Slide 1 / 39 AP Chemistry Slide 2 / 39 Equilibrium Part C : Solubility Equilibrium 2014-10-29 www.njctl.org Table of Contents click on the topic to go to that section Slide 3 / 39 Molar Solubility Calculating Ksp Le-Chatlier and Solubility Equilibrium
Slide 4 / 39 Molar Solubility Return to Table of Contents Solubility Equilibrium Slide 5 / 39 Many shells are made of relatively insoluble calcium carbonate, so the shells are not at huge risk of dissolving in the ocean. Solubility Equilibrium Slide 6 / 39 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) Ca 2+ Ca 2+ CO 3 CO 3 CaCO 3(s) Class Question: Calcium carbonate is a relatively insoluble ionic salt. Would the picture look different for a soluble ionic salt such as Na 2CO 3? Which solution would be the better electrolyte?
Solubility Equilibrium Slide 7 / 39 The degree to which an ionic compound dissociates in water can be determined by measuring it's "K sp" or solubility product equilibrium constant. CaCO 3(s) --> Ca 2+ (aq) + CO 3 (aq) Ksp @ 25 C = 5.0 x 10-9 MgCO 3(s) --> Mg 2+ (aq) + CO 3 (aq) Ksp @ 25 C = 6.8 x 10-6 In both cases above, the equilibrium lies far to the left, meaning relatively few aqueous ions would be present in solution. Class Question: Which saturated solution above would have the higher conductivity and why? Molar Solubility Slide 8 / 39 The molar solubility of an ionic salt is the molar equivalent (mol/l) of the solid that has dissociated into its ions. The molar solubility can be determined either by: 1. Measuring the concentration of ions in solution directly or by... 2. Using the equilibrium constant to first calculate the concentration of ions and thereby use this to find the molar solubility. Class Question: Using the K sp values on the prior slide, which carbonate (MgCO 3 or CaCO 3) would have the higher molar solubility and why? Molar Solubility Slide 9 / 39 Calculating the molar solubility from ion concentrations simply involves using stoichiometrical ratios. Example: What is the molar solubility of a saturated silver carbonate solution in which the [Ag + ] = 2.4 x 10-8? Ag 2CO 3(s) --> 2Ag + (aq) + CO 3 (aq) For every 2 silver ions in solution, 1 Ag 2CO 3 would have been required to dissociate. So... 2.4 x 10-8 M Ag + x 1 M Ag 2CO 3 = 1.2 x 10-8 M 2 M Ag+ Class Question: What would one need to know to find the number of grams of silver carbonate that were dissolved? Then, explain how this would be calculated.
Molar Solubility Slide 10 / 39 Calculating the molar solubility from an equilibrium constant requires writing and using an equilibrium expression. Example: What is the molar solubility of a saturated aqueous solution of PbI 2? (Ksp @25 C = 1.39 x 10-8 ) PbI 2(s) --> Pb 2+ (aq) + 2I - (aq) Ksp = 1.39 x 10-8 = [Pb 2+ ][I - ] 2 Since neither ion concentration is known, we will substitute "x" for the [Pb 2+ ] and "2x" for the [I - ]. 1.39 x 10-8 = (x)(2x) 2 = 4x 3 "x" = [Pb 2+ ] = 1.51 x 10-3 M Since 1 Pb 2+ required 1 PbI 2, the molar solubility of the PbI 2(s) = 1.51 x 10-3 M. 1 When 30 grams of NaCl are dissolved into 100 ml of distilled water, all of the solid NaCl has dissolved. The solution must be saturated and the Ksp for the NaCl must be very high. Slide 11 / 39 2 In two separate beakers with identical volumes of water, 10 grams of CuCO 3 was added to beaker 1 and 10 grams of PbCO 3 was added to beaker 2. After stirring, the contents of each beaker were poured through filter paper and dried. The mass of the the solid retained in the filter paper from beaker 1 was 9.995 grams while that from beaker 2 was 9.992 grams. It can be deduced that PbCO 3 has the smaller K sp. Slide 12 / 39
3 Which of the following ionic salts would have the highest molar solubility? Slide 13 / 39 A NiCO 3(s) Ksp = 6.61 x 10-9 B MnCO 3(s) Ksp = 1.82 x 10-11 C ZnCO 3(s) Ksp = 1.45 x 10-11 D Ag 2CrO 4(s) Ksp = 9.00 x 10-12 E All have the same molar solubility 4 A student places 4.000 grams of PbI 2(s) into distilled water to produce a 200.0 ml solution. After stirring, the student filters the solution and dries the solid. What would be expected mass of the solid PbI 2 on the filter paper after drying? Slide 14 / 39 5 The conductivity of a saturated solution of Ag 2CO 3 would be expected to be less than the conductivity of a saturated solution of CaCO 3. Slide 15 / 39
Slide 16 / 39 Calculating Ksp Return to Table of Contents Calculating Ksp Slide 17 / 39 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? 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 - ] Class Question: If an acid was added and reacted with some of the hydroxide ion, would the Ksp increase, decrease, or remain the same? Common Ion Effect Slide 18 / 39 If one of the ions that is part of the solubility equilibria is either already present or added, the equilibria will shift accordingly thereby altering the equilibrium position. The pictures below represent MgCO 3(s) being added to distilled water and to a solution that is 0.1 M Na 2CO 3. Distilled Water 0.1 M Na 2CO 3 MgCO 3 CO 3 Na + Na + MgCO 3(s) --> Mg 2+ (aq) + CO 3 (aq) MgCO 3(s) --> Mg 2+ (aq) + CO 3 (aq) The presence of the carbonate ion (common ion) in the 0.1 M Na 2CO 3 solution shifts the equilibria to the left, diminishing the solubility of the MgCO 3.
