AP Chemistry Table of Contents: Ksp & Solubility Products Click on the topic to go to that section

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1 Slide 1 / 91 Slide 2 / 91 AP Chemistry Aqueous Equilibria II: Ksp & Solubility Products Table of Contents: K sp & Solubility Products Slide 3 / 91 Click on the topic to go to that section Introduction to Solubility Equilibria Calculating K sp from the Solubility Calculating Solubility from Ksp Factors Affecting Solubility Precipitation Reactions and Separation of Ions

2 Slide 4 / 91 Introduction to Solubility Equilibria Return to the Table of Contents Introduction to Solubility Equilibria Slide 5 / 91 Many shells are made of relatively insoluble calcium carbonate, so the shells are not at huge risk of dissolving in the ocean. Introduction to Solubility Equilibria Slide 6 / 91 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 2- CO 3 2- Answer CaCO 3(s) 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?

3 Introduction to Solubility Equilibria Slide 7 / 91 Consider the equilibrium that exists in a saturated solution of CaCO 3 in water: CaCO 3(s) Ca 2+ (aq) + CO 3 2- (aq) Unlike acid-base equilibria which are homogenous, solubility equilibria are heterogeneous, there is always a solid in the reaction. Introduction to Solubility Equilibria Slide 8 / 91 The equilibrium constant expression for this equilibrium is K sp = [Ca 2+ ] [CO 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. Solubility Equilibrium Slide 9 / 91 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 2- (aq) 25 C = 5.0 x 10-9 MgCO 3(s) --> Mg 2+ (aq) + CO 3 2- (aq) 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. Answer Which saturated solution above would have the higher conductivity and why?

4 1 Which Ksp expression is correct for AgCl? Slide 10 / 91 A [Ag + ]/[Cl - ] B [Ag + ][Cl - ] C [Ag 2+ ] 2 [Cl 2- ] 2 D [Ag + ] 2 [Cl - ] 2 E None of the above. 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? Slide 11 / 91 A K sp = [Zn 2+ ][OH - ] 2 / [Zn(OH) 2 ] B K sp = [Zn(OH) 2 ] / [Zn 2+ ][2OH - ] C K sp = [Zn 2+ ][2OH - ] D K sp= [Zn 2+ ][OH - ] 2 3 Which Ksp expression is correct for Fe 3(PO 4) 2? Slide 12 / 91 A [Fe 2+ ] 3 [PO 3-4 ] 2 B [Fe 2+ ] 3 /[PO 3-4 ] 2 C [Fe 3+ ] 2 3- [PO 4 ] 2 D [Fe 2+ ] 2 /[PO 3-4 ] 2 E None of the above.

5 4 When 30 grams of NaCl are mixed into 100 ml of distilled water all of the solid NaCl dissolves. The solution must be saturated and the K sp for the NaCl must be very high. Slide 13 / 91 True False 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. Justify your answer. Slide 14 / 91 True False Solubility Slide 15 / 91 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.

6 Solubility Slide 16 / 91 Example #1 Consider the slightly soluble compound barium oxalate, BaC 2O 4. The solubility of BaC 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 Ba 2+ can dissolve in one liter. Also, 1.3 x 10-3 moles of C 2O 4 2- can dissolve in one liter. What is the maximum amount (in grams) of BaC 2O 4 that could dissolve in 2.5 L (before a solid precipitate or solid settlement occurs)? Example #1 Solubility What is the maximum amount (in grams) of BaC 2O 4 that could dissolve in 2.5 L (before a precipitate occurs)? The solubility of BaC 2O 4 is 1.3 x 10-3 mol/l. Slide 17 / 91 BaC 2O 4 (s) --> Ba 2+ (aq) + C 2O 4 2- (aq) 1.3 x 10-3 mol BaC 2O x 10-3 g 2.5L x = BaC 2O 4 1 liter 3.25 x 10-3 g x 1 mole = 0.73g BaC 2O 4 BaC 2O g 0.73g is the maximum amount of BaC 2O 4 that could dissolve in 2.5 L before a precipitate forms. Solubility Slide 18 / 91 Example #2 Consider the slightly soluble compound lead chloride, PbCl 2. The solubility of PbCl 2 is mol/l. The ratio of cations to anions is 1:2. This means that moles of Pb 2+ can dissolve in one liter. Twice as much, or 2(0.016) = moles of Cl - can dissolve in one liter.

