Slide 1 / 57 Slide 2 / 57 Solubility Products Consider the equilibrium that exists in a saturated solution of BaSO 4 in water: Aqueous equilibria II Solubility Products BaSO 4 (s) ↔ Ba 2+ (aq) + SO 4 2- (aq) Slide 3 / 57 Slide 4 / 57 1 Which Ksp expression is correct for AgCl? Solubility Products [Ag + ]/[Cl - ] A [Ag + ][Cl - ] B The equilibrium constant expression for this equilibrium is [Ag 2+ ] 2 [Cl 2- ] 2 C [Ag + ] 2 [Cl - ] 2 K sp = [Ba 2+ ] [SO 4 2− ] D None of the above. E 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 5 / 57 Slide 6 / 57 2 Given the reaction at equilibrium: Solubility Products 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 is not the same as solubility. · Solubility is generally expressed as the mass of solute K sp = [Zn 2+ ][OH - ] 2 / [Zn(OH) 2 ] A dissolved in 1 L (g/L) or 100 mL (g/mL) of solution, or in mol/L ( M ). K sp = [Zn(OH) 2 ] / [Zn 2+ ][2OH - ] B K sp = [Zn 2+ ][2OH - ] C K sp = [Zn 2+ ][OH - ] 2 D
Slide 7 / 57 Slide 8 / 57 Solubility Solubility Example #1 Consider the slightly soluble compound barium oxalate, BaC 2 O 4 . The term "solubility" represents the maximum amount of solute that can be dissolved in a certain volume The solubility of BaC 2 O 4 is 1.3 x 10 -3 mol/L. before any precipitate is observed. The ratio of cations to anions is 1:1. The solubility of a substance can be given in terms of · This means that 1.3 x 10 -3 moles of Ba 2+ can dissolve grams per liter g/L in one liter. · Also, 1.3 x 10 -3 moles of C 2 O 4 2- can dissolve in one or in terms of liter. moles per liter mol/L · What is the maximum amount (in grams) of BaC 2 O 4 that could dissolve in 2.5 L (before a solid precipitate The latter is sometimes referred to as "molar or solid settlement occurs)? solubility." Slide 9 / 57 Slide 10 / 57 Solubility Solubility What is the maximum amount (in grams) of BaC 2 O 4 that could dissolve in 2.5 L (before a Example #2 precipitate occurs)? Consider the slightly soluble compound lead (II) The solubility of BaC 2 O 4 is 1.3 x 10 -3 mol/L. chloride, PbCl 2 . BaC 2 O 4 --> Ba 2+ + C 2 O 4 2- The solubility of PbCl 2 is 0.016 mol/L. 1 mol BaC 2 O 4 225.3g -------------------- = ----------- The ratio of cations to anions is 1:2. 1.3 x 10 -3 mol BaC 2 O 4 x g = 0.293g BaC 2 O 4 x 2.5L = 0.73g · This means that 0.016 moles of Pb 2+ can dissolve in 1L one liter. Molar solubility always refers to the ion with the lower molar ratio. This is the maximum amount that could dissolve in 2.5 L before a precipitate occurs. · Twice as much, or 2(0.016) = 0.032 moles of Cl - can dissolve in one liter. 1.3 x 10 -3 mol BaC 2 O 4 2.5 L 225.3 g BaC 2 O 4 x x 1 1 mol BaC 2 O 4 1 L Slide 11 / 57 Slide 12 / 57 Solubility Solubility Example #3 Consider the slightly soluble compound silver sulfate, Remember that molar solubility refers to the ion with the lower mole ratio. It does not always refer to the Ag 2 SO 4 . cation, although in most cases it does. The solubility of Ag 2 SO 4 is 0.015 mol/L. Molar The ratio of cations to anions is 2:1. Compound Solubility of [Cation] [Anion] Compound 2- can dissolve · This means that 0.015 moles of SO 4 1.3 x 10 -3 1.3 x 10 -3 1.3 x 10 -3 in one liter. BaC 2 O 4 mol mol mol · Twice as much, or 2(0.015) = 0.030 moles of Ag + can PbCl 2 0.016 mol/L 0.016 mol/L 0.032 mol/L dissolve in one liter. Ag 2 SO 4 0.015 mol/L 0.030 mol/L 0.015 mol/L
Slide 13 / 57 Slide 14 / 57 If the solubility of barium carbonate, BaCO 3 is 7.1 x 10 -5 M, If the solubility of barium carbonate, BaCO 3 is 7.1 x 10 -5 M, 3 4 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. 7.1 x 10 -5 moles A 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 15 / 57 Slide 16 / 57 If the solubility of Ag 2 CrO 4 is 6.5 x 10 -5 M, this means If the solubility of Ag 2 CrO 4 is 6.5 x 10 -5 M, this means 2- , can that a maximum of _______ Ag + ions can be dissolved 5 6 that a maximum of _______chromate ions, CrO 4 be dissolved per liter of solution. per liter of solution. 6.5 x 10 -5 moles 6.5 x 10 -5 moles A A twice 6.5 x 10 -5 moles B twice 6.5 x 10 -5 moles B half 6.5 x 10 -5 moles C half 6.5 x 10 -5 moles C one-fourth 6.5 x 10 -5 moles D one-fourth 6.5 x 10 -5 moles D four times 6.5 x 10 -5 moles E four times 6.5 x 10 -5 moles E Slide 17 / 57 Slide 18 / 57 7 For the slightly soluble compound, AB, the molar solubility is 3 x 10 -8 moles per liter. Calculate the Calculating K sp from the Solubility K sp for this compound. No calculator. AB <--> A + + B - A 3 x 10 -8 Sample Problem The molar solubility of lead (II) bromide, PbBr 2 is 1.0 x 1/2 (3 x 10 -8 ) B 10 -2 at 25 o C. Calculate the solubility product, K sp , for this compound. (3 x 10 -8 )^1/2 C [Pb 2+ ] = 1.0 x 10 -2 mol/L 2 (3 x 10 -8 ) D [Br - ] = 2.0 x 10 -2 mol/L (3 x 10 -8 )^2 Substitute the molar concentrations into the K sp expression E and solve. K sp = [Pb 2+ ][Br - ] 2 = (1.0 x 10 -2 )(2.0 x 10 -2 ) 2 = 4.0 x 10 -6 AgCl < - - > Ag+ + Cl-