Common Ion Effect Slide 19 / 39 Finding the molar solubility when a common ion is present involves writing an equilibrium expression. Example: What is the molar solubility of AgI in a 0.041 M solution of MgI 2? AgI(s) --> Ag + (aq) + I - (aq) Ksp (look-up) = 1.5 x 10-16 = [Ag+][I-] = [x][0.082 + x] Since the equilibria constant is so small, we will expect the change in the [I-] will be negligible compared to the amount already present. 1.5 x 10-16 = [x][0.082] x = [Ag+] = 1.83 x 10-15 M Since 1 AgI is needed to produce 1 Ag +, the molar solubility of AgI in this solution is 1.83 x 10-15 M. Common Ion Effect Slide 20 / 39 Finding the molar solubility when a common ion is present involves writing an equilibrium expression. Example: What is the molar solubility of AgI in a 0.041 M solution of MgI 2? AgI(s) --> Ag + (aq) + I - (aq) Ksp (look-up) = 1.5 x 10-16 = [Ag+][I-] = [x][0.082 + x] Class Question: If the solution were 0.041 M NaI as opposed to 0.041 M MgI 2, how would the molar solubility be affected? Selective Precipitation Ionic salts that share a common ion will require differing concentrations of that ion to form a precipitate. As a result, ions can be removed selectively from a solution. Consider a solution containing 0.3 M Ag + and 0.3 M Pb 2+. Both precipitate with chloride but require differing amounts of chloride as will be shown below. AgCl(s) --> Ag + (aq) + Cl - (aq) Ksp = 1.8 x 10-10 PbCl 2(s) --> Pb 2+ (aq) + 2Cl - (aq) Ksp = 1.7 x 10-5 Slide 21 / 39
Selective Precipitation Slide 22 / 39 Consider a solution containing 0.3 M Ag + and 0.3 M Pb 2+. Both precipitate with chloride but require differing amounts of chloride as will be shown below. AgCl(s) --> Ag + (aq) + Cl - (aq) Ksp = 1.8 x 10-10 PbCl 2(s) --> Pb 2+ (aq) + 2Cl - (aq) Ksp = 1.7 x 10-5 Since AgCl is much less soluble, less Cl- will be needed to precipitate it. This can be calculated as shown below: [Cl-] needed to ppt. AgCl 1.8 x 10-10 = [0.3][Cl-] [Cl-] = 6.0 x 10-10 M [Cl-] needed to ppt. PbCl 2 1.7 x 10-5 = [0.3][Cl-] 2 [Cl-] = 7.5 x 10-3 M Since far less Cl- is needed to ppt. the AgCl, it can be precipitated and removed prior to precipitating PbCl 2. Selective Precipitation Only salts with Ksp values differing by a few orders of magnitude can be selectively precipitated. The bigger the difference in Ksp's, the better. Slide 23 / 39 Example: A student has a solution that is 0.045 M Mg 2+ (aq) and 0.045 M Ca 2+ (aq). They then attempt to selectively precipitate the ions by added sodium carbonate to the solution. What would be the concentration of Ca 2+ when the Mg 2+ begins to precipitate? MgCO3(s) --> Mg 2+ (aq) + CO3 (aq) Ksp = 6.0 x 10-6 6.0 x 10-6 = [0.045][CO3 ] [CO3 ] = 1.33 x 10-4 M CaCO3(s) --> Ca 2+ (aq) + CO3 (aq) Ksp = 3.0 x 10-9 3.0 x 10-9 = [Ca 2+ ][1.33 x 10-4 ] [Ca 2+ ] = 2.2 x 10-5 M Note: The Ca 2+ ion has been virtually entirely removed before the Mg 2+ starts to precipitate, a good separation. 6 What is the Ksp of Ag 2CrO 4(s) if the chromate concentration in a saturated solution is 3.5 x 10-5 M? Slide 24 / 39
7 A 1 L saturated solution of Cu(OH) 2 is filtered and the filtrate is then reacted with 0.1 M HCl and it was found that 0.0042 L of the HCl was needed to reach the equivalence point. What is the Ksp of the Cu(OH) 2? Slide 25 / 39 8 The molar solubility of BaSO 4 would be higher when dissolved in a 0.3 M BaCl 2 solution than in distilled water. Slide 26 / 39 9 The molar solubility of BaSO 4 would be higher in a 0.3 M Na 2SO 4 than in a 0.3 M BaCl 2 solution. Slide 27 / 39