7 Solubility Slide 19 / 91 Example #3 Consider the slightly soluble compound silver sulfate, Ag 2SO 4. The solubility of Ag 2SO 4 is mol/l. The ratio of cations to anions is 2:1. This means that moles of SO 4 2- can dissolve in one liter. Twice as much, or 2(0.015) = moles of Ag + can dissolve in one liter. 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. Slide 20 / 91 Compound Molar Solubility of [Cation] Compound [Anion] BaC 2O x 10-3 mol 1.3 x 10-3 mol 1.3 x 10-3 mol PbCl 2 Ag 2SO mol/l mol/l mol/l mol/l mol/l mol/l 6 If the solubility of barium carbonate, BaCO 3 is 7.1 x 10-5 M, this means that a maximum of barium ions, Ba 2+ ions can be dissolved per liter of solution. Slide 21 / 91 A B C D E 7.1 x 10-5 moles half of that twice as much one-third as much one-fourth as much

8 7 If the solubility of barium carbonate, BaCO 3 is 7.1 x 10-5 M, this means that a maximum of carbonate ions, CO 3 2- ions can be dissolved per liter of solution. Slide 22 / 91 A B C D E 7.1 x 10-5 moles half of that twice as much one-third as much one-fourth as much 8 If the solubility of Ag 2CrO 4 is 6.5 x 10-5 M, this means that a maximum of silver ions, Ag +, can be dissolved per liter of solution. Slide 23 / 91 A B C D E 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 Slide 24 / 91 Calculating K sp from the Solubility Return to the Table of Contents

9 Calculating K sp from the Solubility Slide 25 / 91 Sample Problem The molar solubility of lead (II) bromide, PbBr 2 is 1.0 x 10-2 at 25 o C. Calculate 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 [Br - ] = 2.0 x 10-2 mol/l Substitute the molar concentrations into the K sp expression and solve. K sp = [Pb 2+ ][Br - ] 2 = (1.0 x 10-2 )(2.0 x 10-2 ) 2 = 4.0 x For the slightly soluble salt, CoS, the molar solubility is 5 x 10-5 M. Calculate the K sp for this compound. Slide 26 / 91 A 5 x 10-5 B 1.0 x 10-4 C 2.5 x 10-4 D 5 x E 2.5 x For the slightly soluble salt, BaF 2, the molar solubility is 3 x 10-4 M. Calculate the solubility-product constant for this compound. Slide 27 / 91 A 9 x 10-4 B 9 x 10-8 C 1.8 x 10-7 D 3.6 x 10-7 E 1.08 x 10-10

10 11 For the slightly soluble salt, La(IO 3) 3, the molar solubility is 1 x 10-4 M. Calculate K sp. Slide 28 / 91 A 3 x B 3 x C 2.7 x D 2.7 x E 1 x For the slightly soluble compound, Ca 3(PO 4) 2, the molar solubility is 3 x 10-8 moles per liter. Calculate the K sp for this compound. Slide 29 / 91 A 9.00 x B 1.08 x C 8.20 x D 1.35 x E 3.0 x The concentration of hydroxide ions in a saturated solution of Al(OH) 3 is 1.58x What is the K sp of Al(OH) 3? Slide 30 / 91

11 14 What is the Ksp of Fe(OH) 3(s) if a saturated solution of it has a ph of 11.3? Slide 31 / 91 A 2.0 x B 1.6 x C 2.1 x D 1.4 x 10-8 E 5.4 x Slide 32 / 91 Calculating Solubility from the K sp Return to the Table of Contents Calculating Solubility from the K sp Slide 33 / 91 Example: What is the molar solubility of a saturated aqueous solution of BaCO 3? C = 5.0 x 10-9 ) BaCO 3(s) --> Ba 2+ (aq) + CO 3 2- (aq) Ksp = 5.0 x 10-9 = [Ba 2+ ][CO 3 2- ] Since neither ion concentration is known, we will substitute "x" for the [Ba 2+ ] and "x" for the [CO 3 2- ]. 5.0 x 10-9 = (x)(x) = x 2 "x" = [Ba 2+ ] = [CO 3 2- ] = 7.07 x 10-5 M Since 1 Ba 2+ or 1 CO 3 2- are required for 1 BaCO 3, the molar solubility of the BaCO 3(s) = 7.07 x 10-5 M.

