AP Chemistry. Introduction to Solubility Equilibria. Slide 1 / 91 Slide 2 / 91. Slide 3 / 91. Slide 4 / 91. Slide 5 / 91.

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1 Slide 1 / 91 Slide 2 / 91 P hemistry queous quilibria II: Ksp & Solubility Products Slide 3 / 91 Slide 4 / 91 Table of ontents: K sp & Solubility Products Introduction to Solubility quilibria alculating K sp from the Solubility alculating Solubility from Ksp lick on the topic to go to that section Introduction to Solubility quilibria Factors ffecting Solubility Precipitation Reactions and Separation of Ions Return to the Table of ontents Slide 5 / 91 Introduction to Solubility quilibria Slide 6 / 91 Introduction to Solubility quilibria 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. saturated solution of ao 3(s) a 2+ a 2+ O 3 O 3 Many shells are made of relatively insoluble calcium carbonate, so the shells are not at huge risk of dissolving in the ocean. ao 3(s) alcium carbonate is a relatively insoluble ionic salt. Would the picture look different for a soluble ionic salt such as Na 2O 3? Which solution would be the better electrolyte?

2 Slide 7 / 91 Introduction to Solubility quilibria onsider the equilibrium that exists in a saturated solution of ao 3 in water: ao 3(s) a 2+ (aq) + O 3 (aq) Unlike acid-base equilibria which are homogenous, solubility equilibria are heterogeneous, there is always a solid in the reaction. Slide 8 / 91 Introduction to Solubility quilibria The equilibrium constant expression for this equilibrium is K sp = [a 2+ ] [O 3 2 ] where the equilibrium constant, K sp, is called the solubility product. There is never any denominator in K sp expressions because pure solids are not included in any equilibrium expressions. Slide 9 / 91 Solubility quilibrium The degree to which an ionic compound dissociates in water can be determined by measuring it's "K sp" or solubility product equilibrium constant. ao 3(s) --> a 2+ (aq) + O 3 (aq) 25 = 5.0 x 10-9 MgO 3(s) --> Mg 2+ (aq) + O 3 (aq) 25 = 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. Slide 10 / 91 1 Which Ksp expression is correct for gl? [g + ]/[l - ] [g + ][l - ] [g 2+ ] 2 [l ] 2 [g + ] 2 [l - ] 2 None of the above. Which saturated solution above would have the higher conductivity and why? Slide 10 () / 91 1 Which Ksp expression is correct for gl? [g + ]/[l - ] [g + ][l - ] [g 2+ ] 2 [l ] 2 [g + ] 2 [l - ] 2 None of the above. gl (s) # g + (aq) + l - (aq) Ksp = [g + ][l - ] Slide 11 / 91 2 Given the reaction at equilibrium: Zn(OH) 2 (s) Zn 2+ (aq) + 2OH - (aq) what is the expression for the solubility product constant, K sp, for this reaction? K sp = [Zn 2+ ][OH - ] 2 / [Zn(OH) 2 ] K sp = [Zn(OH) 2 ] / [Zn 2+ ][2OH - ] K sp = [Zn 2+ ][2OH - ] K sp= [Zn 2+ ][OH - ] 2

3 Slide 11 () / 91 2 Given the reaction at equilibrium: Zn(OH) 2 (s) Zn 2+ (aq) + 2OH - (aq) what is the expression for the solubility product constant, K sp, for this reaction? K sp = [Zn 2+ ][OH - ] 2 / [Zn(OH) 2 ] K sp = [Zn(OH) 2 ] / [Zn 2+ ][2OH - ] K sp= [Zn 2+ ][OH - ] 2 Slide 12 / 91 3 Which Ksp expression is correct for Fe 3(PO 4) 2? [Fe 2+ ] 3 [PO 3-4 ] 2 [Fe 2+ ] 3 3- /[PO 4 ] 2 [Fe 3+ ] 2 [PO 3-4 ] 2 [Fe 2+ ] 2 /[PO 3-4 ] 2 None of the above. K sp = [Zn 2+ ][2OH - ] K sp= [Zn 2+ ][OH - ] 2 Slide 12 () / 91 3 Which Ksp expression is correct for Fe 3(PO 4) 2? [Fe 2+ ] 3 [PO 3-4 ] 2 [Fe 2+ ] 3 3- /[POFe 4 ] 3(PO 2 4) 2(s) # 3Fe (aq) + 2PO 4 (aq) [Fe 3+ ] 2 [PO 4 3- ] 2 [Fe 2+ ] 2 /[PO 4 3- ] 2 None of the above. Ksp = [Fe 2+ ] 3 [PO 4 3- ] 2 Slide 13 / 91 4 When 30 grams of Nal are mixed into 100 ml of distilled water all of the solid Nal dissolves. The solution must be saturated and the K sp for the Nal must be very high. True False Slide 13 () / 91 4 When 30 grams of Nal are mixed into 100 ml of distilled water all of the solid Nal dissolves. The solution must be saturated and the K sp for the Nal must be very high. False The solution in this case is unsaturated. It has the True capacity to dissolve more salt. False Slide 14 / 91 5 The conductivity of a saturated solution of g 2O 3 would be expected to be less than the conductivity of a saturated solution of ao 3. Justify your answer. True False