10 Using a table of K sp, predict which of the pairs of ions would be MOST difficult to separate by selective precipitation using hydroxide ions: Slide 28 / 39 A Cu 2+ and Fe 2+ B Fe 2+ and Fe 3+ C Fe 3+ and Ag + D Ag + and Pb 2+ E The Ksp values play no role in selective precipitation 11 A student has a solution that is 0.087 M I - and 0.067 M Cl -. What concentration of Ag + ions would be needed to first start to precipitate the I - AND the Cl -? Slide 29 / 39 12 Using the information from question 11, what would be the concentration of I - when the chloride ion starts to precipitate? Slide 30 / 39
13 Salts with similar Ksp are easier to selectively precipitate than ones with significantly different Ksp values. Slide 31 / 39 Slide 32 / 39 Le-Chatlier and Solubility Equilibrium Return to Table of Contents Le-Chatelier and Solubility Equilibrium Slide 33 / 39 As with any equilibria, changing the concentrations of reactants and products will shift the equilibrium but only changing the temperature will change the equilibrium constant. Common ways to shift solubility equilibria: 1. Changing the temperature 2. Removing an ion by precipitation 3. Removing an ion by reacting it with an acid 4. Increasing the concentration of an ion by adding it to solution. Class Question: Why doesn't changing the concentration of an ion change the value of the Ksp?
Le-Chatelier and Solubility Equilibrium Slide 34 / 39 If two salts share a common ion, one salt can be made more soluble by having its common ion concentration diminished by the precipitation of the other salt Consider AgI(s) and AgBr(s) AgI(s) Ksp = 1.5 x 10-16 AgBr Ksp = 5.4 x 10-13 If one were to add NaI to a saturated solution of AgBr, the I- would preferentially precipitate with the Ag+, thereby dissolving the AgBr! Magic? No, just chemistry! AgBr(s) + I-(aq) --> AgI(s) + Br-(aq) Class Question: Could AgI be made more soluble be adding aqueous NaBr? Le-Chatelier and Solubility Equilibrium Slide 35 / 39 Ionic salts with a weak base ion can be made more soluble in acidic solutions. CaCO 3(s) + H + (aq) --> Ca 2+ (aq) + HCO 3- (aq) In this case, the bicarbonate ion will react with a second proton and form carbonic acid which will decompose into CO 2(g) and H 2O(l). HCO 3- (aq) + H + (aq) --> H 2CO 3(aq) --> CO 2(g) + H 2O(l) Being a gas, the carbon dioxide leaves and thus shifts the equilibrium farther to the right. This is the process that causes cave formation and the deterioration of limestone statues. 14 When a saturated solution of Fe(OH) 3 is prepared, the temperature of the solution increases. If the solution is cooled, the concentration of Fe 3+ will increase and the K sp will be unaffected. Slide 36 / 39
15 Which of the following would be TRUE if sodium carbonate were added to a saturated solution of MgCO 3 (s): Slide 37 / 39 A The Ksp will decrease B The [Mg 2+ ] will decrease C The conductivity of the solution will increase D Both A and B E None of these 16 Which of the following salts would be made appreciably more soluble if dissolved in acidic solution? Slide 38 / 39 A PbCl 2 B CaSO 4 C Cu(OH) 2 D Hg 2I 2 Slide 39 / 39