AP Chemistry Aqueous Equilibria II: Ksp & Solubility Products - PDF document

Slide 1 / 91 Slide 2 / 91 AP Chemistry Aqueous Equilibria II: Ksp & Solubility Products Slide 3 / 91 Slide 4 / 91 Table of Contents: K sp & Solubility Products Click on the topic to go to that section Introduction to Introduction to Solubility Equilibria · Calculating K sp from the Solubility Solubility Equilibria · Calculating Solubility from Ksp · Factors Affecting Solubility · Return to the Precipitation Reactions and Separation of Ions · Table of Contents Slide 5 / 91 Slide 6 / 91 Introduction to Solubility Equilibria Introduction to Solubility Equilibria 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. Answer A saturated solution of CaCO 3 (s) 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. 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 / 91 Slide 8 / 91 Introduction to Solubility Equilibria Introduction to Solubility Equilibria Consider the equilibrium that exists in a saturated solution of CaCO 3 in water: The equilibrium constant expression for this equilibrium is CaCO 3 (s) ↔ Ca 2+ (aq) + CO 32- (aq) K sp = [Ca 2+ ] [CO 32− ] where the equilibrium constant, K sp , is called the Unlike acid-base equilibria which are homogenous, solubility product. solubility equilibria are heterogeneous, there is always a solid in the reaction. There is never any denominator in K sp expressions because pure solids are not included in any equilibrium expressions. Slide 9 / 91 Slide 10 / 91 Solubility Equilibrium 1 Which Ksp expression is correct for AgCl? [Ag + ]/[Cl - ] The degree to which an ionic compound dissociates in water can be A determined by measuring it's "K sp " or solubility product equilibrium [Ag + ][Cl - ] B constant. [Ag 2+ ] 2 [Cl 2- ] 2 C CaCO 3 (s) --> Ca 2+ (aq) + CO 32- (aq) Ksp @ 25 C = 5.0 x 10 -9 [Ag + ] 2 [Cl - ] 2 Answer D MgCO 3 (s) --> Mg 2+ (aq) + CO 32- (aq) Ksp @ 25 C = 6.8 x 10 -6 None of the above. E In both cases above, the equilibrium lies far to the left, meaning relatively few aqueous ions would be present in solution. Which saturated solution above would have the higher conductivity and why? Slide 11 / 91 Slide 12 / 91 2 Given the reaction at equilibrium: 3 Which Ksp expression is correct for Fe 3 (PO 4 ) 2 ? Zn(OH) 2 (s) Zn 2+ (aq) + 2OH - (aq) what is the expression for the solubility product A [Fe 2+ ] 3 [PO 43- ] 2 constant, K sp , for this reaction? [ Fe 2+ ] 3 /[ PO 43- ] 2 B C [ Fe 3+ ] 2 [ PO 43- ] 2 K sp = [Zn 2+ ][OH - ] 2 / [Zn(OH) 2 ] A D [ Fe 2+ ] 2 /[ PO 43- ] 2 None of the above. K sp = [Zn(OH) 2 ] / [Zn 2+ ][2OH - ] E B K sp = [Zn 2+ ][2OH - ] C K sp = [Zn 2+ ][OH - ] 2 D

Slide 13 / 91 Slide 14 / 91 4 When 30 grams of NaCl are mixed into 100 mL of 5 The conductivity of a saturated solution of Ag 2 CO 3 would distilled water all of the solid NaCl dissolves. The solution be expected to be less than the conductivity of a must be saturated and the K sp for the NaCl must be very saturated solution of CaCO 3 . Justify your answer. high. True True False False Slide 15 / 91 Slide 16 / 91 Solubility Solubility Example #1 The term solubility represents the maximum amount of solute that can be dissolved in a certain volume before any precipitate is Consider the slightly soluble compound barium oxalate, observed. BaC 2 O 4 . The solubility of BaC 2 O 4 is 1.3 x 10 -3 mol/L. The solubility of a substance can be given in terms of grams per liter g/L The ratio of cations to anions is 1:1. or in terms of This means that 1.3 x 10 -3 moles of Ba 2+ can dissolve in one liter. moles per liter mol/L Also, 1.3 x 10 -3 moles of C 2 O 42- can dissolve in one liter. What is the maximum amount (in grams) of BaC 2 O 4 that could The latter is sometimes referred to as molar solubility. For any slightly dissolve in 2.5 L (before a solid precipitate or solid settlement soluble salt the molar solubility always refers to the ion with the lower molar ratio. occurs)? Slide 17 / 91 Slide 18 / 91 Solubility Solubility Example #1 Example #2 What is the maximum amount (in grams) of BaC 2 O 4 that could Consider the slightly soluble compound lead chloride, dissolve in 2.5 L (before a precipitate occurs)? PbCl 2 . The solubility of BaC 2 O 4 is 1.3 x 10 -3 mol/L. The solubility of PbCl 2 is 0.016 mol/L. BaC 2 O 4 (s) --> Ba 2+ (aq) + C 2 O 42-(aq) The ratio of cations to anions is 1:2. 1.3 x 10 -3 mol BaC 2 O 4 3.25 x 10 - 3 g 2.5L x -------------------- = BaC 2 O 4 1 liter This means that 0.016 moles of Pb 2+ can dissolve in one liter. 3.25 x 10 - 3 g x 1 mole = 0.73g BaC 2 O 4 BaC 2 O 4 225.3 g Twice as much, or 2(0.016) = 0.032 moles of Cl - can 0.73g is the maximum amount of BaC 2 O 4 that could dissolve dissolve in one liter. in 2.5 L before a precipitate forms.