4 Slide 14 () / 91 5 The conductivity of a saturated solution of g 2O 3 would be expected to be less than the conductivity of a saturated solution of ao 3. Justify your answer. True False False For solutions of the same concentration, g 2O 3 would dissociate into more ions so therefore it would have a greater conductivity Slide 15 / 91 Solubility The term solubility represents the maximum amount of solute that can be dissolved in a certain volume before any precipitate is observed. The solubility of a substance can be given in terms of grams per liter g/l or in terms of moles per liter mol/l The latter is sometimes referred to as molar solubility. For any slightly soluble salt the molar solubility always refers to the ion with the lower molar ratio. xample #1 Slide 16 / 91 Solubility onsider the slightly soluble compound barium oxalate, a 2O 4. The solubility of a 2O 4 is 1.3 x 10-3 mol/l. The ratio of cations to anions is 1:1. This means that 1.3 x 10-3 moles of a 2+ can dissolve in one liter. lso, 1.3 x 10-3 moles of 2O 4 can dissolve in one liter. What is the maximum amount (in grams) of a 2O 4 that could dissolve in 2.5 L (before a solid precipitate or solid settlement occurs)? xample #1 Slide 17 / 91 Solubility What is the maximum amount (in grams) of a 2O 4 that could dissolve in 2.5 L (before a precipitate occurs)? The solubility of a 2O 4 is 1.3 x 10-3 mol/l. a 2O 4 (s) --> a 2+ (aq) + 2O 4 (aq) 1.3 x 10-3 mol a 2O x 10-3 g 2.5L x = a 2O 4 1 liter 3.25 x 10-3 g x 1 mole = 0.73g a 2O 4 a 2O g 0.73g is the maximum amount of a 2O 4 that could dissolve in 2.5 L before a precipitate forms. xample #2 Slide 18 / 91 Solubility onsider the slightly soluble compound lead chloride, Pbl 2. The solubility of Pbl 2 is mol/l. xample #3 Slide 19 / 91 Solubility onsider the slightly soluble compound silver sulfate, g 2SO 4. The solubility of g 2SO 4 is mol/l. The ratio of cations to anions is 1:2. The ratio of cations to anions is 2:1. This means that moles of Pb 2+ can dissolve in one liter. This means that moles of SO 4 can dissolve in one liter. Twice as much, or 2(0.016) = moles of l - can dissolve in one liter. Twice as much, or 2(0.015) = moles of g + can dissolve in one liter.

5 Slide 20 / 91 Solubility Remember that molar solubility refers to the ion with the lower mole ratio. It does not always refer to the cation, although in most cases it does. ompound a 2O 4 Pbl 2 Molar Solubility of [ation] ompound 1.3 x 10-3 mol 1.3 x 10-3 mol [nion] 1.3 x 10-3 mol mol/l mol/l mol/l Slide 21 / 91 6 If the solubility of barium carbonate, ao 3 is 7.1 x 10-5 M, this means that a maximum of barium ions, a 2+ ions can be dissolved per liter of solution. 7.1 x 10-5 moles half of that twice as much one-third as much one-fourth as much g 2SO mol/l mol/l mol/l Slide 21 () / 91 6 If the solubility of barium carbonate, ao 3 is 7.1 x 10-5 M, this means that a maximum of barium ions, a 2+ ions can be dissolved per liter of solution. 7 Slide 22 / 91 If the solubility of barium carbonate, ao 3 is 7.1 x 10-5 M, this means that a maximum of carbonate ions, O 3 ions can be dissolved per liter of solution. 7.1 x 10-5 moles half of that The ratio of ions is 1:1 the maximum amount of a 2+ is twice as much 7.1 x 10-5 moles per 1 liter. one-third as much 7.1 x 10-5 moles half of that twice as much one-third as much one-fourth as much one-fourth as much Slide 22 () / 91 Slide 23 / 91 7 If the solubility of barium carbonate, ao 3 is 7.1 x 10-5 M, this means that a maximum of carbonate ions, O 3 ions can be dissolved per liter of solution. 8 If the solubility of g 2rO 4 is 6.5 x 10-5 M, this means that a maximum of silver ions, g +, can be dissolved per liter of solution. 7.1 x 10-5 moles half of that The ratio of ions is 1:1 the maximum amount of O twice as much 3 is 7.1 x 10-5 moles per 1 liter. one-third as much one-fourth as much 6.5 x 10-5 moles twice 6.5 x 10-5 moles half 6.5 x 10-5 moles one-fourth 6.5 x 10-5 moles four times 6.5 x 10-5 moles

6 8 Slide 23 () / 91 If the solubility of g 2rO 4 is 6.5 x 10-5 M, this means that a maximum of silver ions, g +, can be dissolved per liter of solution. The ratio of ions is 2:1. The 6.5 x 10-5 moles molar solubility always refers twice 6.5 x 10-5 to the ion of the lowest mole moles ratio, in this case ro 4. half 6.5 x 10-5 moles Therefore, the maximum amount of g + is twice one-fourth 6.5 x x moles 10-5 moles per 1 liter. four times 6.5 x 10-5 moles Slide 24 / 91 alculating K sp from the Solubility Return to the Table of ontents Slide 25 / 91 Slide 26 / 91 alculating K sp from the Solubility Sample Problem The molar solubility of lead (II) bromide, Pbr 2 is 1.0 x 10-2 at 25 o. alculate the solubility product, K sp, for this compound. The molar solubility always refers to the ion of the lower molar ratio, therefore [Pb 2+ ] = 1.0 x 10-2 mol/l and [r - ] = 2.0 x 10-2 mol/l Substitute the molar concentrations into the K sp expression and solve. K sp = [Pb 2+ ][r - ] 2 = (1.0 x 10-2 )(2.0 x 10-2 ) 2 = 4.0 x For the slightly soluble salt, os, the molar solubility is 5 x 10-5 M. alculate the K sp for this compound. 5 x x x x x 10-9 Slide 26 () / 91 9 For the slightly soluble salt, os, the molar solubility is 5 x 10-5 M. alculate the K sp for this compound. Slide 27 / For the slightly soluble salt, af 2, the molar solubility is 3 x 10-4 M. alculate the solubility-product constant for this compound. 5 x x x x x 10-9 K sp = [o 2+ ][S ] K sp = (5 x 10-5 M) 2 K sp = 2.5 x x x x x x 10-10