Slide 19 / 57 Slide 20 / 57 8 For the slightly soluble compound, XY, the molar 9 For the slightly soluble compound, MN, the molar solubility is 5 x 10 -5 M. Calculate the K sp for this solubility is 4 x 10 -6 M. Calculate the K sp for this compound. No calculator. compound. No calculator. MN <--> M + + N - XY <--> X + + Y - A 4 x 10 -6 A 5 x 10 -5 10 x 10 -5 B 16 x 10 -6 B 25 x 10 -5 C 16 x 10 -12 C 5 x 10 -10 D 16 x 10 -36 D 25 x 10 -10 E BaCO 3 < - - > Ba 2+ + CO 3 2- Slide 21 / 57 Slide 22 / 57 10 For the slightly soluble compound, AB 2 , the molar 11 For the slightly soluble compound, X 3 Y, the molar solubility is 3 x 10 -4 M. Calculate the solubility- solubility is 1 x 10 -4 M. Calculate the solubility product constant for this compound. product for this compound. No calculator. No calculator. AB 2 <--> A 2+ + 2B - X 3 Y <-> 3X + + Y 3- A 9 x 10 -4 A 3 x 10 -4 9 x 10 -8 3 x 10 -8 B B 18 x 10 -8 27 x 10 -12 C C 36 x 10 -8 27 x 10 -16 D D 108 x 10 -12 E Fe(OH) 3 < - - > Fe 3+ + 3(OH) - Na 3 P <--> 3Na+ + P 3- PbCl 2 < - - > Pb 2+ + 2Cl - Slide 23 / 57 Slide 24 / 57 Calculating Solubility from the K sp Calculating Solubility from the K sp a) pure water CaF 2 < - - > Ca 2+ + 2F - If we assume x as the dissociation then, Sample Problem Ca 2+ ions = x Calculate the solubility of CaF 2 in grams per liter in and [F - ] = 2x a) pure water K sp = [Ca 2+ ] [F - ] 2 = ( x )(2 x ) 2 b) a 0.15 M KF solution c) a 0.080 M Ca(NO 3 ) 2 solution K sp = 3.9 x 10 -11 = 4 x 3 So x = 2.13 x 10 -4 mol/L x (78 g/mol CaF 2 ) The solubility product for calcium fluoride, CaF 2 is 3.9 x 10 -11 Solubility is 0.0167 g/L 2.13 x10 -4 Ca 2+ mol/L x 1mol/L CaF 2 78g/L --------------------- x ------------- 1mol/L Ca 2+ ions 1 mol CaF 2
Slide 25 / 57 Slide 26 / 57 Calculating Solubility from the K sp Calculating Solubility from the K sp b) a 0.15 M KF solution Calculate the solubility of CaF 2 in grams per liter in remember KF is a strong electrolyte, is completely ionized. c) a 0.080 M Ca(NO 3 ) 2 solution the major source of F- ions, then [F - ] =0.15M [Ca 2+ ] = 0.08M The solubility product for calcium fluoride, CaF 2 is 3.9 x 10 -11 The solubility product for calcium fluoride,CaF 2 is 3.9 x 10 -11 [ F - ] = 0.15M Ksp = [Ca 2+ ] [F - ] 2 = ( x )(0.15) 2 Ca 2+ 2 F - CaF 2 (s) <---> + (aq) (aq) Ksp = 3.9 x 10 -11 = 0.0225 x Ksp = [Ca 2+ ] [F-] 2 = (0.080)( x ) 2 So x = ______ mol/L Solubility is = ______ x (78 g/mol CaF 2 ) = ______ g/L Ksp = 3.9 x 10 -11 = 0.080 x 2 So x = 2.2 x 10 -5 mol/L * (78 g/mol CaF 2 )/ 2 1.73 x10 -9 Ca 2+ mol/L x 1mol/L CaF 2 78g/L --------------------- x ------------- Solubility is 0.000858 g/L 1mol/L Ca 2+ ions 1 mol CaF 2 Slide 27 / 57 Slide 28 / 57 Calculating Solubility from the K sp 12 Calculate the concentration of silver ion when the solubility product constant of AgI is 10 -16 . Recall from the Common-Ion Effect that adding a 0.5 (1 x 10 -16 ) strong electrolyte to a weakly soluble solution with a A common ion will decrease the solubility of the weak 2 (1 x 10 -16 ) electrolyte. B (1 x 10 -16 ) 2 C Compare the solubilities from the previous Sample Problem Ca 2+ 2 F - CaF 2 (s) <---> + (1 x 10 -16 ) D (aq) (aq) CaF 2 dissolved with: Solubility of CaF 2 pure water 0.016 g/L 0.015 M KF 0.080 M Ca(NO 3 ) 2 0.0017 g/L These results support Le Chatelier's Principle that increasing a product concentration will shift equilibrium to the left. Slide 29 / 57 Slide 30 / 57 The Ksp of a compound of formula AB 3 is 1.8 x 10 -18. 13 14 The Ksp of a compound of formula AB3 is 1.8 x 10-18. The molar solubility of the compound is ---- The solubility of the compound is ---- The molar mass is 210g/mol
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