Slide 19 / 91 Slide 20 / 91 Solubility Solubility Example #3 Remember that molar solubility refers to the ion with the lower mole ratio . It does not always refer to the Consider the slightly soluble compound silver sulfate, cation, although in most cases it does. Ag 2 SO 4 . The solubility of Ag 2 SO 4 is 0.015 mol/L. Molar Compound Solubility of [Cation] [Anion] Compound The ratio of cations to anions is 2:1. 1.3 x 10 -3 1.3 x 10 -3 1.3 x 10 -3 BaC 2 O 4 mol mol mol This means that 0.015 moles of SO 42- can dissolve in one liter. PbCl 2 0.016 mol/L 0.032 mol/L 0.016 mol/L Ag 2 SO 4 0.030 mol/L 0.015 mol/L 0.015 mol/L Twice as much, or 2(0.015) = 0.030 moles of Ag + can dissolve in one liter. Slide 21 / 91 Slide 22 / 91 If the solubility of barium carbonate, BaCO 3 is 7.1 x 10 -5 M, 6 If the solubility of barium carbonate, BaCO 3 is 7.1 x 10 -5 M, 7 this means that a maximum of _______carbonate ions, this means that a maximum of _______barium ions, Ba 2+ 2- ions can be dissolved per liter of solution. CO 3 ions can be dissolved per liter of solution. A 7.1 x 10 -5 moles A 7.1 x 10 -5 moles B half of that B half of that C twice as much C twice as much D one-third as much D one-third as much E one-fourth as much E one-fourth as much Slide 23 / 91 Slide 24 / 91 If the solubility of Ag 2 CrO 4 is 6.5 x 10 -5 M, this means 8 that a maximum of _______silver ions, Ag + , can be dissolved per liter of solution. A 6.5 x 10 -5 moles Calculating K sp from B twice 6.5 x 10 -5 moles the Solubility C half 6.5 x 10 -5 moles D one-fourth 6.5 x 10 -5 moles E four times 6.5 x 10 -5 moles Return to the Table of Contents

Slide 25 / 91 Slide 26 / 91 9 For the slightly soluble salt, CoS, the molar Calculating K sp from the Solubility solubility is 5 x 10 -5 M. Calculate the K sp for this compound. Sample Problem The molar solubility of lead (II) bromide, PbBr 2 is 1.0 x 10 -2 at A 5 x 10 -5 25 o C. Calculate the solubility product, K sp , for this compound. B 1.0 x 10 -4 The molar solubility always refers to the ion of the lower molar ratio, therefore [Pb 2+ ] = 1.0 x 10 -2 mol/L and 2.5 x 10 -4 C [Br - ] = 2.0 x 10 -2 mol/L D 5 x 10 -10 Substitute the molar concentrations into the K sp expression and solve. 2.5 x 10 -9 E K sp = [Pb 2+ ][Br - ] 2 = (1.0 x 10 -2 )(2.0 x 10 -2 ) 2 = 4.0 x 10 -6 Slide 27 / 91 Slide 28 / 91 10 For the slightly soluble salt, BaF 2 , the molar solubility 11 For the slightly soluble salt, La(IO 3 ) 3 , the molar is 3 x 10 -4 M. Calculate the solubility-product solubility is 1 x 10 -4 M. Calculate K sp . constant for this compound. A 3 x 10 -12 A 9 x 10 -4 B 3 x 10 -16 B 9 x 10 -8 2.7 x 10 -11 C C 1.8 x 10 -7 D 2.7 x 10 -15 3.6 x 10 -7 D 1.08 x 10 -10 E E 1 x 10 -12 Slide 29 / 91 Slide 30 / 91 12 For the slightly soluble compound, Ca 3 (PO 4 ) 2 , the 13 The concentration of hydroxide ions in a saturated molar solubility is 3 x 10 -8 moles per liter. Calculate solution of Al(OH) 3 is 1.58x10 -15 . What is the K sp of the K sp for this compound. Al(OH) 3 ? 9.00 x 10 -16 A B 1.08 x 10 -38 C 8.20 x 10 -32 D 1.35 x 10 -13 E 3.0 x 10 -20