7 Slide 27 () / For the slightly soluble salt, af 2, the molar solubility is 3 x 10-4 M. alculate the solubility-product constant for this compound. Slide 28 / For the slightly soluble salt, La(IO 3) 3, the molar solubility is 1 x 10-4 M. alculate K sp. 3 x x x x x x K sp = [a 2+ ][F-] 2 [a 2+ ] = 3 x 10-4, [F - ] = 6 x 10-4 K sp = (3 x 10-4 )(6 x 10-4 ) 2 K sp = 1.08 x x x x x Slide 28 () / For the slightly soluble salt, La(IO 3) 3, the molar solubility is 1 x 10-4 M. alculate K sp. 3 x x K sp = [La 3+ ][IO 3- ] 3 [La 2.7 x ] = 1 x 10-4, [IO 3- ] = 3(1x 10-4 ) K sp = (1 x 10-4 )(3 x 10-4 ) x 10K -15 sp = 3 x x [This object is a pull Slide 29 / For the slightly soluble compound, a 3(PO 4) 2, the molar solubility is 3 x 10-8 moles per liter. alculate the K sp for this compound x x x x x Slide 29 () / For the slightly soluble compound, a 3(PO 4) 2, the molar solubility is 3 x 10-8 moles per liter. alculate the K sp for this compound. Slide 30 / The concentration of hydroxide ions in a saturated solution of l(oh) 3 is 1.58x What is the K sp of l(oh) 3? 9.00 x K sp = [a 2+ ] 3 3- [PO 4 ] x x x x [PO 4 3- ] = 3 x 10-8 The ratio of a 2+ to PO 4 3-3:2 [a 2+ ] = 3 x 10-8 x 3/2 = 4.5 x 10-6 K sp = (4.5 x 10-6 ) 3 (3 x 10-8 ) 2 K sp = 8.20 x 10-32

8 Slide 30 () / The concentration of hydroxide ions in a saturated solution of l(oh) 3 is 1.58x What is the K sp of l(oh) 3? Slide 31 / What is the Ksp of Fe(OH) 3(s) if a saturated solution of it has a ph of 11.3? l(oh) 3(s) # l 3+ (aq) + 3OH - (aq) Ksp = [l 3+ ][OH - ] 3 The ratio of l 3+ to OH - is 1 to 3. [l 3+ ] = (1/3) 1.58x x = 5.3 x Ksp= (5.3 x )(1.58 x ) x x Ksp = 2.1 x x x x Slide 31 () / 91 Slide 32 / What is the Ksp of Fe(OH) 3(s) if a saturated solution of it has a ph of 11.3? 2.0 x x Fe(OH)3 (s) # Fe 3+ (aq) + 3OH - (aq) Ksp = [Fe 3+ ][OH - ] 3 ph = 11.3, poh = 3.7 [OH - ] = = 2.0 x 10-4 The ratio of [Fe 3+ ] to [OH - ] to = 1:3 alculating Solubility from the K sp 2.1 x [Fe 3+ ] = 1/3 x 2.0 x 10-4 = 6.7 x 10-5 Ksp = (6.7 x 10-5 )(2.0 x 10-4 ) x 10-8 Ksp = 5.4 x Return to the Table of ontents 5.4 x Slide 33 / 91 Slide 34 / 91 alculating Solubility from the K sp alculating Solubility from the K sp xample: What is the molar solubility of a saturated aqueous solution of ao 3? = 5.0 x 10-9 ) ao 3(s) --> a 2+ (aq) + O 3 (aq) Ksp = 5.0 x 10-9 = [a 2+ ][O 3 ] Since neither ion concentration is known, we will substitute "x" for the [a 2+ ] and "x" for the [O 3 ]. 5.0 x 10-9 = (x)(x) = x 2 xample: What is the molar solubility of a saturated aqueous solution of PbI 2? = 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 - ] x 10-8 = (x)(2x) 2 = 4x 3 "x" = [a 2+ ] = [O 3 ] = 7.07 x 10-5 M Since 1 a 2+ or 1 O 3 are required for 1 ao 3, the molar solubility of the ao 3(s) = 7.07 x 10-5 M. "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.

9 Slide 35 / alculate the concentration of silver ion when the solubility product constant of gi is 1 x Slide 35 () / alculate the concentration of silver ion when the solubility product constant of gi is 1 x (1 x ) 2 (1 x ) (1 x ) (1 x ) 2 (1 x ) (1 x ) 2 K sp = [g + ][I - ] 1 x = x 2 (1 x ) (1 x ) x = 1x10-16 Slide 36 / alculate the molar solubility of PbF 2 that has a K sp at 25 = 3.6 x Students type their answers here Slide 36 () / alculate the molar solubility of PbF 2 that has a K sp at 25 = 3.6 x Students type their answers here PbF 2(s) # Pb 2+ (aq) + 2F - (aq) K sp = [Pb 2+ ][F - ] 2 K sp = x(2x) x 10-6 = 4x 3 x = 3.6 x 10-6 /4 x = 9.7 x 10-3 mol/liter Slide 37 / The K sp of a compound of formula 3 is 1.8 x What is the molar solubility of the compound? Slide 37 () / The K sp of a compound of formula 3 is 1.8 x What is the molar solubility of the 3(s) # compound? 3+ (aq) (aq) K sp = [ 3+ ][ - ] 3 K sp = x(3x) x = 27x 4 x = 6.7 x x = 1.6 x 10-5 mol/liter

10 Slide 38 / The K sp of a compound of formula 3 is 1.8 x The molar mass is 280g/mol. What is the solubility? Slide 38 () / The K sp of a compound of formula 3 is 1.8 x The molar mass is 280g/mol. What is the solubility? The molar solubility is 1.6 x 10-5 mol/liter x 280g/mol = 4.6 x 10-3 g/liter Slide 39 / Which of the following ionic salts would have the highest molar solubility? Slide 39 () / Which of the following ionic salts would have the highest molar solubility? NiO 3(s) Ksp = 6.61 x 10-9 MnO 3(s) Ksp = 1.82 x ZnO 3(s) Ksp = 1.45 x g 2rO 4(s) Ksp = 9.00 x ll have the same molar solubility NiO 3(s) Ksp = 6.61 x 10-9 g 2rO 4 would have the highest MnO 3(s) Ksp = 1.82 x molar solubility. The ratio of ions is 2:1, this results ZnO 3(s) Ksp = 1.45 x in 10 K -11 sp = 4x 3. When you take the cube root g 2rO 4(s) Ksp = 9.00 of x K 10 sp you -12 get a molar solubility that is larger than the other salts ll have the same molar listed. solubility Slide 40 / 91 Slide 41 / 91 ommon Ion ffect Factors ffecting Solubility onsider a saturate solution of barium sulfate: aso 4(s) a 2+ (aq) + SO 4 (aq) If one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease. Return to the Table of ontents So adding any soluble salt containing either a 2+ or SO 4 ions will decrease the solubility of barium sulfate.

