PbI 2 on the filter paper be expected mass of the solid PbI 2 on the filter paper after drying? after drying? Answer 3.86 g [This object is a pull tab]
Slide 15 / 39 Slide 15 (Answer) / 39 5 The conductivity of a saturated solution of Ag 2 CO 3 would 5 The conductivity of a saturated solution of Ag 2 CO 3 would be expected to be less than the conductivity of a be expected to be less than the conductivity of a saturated solution of CaCO 3 . saturated solution of CaCO 3 . True True False False Answer False [This object is a pull tab] Slide 16 / 39 Slide 17 / 39 Calculating Ksp To find the Ksp, the concentrations of at least one of the ions must be known and an equilibrium expression must be used. Example: What is the Ksp of Fe(OH) 3 (s) if a saturated solution of it has a pH of 11.3? Calculating Ksp Answer Fe(OH) 3 (s) --> Fe 3+ (aq) + 3OH - (aq) pH = 11.3 --> pOH = 2.7 --> [OH-] = 1.99 x 10 -3 Ksp = [Fe 3+ ][OH - ] 3 = (6.65 x 10 -4 )(1.99 x 10 -3 ) 3 = 5.24 x 10 -12 Note that the [Fe 3+ ] is 1/3 that of the [OH - ] Return to Table of Class Question: If an acid was added and reacted with some of Contents the hydroxide ion, would the Ksp increase, decrease, or remain the same? Slide 18 / 39 Slide 19 / 39 Common Ion Effect Common Ion Effect If one of the ions that is part of the solubility equilibria is either Finding the molar solubility when a common ion is present involves already present or added, the equilibria will shift accordingly thereby writing an equilibrium expression. altering the equilibrium position. Example: The pictures below represent MgCO 3 (s) being added to distilled water What is the molar solubility of AgI in a 0.041 M solution of MgI 2 ? and to a solution that is 0.1 M Na 2 CO 3 . AgI(s) --> Ag + (aq) + I - (aq) Distilled Water 0.1 M Na 2 CO 3 MgCO 3 Ksp (look-up) = 1.5 x 10 -16 = [Ag+][I-] = [x][0.082 + x] CO 32- Na + Since the equilibria constant is so small, we will expect the change Na + in the [I-] will be negligible compared to the amount already present. MgCO 3 (s) --> Mg 2+ (aq) + CO 32- (aq) MgCO 3 (s) --> Mg 2+ (aq) + CO 32- (aq) 1.5 x 10 -16 = [x][0.082] x = [Ag+] = 1.83 x 10 -15 M The presence of the carbonate ion (common ion) in the 0.1 M Na 2 CO 3 solution shifts the equilibria to the left, diminishing the Since 1 AgI is needed to produce 1 Ag + , the molar solubility of AgI solubility of the MgCO 3 . in this solution is 1.83 x 10 -15 M.
Recommend
More recommend