12 Calculating Solubility from the K sp Slide 34 / 91 Example: What is the molar solubility of a saturated aqueous solution of PbI 2? 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 - ] 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. 15 Calculate the concentration of silver ion when the solubility product constant of AgI is 1 x Slide 35 / 91 A 0.5 (1 x ) B 2 (1 x ) C (1 x ) 2 D (1 x ) 16 Calculate the molar solubility of PbF 2 that has a K sp at 25 = 3.6 x Students type their answers here Slide 36 / 91

13 17 The K sp of a compound of formula AB 3 is 1.8 x What is the molar solubility of the compound? Slide 37 / The K sp of a compound of formula AB 3 is 1.8 x The molar mass is 280g/mol. What is the solubility? Slide 38 / Which of the following ionic salts would have the highest molar solubility? Slide 39 / 91 A NiCO 3(s) Ksp = 6.61 x 10-9 B MnCO 3(s) Ksp = 1.82 x C ZnCO 3(s) Ksp = 1.45 x D Ag 2CrO 4(s) Ksp = 9.00 x E All have the same molar solubility

14 Slide 40 / 91 Factors Affecting Solubility Return to the Table of Contents Common Ion Effect Slide 41 / 91 Consider a saturate solution of barium sulfate: BaSO 4(s) Ba 2+ (aq) + SO 4 2- (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. So adding any soluble salt containing either Ba 2+ or SO 4 2- ions will decrease the solubility of barium sulfate. Common Ion Effect Slide 42 / 91 Sample Problem Calculate the solubility of CaF 2 in grams per liter in a) pure water b) a 0.15 M KF solution c) a M Ca(NO 3) 2 solution The solubility product for calcium fluoride, CaF 2 is 3.9 x 10-11

15 Common Ion Effect Slide 43 / 91 a) pure water CaF 2(s) Ca 2+ (aq) + 2F - (aq) If we assume x as the dissociation then, Ca 2+ ions = x and [F - ] = 2x K sp = [Ca 2+ ] [F - ] 2 = (x)(2x) 2 Note K sp = 3.9 x = 4x 3 So x = 2.13 x 10-4 mol/l x (78 g/mol CaF 2) Solubility is g/l Common Ion Effect 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, CaF 2 is 3.9 x [ F - ] = 0.15M Ksp = [Ca 2+ ] [F - ] 2 = (x)(0.15) 2 Note Ksp = 3.9 x = x So x = mol/l Solubility is = x (78 g/mol CaF 2) = g/l Common Ion Effect Slide 45 / 91 Calculate the solubility of CaF 2 in grams per liter in c) a M Ca(NO 3) 2 solution [Ca 2+ ] = 0.08M The solubility product for calcium fluoride,caf 2 is 3.9 x CaF 2 (s) Ca 2+ (aq) + 2 F - (aq) Ksp = [Ca 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 CaF 2)/ 2 Solubility is g/l

16 Common Ion Effect Slide 46 / 91 Recall from the Common-Ion Effect that adding a strong electrolyte to a weakly soluble solution with a common ion will decrease the solubility of the weak electrolyte. Compare the solubilities from the previous Sample Problem CaF 2 (s) Ca 2+ (aq) + 2 F - (aq) CaF 2 dissolved with: pure water Solubility of CaF g/l M KF 1.35x10-7 g/l M Ca(NO 3) g/l These results support Le Chatelier's Principle that increasing a product concentration will shift equilibrium to the left. # 20 What is the molar solubility of a saturated solution of Ag 2CrO 4? K sp at 25 is = 1.2 x Slide 47 / 91 A 1.1 x 10-4 B 6.7 x 10-5 C 8.4 x 10-5 D 5.5 x 10-7 E 2.2 x What is the molar solubility of a saturated solution of Ag 2CrO 4 in 0.100M K 2CrO 4? K sp at 25 is = 1.2 x Slide 48 / 91 A 3.0 x B 6.3 x 10-5 C 5.1 x 10-8 D 3.5 x 10-7 E 1.7 x 10-6

17 22 What is the molar solubility of a saturated solution of Ag 2CrO 4 in 0.200M AgCl? K sp at 25 is = 1.2 x Slide 49 / 91 A 3.0 x B 6.3 x 10-5 C 3.11 x D 3.5 x 10-7 E 6.7 x 10-6 Changes in ph Slide 50 / 91 The solubility of almost any ionic compound is affected by changes in ph. Consider 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 Châtelier s Principle.) Changes in ph Slide 51 / 91 Changes in ph can also affect the solubility of salts that contain the conjugate base of a weak acid. Consider the dissociation of the salt calcium fluoride: CaF 2 (s) Ca 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 CaF 2 be more or less soluble?