11 Slide 42 / 91 ommon Ion ffect Slide 43 / 91 ommon Ion ffect Sample Problem alculate the solubility of af 2 in grams per liter in a) pure water b) a 0.15 M KF solution c) a M a(no 3) 2 solution The solubility product for calcium fluoride, af 2 is 3.9 x a) pure water af 2(s) a 2+ (aq) + 2F - (aq) If we assume x as the dissociation then, a 2+ ions = x and [F - ] = 2x K sp = [a 2+ ] [F - ] 2 = (x)(2x) 2 K sp = 3.9 x = 4x 3 So x = 2.13 x 10-4 mol/l x (78 g/mol af 2) Solubility is g/l Note Slide 44 / 91 b) a 0.15 M KF solution Remember KF, a strong electrolyte, is completely ionized and the major source of F- ions. [F - ] =0.15M The solubility product for calcium fluoride, af 2 is 3.9 x [ F - ] = 0.15M Ksp = [a 2+ ] [F - ] 2 = (x)(0.15) 2 Ksp = 3.9 x = x So x = mol/l ommon Ion ffect Solubility is = x (78 g/mol af 2) = g/l Note Slide 45 / 91 ommon Ion ffect alculate the solubility of af 2 in grams per liter in c) a M a(no 3) 2 solution [a 2+ ] = 0.08M The solubility product for calcium fluoride,af 2 is 3.9 x af 2 (s) a 2+ (aq) + 2 F - (aq) Ksp = [a 2+ ] [F-] 2 = (0.080)(x) 2 Ksp = 3.9 x = 0.080x 2 So x = 2.2 x 10-5 mol/l * (78 g/mol af 2)/ 2 Solubility is g/l Slide 46 / 91 ommon Ion ffect Recall from the ommon-ion ffect that adding a strong electrolyte to a weakly soluble solution with a common ion will decrease the solubility of the weak electrolyte. ompare the solubilities from the previous Sample Problem af 2 (s) a 2+ (aq) + 2 F - (aq) af 2 dissolved with: pure water Solubility of af g/l M KF 1.35x10-7 g/l M a(no 3) g/l These results support Le hatelier's Principle that increasing a product concentration will shift equilibrium to the left. Slide 47 / What is the molar solubility of a saturated solution of # g 2rO 4? K sp at 25 is = 1.2 x x x x x x 10-8

12 Slide 47 () / 91 # 20 What is the molar solubility of a saturated solution of g 2rO 4? K sp at 25 is = 1.2 x Slide 48 / What is the molar solubility of a saturated solution of g 2rO 4 in 0.100M K 2rO 4? K sp at 25 is = 1.2 x x x x x x 10-8 g 2rO 4(s) 2g + (aq) + ro # 4 (aq) K sp = [g + ] 2 [ro 4 ] = (2x) 2 (x) 1.2 x = 4x 3 x = 1.2 x /4 x = 6.7 x x x x x x 10-6 Slide 48 () / What is the molar solubility of a saturated solution of g 2rO 4 in 0.100M K 2rO 4? K sp at 25 is = 1.2 x Slide 49 / What is the molar solubility of a saturated solution of g 2rO 4 in 0.200M gl? K sp at 25 is = 1.2 x x x x x x 10-6 g 2rO 4(s) 2g + (aq) + ro # 4 (aq) K sp = [g + ] 2 [ro 4 ] = (2x) 2 (0.100M) 1.2 x = x 2 x = 1.2 x /0.400 x = 1.7 x x x x x x 10-6 Slide 49 () / What is the molar solubility of a saturated solution of g 2rO 4 in 0.200M gl? K sp at 25 is = 1.2 x x x x x x 10-6 g 2rO 4(s) 2g + (aq) + ro # 4 (aq) K sp = [g + ] 2 [ro 4 ] = (0.200) 2 (x) 1.2 x = x x = 3.11 x Slide 50 / 91 hanges in ph The solubility of almost any ionic compound is affected by changes in ph. onsider dissociation equation for magnesium hydroxide: Mg(OH) (s) # Mg 2+ (aq) + 2OH - (aq) What do you expect will happen to the equilibrium if the ph of this system is lowered by adding a strong acid? Will one of the substances in the equilibrium interact with the strong acid? Would the Mg(OH) 2 be more or less soluble? (Think Le hâtelier s Principle.)

13 Slide 50 () / 91 hanges in ph The solubility of almost any ionic compound is affected by changes in ph. The added H + will react with the onsider dissociation OH equation - and take for magnesium it out of solution. hydroxide: Mg(OH) (s) # Mg 2+ (aq) + 2OH - (aq) The equilibrium will shift to the right and the salt, Mg(OH) 2, will be more soluble. What do you expect will happen to the equilibrium if the ph of this system is lowered by adding a strong acid? Will one of the substances in the [This equilibrium object is a pull interact with the strong acid? Would the Mg(OH) 2 be more or less soluble? (Think Le hâtelier s Principle.) Slide 51 / 91 hanges in ph hanges in ph can also affect the solubility of salts that contain the conjugate base of a weak acid. onsider the dissociation of the salt calcium fluoride: af 2 (s) a 2+ (aq) + 2F - (aq) What do you expect will happen to the equilibrium if the ph of this system is lowered by adding a strong acid? Will one of the substances in the equilibrium interact with the strong acid? Would the af 2 be more or less soluble? Slide 51 () / 91 hanges in ph hanges in ph can also affect the solubility of salts that contain the conjugate base of a weak The acid. F - will react with H + forming hydrofluoric acid and some of it will leave onsider the dissociation of the salt calcium fluoride: from solution. ccording to Le hâtelier s Principle af 2 (s) athe 2+ (aq) equilibrium + 2F - (aq) will shift to the right and af 2 will become more soluble. What do you expect will happen to the equilibrium if the ph of this system is lowered by adding a strong acid? Will one of the substances in the equilibrium interact with the strong acid? Would the af 2 be more or less soluble? Slide 52 / 91 hanges in ph Sample Problem alculate the molar solubility of Mn(OH) 2 in a) in a solution buffered at ph=9.5 b) in a solution buffered at ph=8.0 c) pure water The solubility product for Mn(OH) 2 at 25 is 1.6 x Slide 53 / 91 In a solution buffered at ph=9.5, the [H + ] = 3.2 x 10-10, the [OH - ] = 3.2x The solubility product for Mn(OH) 2, is 1.6 x [ OH - ] = 3.2 x 10-5 M hanges in ph Sample Problem alculate the molar solubility of Mn(OH) 2 in a) in a solution buffered at ph = x = [Mn 2+ ] [OH - ] 2 = (x)(3.2 x 10-5 ) 2 x = 1.6 x /(3.2 x 10-5 ) 2 = 1.56 x 10-4 mol/l Slide 54 / 91 In a solution buffered at ph=8.0, the [H + ] = 1 x 10-8, the [OH - ] = 1x The solubility product for Mn(OH) 2, is 1.6 x [ OH - ] = 1 x 10-6 M 1.6 x = [Mn 2+ ] [OH - ] 2 = (x)(1 x 10-6 ) 2 x = 1.6 x / (1 x 10-6 ) 2 = So x = 0.16 mol/l hanges in ph Sample Problem alculate the molar solubility of Mn(OH) 2 in b) in a solution buffered at ph = 8.0 Note

14 Slide 55 / 91 In pure water the ph=7.0, the [H + ] = 1 x 10-7, the [OH - ] = 1x The solubility product for Mn(OH) 2, is 1.6 x [ OH - ] = 1 x 10-7 M 1.6 x = [Mn 2+ ] [OH - ] 2 = (x)(1 x 10-7 ) 2 x = 1.6 x / (1 x 10-7 ) 2 = So x = 16 mol/liter hanges in ph Sample Problem alculate the molar solubility of Mn(OH) 2 in c) in pure water Note Slide 56 / 91 hanges in ph If a substance has a basic anion, it will be more soluble in an acidic solution. If a substance has an acidic cation, it will be more soluble in basic solutions. We will discuss in a little while the affect of ph changes on substances that are amphoteric. o you remember what it means when a substance is amphoteric? Slide 57 / Given the system at equilibrium gl (s) g + (aq) + l - (aq) When 0.01 M Hl is added to the sytem, the point of equilibrium will shift to the. right and the concentration of g + will decrese right and the concentration of g + will increase left and the concentration of g + will decrease left and the concentration of g + will increase Slide 57 () / Given the system at equilibrium gl (s) g + (aq) + l - (aq) When 0.01 M Hl is added When to 0.01M the Hl sytem, is added, the point of according to Le hatelier's equilibrium will shift to the. right and the concentration of g + will decrese right and the concentration of g + will increase left and the concentration of g + will decrease left and the concentration of g + will increase principle the equilibrium will shift away from the additional l - to the left and the concentration of g + will decrease. Slide 58 / Which of the following substances are more soluble in acidic solution than in basic solution? Select all that apply. Slide 58 () / Which of the following substances are more soluble in acidic solution than in basic solution? Select all that apply. Pbl 2 Pbl 2,, ao 3 gi Fe(OH) 3 ao 3 gi Fe(OH) 3 ll these ionic compounds contain a basic anion that would react with the added H + causing a shift to the right and resulting in a more soluble salt. MgF 2 MgF 2

15 Slide 59 / What is the solubility of Zn(OH) 2 in a solution that is buffered at ph = 8.5? K sp = 3.0 x Students type their answers here Slide 59 () / What is the solubility of Zn(OH) 2 in a solution that is buffered at ph = 8.5? K sp = 3.0 x Students type their answers here The solubility product for Zn(OH) 2, s 3.0 x ph = 8.5, poh = 5.5 and [ OH - ] = 3.2 x 10-6 M K sp = [Zn 2+ ] [OH - ] 2 = (x)(3.2 x 10-6 ) 2 x = 3.0 x /(3.2 x 10-8 ) 2 x = 3.0 x 10-5 mol/l Slide 60 / Will the solubility of Zn(OH) 2 in a solution that is buffered at ph = 11.0 be greater than in a solution buffered at 8.5? xplain. Yes No Slide 60 () / Will the solubility of Zn(OH) 2 in a solution that is buffered at ph = 11.0 be greater than in a solution buffered at No 8.5? xplain. The solubility of Zn(OH) 2 increases in more acidic solutions. When the ph Yes increases, as in this case, from 8.5 to 11.0 the solubility will No decrease. Slide 61 / The molar solubility of NH 4l increases as ph. increases decreases is unaffected by changes in ph Slide 61 () / The molar solubility of NH 4l increases as ph. increases decreases NH 4l has an acidic cation, NH 4+. s ph increases the solution becomes more basic, the equilibrium shifts to the right and the salt becomes more soluble. is unaffected by changes in ph