18 Changes in ph Slide 52 / 91 Sample Problem Calculate 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 Changes in ph Sample Problem Calculate the molar solubility of Mn(OH) 2 in a) in a solution buffered at ph = 9.5 In a solution buffered at ph=9.5, the [H + ] = 3.2 x 10-10, the [OH - ] = 3.2x Slide 53 / 91 The solubility product for Mn(OH) 2, is 1.6 x [ OH - ] = 3.2 x 10-5 M 1.6 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 Changes in ph Sample Problem Calculate the molar solubility of Mn(OH) 2 in b) in a solution buffered at ph = 8.0 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 Note 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

19 Changes in ph Sample Problem Calculate the molar solubility of Mn(OH) 2 in c) in pure water 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 Note 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 Changes in ph Slide 56 / 91 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. Do you remember what it means when a substance is amphoteric? 23 Given the system at equilibrium AgCl (s) Ag + (aq) + Cl - (aq) Slide 57 / 91 When 0.01 M HCl is added to the sytem, the point of equilibrium will shift to the. A B C D right and the concentration of Ag + will decrese right and the concentration of Ag + will increase left and the concentration of Ag + will decrease left and the concentration of Ag + will increase

20 24 Which of the following substances are more soluble in acidic solution than in basic solution? Select all that apply. Slide 58 / 91 A PbCl 2 B BaCO 3 C AgI D Fe(OH) 3 E MgF 2 25 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 / 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? Explain. Slide 60 / 91 A Yes B No

21 27 The molar solubility of NH 4Cl increases as ph. Slide 61 / 91 A increases B decreases C is unaffected by changes in ph 28 The molar solubility of Na 2CO 3 increases as ph. Slide 62 / 91 A increases B decreases C is unaffected by changes in ph Complex 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. Slide 63 / 91 For example, the addition of excess ammonia to AgCl will cause the AgCl to dissolve. This process is the sum of two reactions resulting in: AgCl (s) + 2NH 3(aq) Ag(NH 3) 2 + (aq) + Cl - (aq) Added NH 3 reacts with Ag + forming Ag(NH 3) 2+. Adding enough NH 3 results in the complete dissolution of AgCl.

22 Amphoterism Slide 64 / 91 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. Examples of such these substances are oxides and hydroxides of Al 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: Al(OH) 3(s) # Al 3+ (aq) + 3 OH - (aq) # Amphoterism Slide 65 / 91 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. Aluminum hydroxide reacts with OH - to form a complex ion in the following reaction: Al(OH) 3(s) + OH - (aq) # Al(OH) 4 - (aq) As a result of the formation of the complex ion, Al(OH) 4 -, aluminum hydroxide is more soluble. Many metal hydroxides only react in strongly acidic solutions. Ca(OH) 2, Fe(OH) 2 and Fe(OH) 3 are only more soluble in acidic solution they are not amphoteric. 29 Which of the following factors affect solubility? Slide 66 / 91 A B C D E ph Formation of Complex Ions Common-Ion Effect A and C A, B, and C

23 Slide 67 / 91 Precipitation Reactions and Separation of Ions Return to the Table of Contents Precipitation Reactions and Separation of Ions Slide 68 / 91 Do you remember the solubility rules? 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: Any salt made with a Group I metal is soluble. All salts containing nitrate ion are soluble. All salts containing ammonium ion are soluble. Do you remember what metal cations tended to be insoluble? Ag +, Pb 2+, and Hg 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? Slide 69 / 91 A B C D E sodium silver sodium nitrate chloride nitrate silver chloride Not enough information

24 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? Slide 70 / 91 A B C D E potassium bromide calcium carbonate potassium calcium carbonate bromide Not enough information 32 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? Slide 71 / 91 A PbSO 4 B Pb(SO 4 ) 2 C Pb 2 SO 4 D Mg(NO 3 ) 2 E Not enough information 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? Slide 72 / 91 A B C D E 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

25 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? Slide 73 / 91 A B C D A zinc carbonate precipitate forms, since Q>K. A zinc carbonate precipitate forms, since Q<K. A sodium nitrate precipitate forms, since Q>K. 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. Separation of Ions Slide 75 / 91 Imagine, you have a test tube that contains Ag +, Pb 2+ and Cu 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.