16 Slide 62 / The molar solubility of Na 2O 3 increases as ph. increases decreases is unaffected by changes in ph Slide 62 () / The molar solubility of Na 2O 3 increases as ph. increases decreases is unaffected salt by becomes changes less in ph soluble. Na 2O 3 has a basic anion, O 3. s ph increases the solution becomes more basic, the equilibrium shifts to the left and the Slide 63 / 91 omplex Ions Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent. The formation of complex ions particularly with transitional metals can dramatically affect the solubility of a metal salt. For example, the addition of excess ammonia to gl will cause the gl to dissolve. This process is the sum of two reactions resulting in: gl (s) + 2NH 3(aq) g(nh 3) 2 + (aq) + l - (aq) dded NH 3 reacts with g + forming g(nh 3) 2+. dding enough NH 3 results in the complete dissolution of gl. Slide 64 / 91 mphoterism Some metal oxides and hydroxides are soluble in strongly acidic and in strongly basic solutions because they can act either as acids or bases. These substances are said to be amphotheric. xamples of such these substances are oxides and hydroxides of l 3+, Zn 2+, and Sn 2+. They dissolve in acidic solutions because their anion is protonated by the added H + and is pulled from solution causing a shift in the equilibrium to the right. For example: l(oh) 3(s) # l 3+ (aq) + 3 OH - (aq) Slide 65 / 91 mphoterism # Slide 66 / Which of the following factors affect solubility? However these oxides and hydroxides also dissolve in strongly basic solutions. This is because they form complex ions containing several typically four hydroxides bound to the metal ion. luminum hydroxide reacts with OH - to form a complex ion in the following reaction: l(oh) 3(s) + OH - (aq) # l(oh) 4 - (aq) s a result of the formation of the complex ion, l(oh) 4 -, aluminum hydroxide is more soluble. Many metal hydroxides only react in strongly acidic solutions. a(oh) 2, Fe(OH) 2 and Fe(OH) 3 are only more soluble in acidic solution they are not amphoteric. ph Formation of omplex Ions ommon-ion ffect and,, and

17 Slide 66 () / 91 Slide 67 / Which of the following factors affect solubility? ph Formation of omplex Ions ommon-ion ffect and,, and, and are correct. Precipitation Reactions and Separation of Ions Return to the Table of ontents Slide 68 / 91 Slide 69 / 91 Precipitation Reactions and Separation of Ions o you remember the solubility rules? 30 What is the name of the solid precipitate that is formed when a solution of sodium chloride is mixed with a solution of silver nitrate? They were useful before when we were trying to qualitatively determine if a given reaction would produce a precipitate. They will be useful now for the same reason however now we are going to add a quantitative component that we will discuss soon. In general, soluble salts were: ny salt made with a Group I metal is soluble. ll salts containing nitrate ion are soluble. ll salts containing ammonium ion are soluble. o you remember what metal cations tended to be insoluble? g +, Pb 2+, and Hg 2+ sodium silver sodium nitrate chloride nitrate silver chloride Not enough information Slide 69 () / 91 Slide 70 / What is the name of the solid precipitate that is formed when a solution of sodium chloride is mixed with a solution of silver nitrate? 31 What is the name of the solid precipitate that is formed when a solution of potassium carbonate is mixed with a solution of calcium bromide? sodium silver potassium bromide sodium nitrate calcium carbonate chloride nitratenal (aq) + gno 3(aq) silver chloride # NaNO 3(aq) + gl (s) potassium calcium carbonate bromide Not enough information Not enough information

18 Slide 70 () / What is the name of the solid precipitate that is formed when a solution of potassium carbonate is mixed with a solution of calcium bromide? Slide 71 / What is the name of the solid precipitate that is formed when a solution of lead (IV) nitrate is mixed with a solution of magnesium sulfate? potassium bromide K calcium carbonate 2O 3 (aq) + ar 2 (aq) # 2Kr (aq) + ao 3(s) potassium calcium carbonate bromide Not enough information PbSO 4 Pb(SO 4 ) 2 Pb 2 SO 4 Mg(NO 3 ) 2 Not enough information Slide 71 () / 91 Slide 72 / What is the name of the solid precipitate that is formed when a solution of lead (IV) nitrate is mixed with a solution of magnesium sulfate? 33 The K sp for Zn(OH) 2 is 5.0 x Will a precipitate form in a solution whose solubility is 8.0x10-2 mol/l Zn(OH) 2? PbSO 4 Pb(SO 4 ) 2 Pb 2 SO 4 Mg(NO 3 ) 2 Pb(NO 3) 4(aq) + MgSO 4(aq) Mg(NO 3) 2 (aq) + PbSO 4(s) # Not enough information yes, because Q sp < K sp yes, because Q sp > K sp no, because Q sp = K sp no, because Q sp < K sp no, because Q sp > K sp Slide 72 () / 91 Slide 73 / The K sp for Zn(OH) 2 is 5.0 x Will a precipitate form in a solution whose Q = [Znsolubility 2+ ][OH - ] is 8.0x10-2 mol/l Zn(OH) 2? Q = x(2x) 2 yes, because Q = sp < 2.0 Kx 10-3 sp Q>K sp yes, because Q sp > K sp When Q>Ksp, a precipitate will no, because Qform. sp = K sp no, because Q sp < K sp no, because Q sp > K sp Q = (8.0 x 10-2 )(1.6 x 10-1 ) 2 34 The K sp for zinc carbonate is 1 x If equivalent amounts 0.2M sodium carbonate and 0.1M zinc nitrate are mixed, what happens? zinc carbonate precipitate forms, since Q>K. zinc carbonate precipitate forms, since Q<K. sodium nitrate precipitate forms, since Q>K. No precipitate forms, since Q=K.

19 Slide 73 () / The K sp for zinc carbonate is 1 x If equivalent amounts 0.2M sodium carbonate and 0.1M zinc nitrate are mixed, what happens? Na2O3 (aq)+ Zn(NO3)(aq) # ZnO3 (s) + NaNO3 (aq) zinc carbonate precipitate forms, since Q>K. zinc carbonate ZnO3(s) # Znprecipitate 2+ (aq) + O3 (aq) forms, since Q<K. The salt we are interested in is Q = [Zn 2+ ][O3 ] sodium Q= nitrate (0.2)(0.1)= precipitate 2 x 1-2 forms, since Q>K. Q>Ksp, therefore ZnO3 will precipitate. No precipitate forms, since Q=K. Slide 74 / 91 Separation of Ions When metals are found in natural they are usually found as metal ores. The metal contained in these ores are in the form of insoluble salts. To make extraction even more difficult the ores often contain several metal salts. In order to separate out the metals, one can use differences in solubilities of salts to separate ions in a mixture. Slide 75 / 91 Separation of Ions Imagine, you have a test tube that contains g +, Pb 2+ and u 2+ ions and you want to selectively remove each ion and place them into separate test tubes. What reagent could you add to the test tube that will form a precipitate with one or move of the cations and leave the others in solution? You can use your knowledge of the solubility rules or Ksp values for various metal salts to help you accomplish this goal. Slide 76 / 91 Separation of Ions You should remember that g + and Pb 2+ readily form insoluble salts and that u 2+ does not form insoluble salts as readily. Looking at some solubility product values, you will find the following: Salt Ksp g 2S 6 x PbS 3 x us 6 x gl 1.8 x Pbl x 10-5 You will notice that ul 2 is not to be found. This means ul 2 is a soluble salt! Slide 77 / 91 Separation of Ions dding l - should precipitate the g + and Pb 2+ ions but not the u 2+ ions. We can remove g + and Pb 2+ from the test tube. Now, how can we separate the g + and Pb 2+ ions? Salt Ksp g 2S 6 x PbS 3 x us 6 x gl 1.8 x Pbl x 10-5 o you notice the significant difference between the K sp values for g 2S and PbS? Maybe we can precipitate one of the salts out before the other if we control the concentration of S added. Which salt g 2S and PbS should precipitate first when we begin to add S? Slide 78 / 91 Separation of Ions If we have 0.100M concentrations of g + and Pb 2+ and we begin to add 0.200M K 2S the g 2S should precipitate first. For g 2S: K sp = 6 x = [g + ] 2 [S ] = (0.100) 2 (x) x = [S ] = 6 x M. If this concentration of S is added g 2S will precipitate. For PbS: Ksp = 3 x = [Pb 2+ ][S ] = 0.100(x) x = [S ] = 3 x M. greater amount of S is needed to precipitate the PbS. Therefore, g 2S will precipitate first.

20 Slide 79 / 91 Separation of Ions Problems Most of the problems in this section ask you whether a precipitate will form after mixing certain solutions together. General Problem-Solving Strategy Step 1 - etermine which of the products is the precipitate. Write the K sp expression for this compound. Step 2 - alculate the cation concentration of this slightly soluble compound. Step 3 - alculate the anion concentration of this slightly soluble compound. Step 4 - Substitute the values into the reaction quotient (Q) expression. Recall that this is the same expression as K. Step 5 - ompare Q to K to determine whether a precipitate will form. Sample Problem Slide 81 / 91 Separation of Ions Problems Will a precipitate form if you mix 50.0 ml of 0.20 M barium chloride, al 2, and 50.0 ml of 0.30 M sodium sulfate, Na 2SO 4? Step 1 - etermine which of the products is the precipitate. Write the K sp expression for this compound. aso 4 (s) a 2+ (aq) + SO 4 (aq) Step 2 - alculate the cation concentration of this slightly soluble compound. M 1V 1 =M 2V 2 M 2 = (M 1V 1) / V 2 M 2= (0.20M*50.0mL) / 100 ml M 2 = 0.10 M al 2 [a 2+ ] = 0.10 M Slide 80 / 91 Separation of Ions Problems In order for a precipitate to form the equilibrium that exists between the solution and the insoluble salt must reside on the left. We can determine to which side the equilibrium will shift using Q, the Reaction Quotient. If Q = Ksp If Q > Ksp If Q < Ksp then you have an exactly perfect saturated solution with not one speck of undissolved solid. then YS you will observe a precipitate; the number of cations and anions exceeds the solubility then NO precipitate will form; there are so few cations and anions that they all remain dissolved In a solution, If Q = K sp, the system is at equilibrium and the solution is saturated. If Q > K sp, the salt will precipitate until Q = K sp. If Q < K sp, more solid can dissolve until Q = K sp. Slide 82 / 91 Separation of Ions Problems Sample Problem - s (con't) Will a precipitate form if you mix 50.0 ml of 0.20 M barium chloride, al 2, and 50.0 ml of 0.30 M sodium sulfate, Na 2SO 4? Step 3 - alculate the anion concentration of this slightly soluble compound. M 1V 1 =M 2V 2 M 2 = (M 1V 1) / V 2 M 2= (0.30M*50.0mL) / 100 ml M 2 = 0.15 M Na 2SO 4 [SO 4 ] = 0.15 M Step 4 - Substitute the values into the reaction quotient (Q) expression. Recall that this is the same expression as K. Q = [a 2+ ] [SO 4 ] = (0.10) (0.15) = Slide 83 / 91 Separation of Ions Problems Sample Problem - s (con't) Will a precipitate form if you mix 50.0 ml of 0.20 M barium chloride, al 2, and 50.0 ml of 0.30 M sodium sulfate, Na 2SO 4? Step 5 - ompare Q to K to determine whether a precipitate will form. The K sp for barium sulfate is 1 x Therefore, since Q > K, there will be a precipitate formed when you mix equal amounts of 0.20 M al 2, and 0.30 M Na 2SO 4. Slide 84 / 91 Separation of Ions Problems In summary, to selectively precipitate metal ions from a solution that contains a number of metal ions you should use the solubility rules and K sp values to determine an experimental strategy. The solubility rules may lead you to the identity of an anion that will result in separation of certain metal ions however, at other times the quantity of the added anion will be instrumental in the separation given that metal salts have different degrees of solubility as seen in their K sp values.