26 Separation of Ions Slide 76 / 91 You should remember that Ag + and Pb 2+ readily form insoluble salts and that Cu 2+ does not form insoluble salts as readily. Looking at some solubility product values, you will find the following: Salt Ksp Ag 2S 6 x PbS 3 x CuS 6 x AgCl 1.8 x PbCl x 10-5 You will notice that CuCl 2 is not to be found. This means CuCl 2 is a soluble salt! Separation of Ions Slide 77 / 91 Adding Cl - should precipitate the Ag + and Pb 2+ ions but not the Cu 2+ ions. We can remove Ag + and Pb 2+ from the test tube. Now, how can we separate the Ag + and Pb 2+ ions? Salt Ksp Ag 2S 6 x PbS 3 x CuS 6 x AgCl 1.8 x PbCl x 10-5 Do you notice the significant difference between the K sp values for Ag 2S and PbS? Maybe we can precipitate one of the salts out before the other if we control the concentration of S 2- added. Which salt Ag 2S and PbS should precipitate first when we begin to add S 2-? Separation of Ions Slide 78 / 91 If we have 0.100M concentrations of Ag + and Pb 2+ and we begin to add 0.200M K 2S the Ag 2S should precipitate first. For Ag 2S: K sp = 6 x = [Ag + ] 2 [S 2- ] = (0.100) 2 (x) x = [S 2- ] = 6 x M. If this concentration of S 2- is added Ag 2S will precipitate. For PbS: Ksp = 3 x = [Pb 2+ ][S 2- ] = 0.100(x) x = [S 2- ] = 3 x M. A greater amount of S 2- is needed to precipitate the PbS. Therefore, Ag 2S will precipitate first.

27 Separation of Ions Problems Slide 79 / 91 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 - Determine which of the products is the precipitate. Write the K sp expression for this compound. Step 2 - Calculate the cation concentration of this slightly soluble compound. Step 3 - Calculate 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 - Compare Q to K to determine whether a precipitate will form. Separation of Ions Problems Slide 80 / 91 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 YES 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. Sample Problem Separation of Ions Problems Slide 81 / 91 Will a precipitate form if you mix 50.0 ml of 0.20 M barium chloride, BaCl 2, and 50.0 ml of 0.30 M sodium sulfate, Na 2SO 4? Step 1 - Determine which of the products is the precipitate. Write the K sp expression for this compound. BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) Step 2 - Calculate 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 BaCl 2 [Ba 2+ ] = 0.10 M

28 Separation of Ions Problems Sample Problem - Answers (con't) Will a precipitate form if you mix 50.0 ml of 0.20 M barium chloride, BaCl 2, and 50.0 ml of 0.30 M sodium sulfate, Na 2SO 4? Slide 82 / 91 Step 3 - Calculate 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 2- ] = 0.15 M Step 4 - Substitute the values into the reaction quotient (Q) expression. Recall that this is the same expression as K. Q = [Ba 2+ ] [SO 4 2- ] = (0.10) (0.15) = Separation of Ions Problems Slide 83 / 91 Sample Problem - Answers (con't) Will a precipitate form if you mix 50.0 ml of 0.20 M barium chloride, BaCl 2, and 50.0 ml of 0.30 M sodium sulfate, Na 2SO 4? Step 5 - Compare 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 BaCl 2, and 0.30 M Na 2SO 4. Separation of Ions Problems Slide 84 / 91 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.

29 35 A solution contains 2.0 x 10-5 M barium ions and 1.8 x 10-4 M lead (II) ions. If Na 2CrO 4 is added, which will precipitate first from solution? The K sp for BaCrO 4 is 2.1 x and the K sp for PbCrO 4 is 2.8 x Slide 85 / 91 A BaCrO 4 B PbCrO 4 C They will precipitate at the same time. D It's impossible to determine with the information provided. 36 A solution contains 2.0 x 10-4 M Ag + and 2.0 x 10-4 M Pb 2+. If NaCl is added, will AgCl (K sp = 1.8 x ) or PbCI 2 (K sp = 1.7 x 10-5 ) precipitate first? Slide 86 / 91 A AgCl B PbCl 2 C They will precipitate at the same time. D It is impossible to determine with the information provided. 37 A solution contains 2.0 x 10-4 M Ag + and 1.7 x 10-3 M Pb 2+. If NaCl is added. What concentration of Cl - is needed to Students type their answers here begin precipitation. AgCl (K sp = 1.8 x ) and PbCI 2 (K sp = 1.7 x 10-5 ) Slide 87 / 91

30 38 A solution contains 2.0 x 10-4 M Ag + and 1.7 x 10-3 M Pb 2+. If NaCl 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? AgCl (Ksp = 1.8 x ) and PbCI 2 (Ksp = 1.7 x 10-5 ) Slide 88 / Will Co(OH) 2 precipitate from solution if the ph of a 0.002M solution of Co(NO 3) 2 is adjusted to 8.4? K sp for Co(OH) 2 is 2.5 x A Yes Slide 89 / 91 B No 40 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 Slide 90 / 91 Students type their answers here

31 Slide 91 / 91