21 Slide 85 / solution contains 2.0 x 10-5 M barium ions and 1.8 x 10-4 M lead (II) ions. If Na 2rO 4 is added, which will precipitate first from solution? The K sp for aro 4 is 2.1 x and the K sp for PbrO 4 is 2.8 x aro 4 PbrO 4 They will precipitate at the same time. It's impossible to determine with the information provided. Slide 85 () / solution contains 2.0 x 10-5 M barium ions and 1.8 x 10-4 M lead (II) ions. If Na 2rO 4 is added, which will precipitate first from solution? The K sp for aro 4 is 2.1 x and the K sp for PbrO 4 is 2.8 x aro4(s) # a 2+ (aq) + ro4 (aq) aro 4 PbrO 4 Ksp =[a 2+ ][ro4 ] = 2.1 x In order for this salt to precipitate Q>Ksp therefore [ro4 ] > Ksp/[a 2+ ] [ro4 ] > 2.10 x /2.0 x When the [ro4 ] 1.05 x 10-5 aro4 will precipitate. ompleting the same calculation for PbrO4, we find that They will precipitate at the same time. It's impossible to determine with the information provided. [ro4 ] > 2.8 x /1.8 x PbrO4 will precipitate when [ro4 ] > 1.56 x It takes much [This less object ro4 is a pull to precipitate the PbrO4 so it will precipitate first. Slide 86 / solution contains 2.0 x 10-4 M g + and 2.0 x 10-4 M Pb 2+. If Nal is added, will gl (K sp = 1.8 x ) or PbI 2 (K sp = 1.7 x 10-5 ) precipitate first? gl Pbl 2 They will precipitate at the same time. It is impossible to determine with the information provided. Slide 86 () / solution contains 2.0 x 10-4 M g + and 2.0 x 10-4 M Pb 2+. If Nal is added, will gl (K sp = 1.8 x ) or PbI 2 (K sp = 1.7 x 10-5 gl(s) )# precipitate g first? + (aq) + l - (aq) gl Pbl 2 Ksp =[g + ][l - ] = 1.8 x In order for this salt to precipitate Q>Ksp therefore [l-] > Ksp/[g + ] [l - ] > 1.8 x /2.0 x When the [l - ] > 9.0 x 10-7 gl will precipitate. ompleting the same calculation They will for Pbl2, precipitate we find that the same time. [l - ] > 1.7 x 10-5 /2.0 x Pbl2 will precipitate when [l - ] > 8.5 x It takes much less of l - to precipitate the gl so it will precipitate first. It is impossible to determine with the information [This provided. object is a pull Slide 87 / solution contains 2.0 x 10-4 M g + and 1.7 x 10-3 M Pb 2+. If Nal is added. What concentration of l - is needed to Students type their answers here begin precipitation. gl (K sp = 1.8 x ) and PbI 2 (K sp = 1.7 x 10-5 ) Slide 87 () / solution contains 2.0 x 10-4 M g + and 1.7 x 10-3 M Pb 2+. If Nal is added. What concentration of l - is needed to Students type their answers here begin precipitation. gl (K sp = 1.8 x ) and PbI 2 (K sp = 1.7 x 10-5 ) When the [l - ] > 9.0 x 10-7, gl will begin to precipitate.

22 Slide 88 / solution contains 2.0 x 10-4 M g + and 1.7 x 10-3 M Pb 2+. If Nal is added. What will be the concentration of the Students type their answers here first ion to precipitate when the second ion begins to precipitate? gl (Ksp = 1.8 x ) and PbI 2 (Ksp = 1.7 x 10-5 ) Slide 88 () / solution contains 2.0 x 10-4 M g + and 1.7 x 10-3 M Pb 2+. If Nal is added. What will be the concentration of the Students type their answers here first ion to precipitate when the second ion begins to precipitate? gl (Ksp = 1.8 x ) and PbI 2 (Ksp = 1.7 x 10-5 ) Pbl 2 begins to precipitate when the [l - ] > 8.5 x We can solve for [g + ] at this concentration of l -. K sp (gl) = 1.8 x = [g + ][l - ] [g + ] = 1.8 x / 8.5 x 10-2 [g + ] = 2.1 X 10-9 Slide 89 / Will o(oh) 2 precipitate from solution if the ph of a 0.002M solution of o(no 3) 2 is adjusted to 8.4? K sp for o(oh) 2 is 2.5 x Yes No Slide 89 () / Will o(oh) 2 precipitate from solution if the ph of a 0.002M solution o(oh) of o(no 2 (s) # o 3) 2 2+ is (aq) adjusted + 2OH - (aq) to 8.4? K sp for o(oh) 2 is 2.5 x o(oh) will precipitate if Q>K sp. Yes No ph = 8.4, poh=5.6, [OH - ] = 2.5 x 10-6 Q = [o 2+ ][OH - ] 2 = (0.002)(2.5 x 10-6 ) 2 Q = 1.26 x Q<K sp, therefore a precipitate will not form. Slide 90 / Will a precipitate form if you mix 25.0 ml of M calcium chloride, and 50.0 ml of M lithium chromate? The K sp for calcium chromate is 4.5 x Students type their answers here Slide 90 () / Will a precipitate form if you mix 25.0 ml of M aro4(s) # a 2+ (aq) + ro4 (aq) calcium chloride, In order and to solve 50.0 this ml problem of you have to M lithium determine the concentrations of a chromate? The K sp for calcium chromate 2+ and is 4.5 x ro4 in solution after mixing. Then you will need to calculate Q and then compare it to Ksp. [a 2+ ]: M2= M1V1/V2 = (0.250M)(0.025 L)/0.075L M2= Students type their answers here [ro4 ]: M2= M1V1/V2 = (0.155M)(0.050 L)/ 0.075L M2= Q = (0.0833)(0.1033) = 8.61 x 10-3 Q>Ksp therefore [This object aro4 is a pull will precipitate.

23 Slide 91 / 91