Slide 1 / 57 Aqueous equilibria II Solubility Products
Slide 2 / 57 Solubility Products Consider the equilibrium that exists in a saturated solution of BaSO 4 in water: BaSO 4 (s) ↔ Ba 2+ (aq) + SO 4 2- (aq)
Slide 3 / 57 Solubility Products The equilibrium constant expression for this equilibrium is K sp = [Ba 2+ ] [SO 4 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 4 / 57 1 Which Ksp expression is correct for AgCl? [Ag + ]/[Cl - ] A [Ag + ][Cl - ] B [Ag 2+ ] 2 [Cl 2- ] 2 C [Ag + ] 2 [Cl - ] 2 D None of the above. E
Slide 5 / 57 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 ] A K sp = [Zn(OH) 2 ] / [Zn 2+ ][2OH - ] B K sp = [Zn 2+ ][2OH - ] C K sp = [Zn 2+ ][OH - ] 2 D
Slide 6 / 57 Solubility Products · K sp is not the same as solubility. · Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L) or 100 mL (g/mL) of solution, or in mol/L ( M ).
Slide 7 / 57 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."
Slide 8 / 57 Solubility Example #1 Consider the slightly soluble compound barium oxalate, BaC 2 O 4 . The solubility of BaC 2 O 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 2 O 4 2- can dissolve in one liter. · What is the maximum amount (in grams) of BaC 2 O 4 that could dissolve in 2.5 L (before a solid precipitate or solid settlement occurs)?
Slide 9 / 57 Solubility What is the maximum amount (in grams) of BaC 2 O 4 that could dissolve in 2.5 L (before a precipitate occurs)? The solubility of BaC 2 O 4 is 1.3 x 10 -3 mol/L. BaC 2 O 4 --> Ba 2+ + C 2 O 4 2- 1 mol BaC 2 O 4 225.3g -------------------- = ----------- 1.3 x 10 -3 mol BaC 2 O 4 x g = 0.293g BaC 2 O 4 x 2.5L = 0.73g 1L This is the maximum amount that could dissolve in 2.5 L before a precipitate occurs. 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 10 / 57 Solubility Example #2 Consider the slightly soluble compound lead (II) chloride, PbCl 2 . The solubility of PbCl 2 is 0.016 mol/L. The ratio of cations to anions is 1:2. · This means that 0.016 moles of Pb 2+ can dissolve in one liter. Molar solubility always refers to the ion with the lower molar ratio. · Twice as much, or 2(0.016) = 0.032 moles of Cl - can dissolve in one liter.
Slide 11 / 57 Solubility Example #3 Consider the slightly soluble compound silver sulfate, Ag 2 SO 4 . The solubility of Ag 2 SO 4 is 0.015 mol/L. The ratio of cations to anions is 2:1. 2- can dissolve · This means that 0.015 moles of SO 4 in one liter. · Twice as much, or 2(0.015) = 0.030 moles of Ag + can dissolve in one liter.
Slide 12 / 57 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. Molar Compound Solubility of [Cation] [Anion] Compound 1.3 x 10 -3 1.3 x 10 -3 1.3 x 10 -3 BaC 2 O 4 mol mol mol PbCl 2 0.016 mol/L 0.016 mol/L 0.032 mol/L Ag 2 SO 4 0.015 mol/L 0.030 mol/L 0.015 mol/L
Slide 13 / 57 If the solubility of barium carbonate, BaCO 3 is 7.1 x 10 -5 M, 3 this means that a maximum of _______barium ions, Ba 2+ ions can be dissolved per liter of solution. A 7.1 x 10 -5 moles B half of that C twice as much D one-third as much E one-fourth as much
Slide 14 / 57 If the solubility of barium carbonate, BaCO 3 is 7.1 x 10 -5 M, 4 this means that a maximum of _______carbonate ions, 2- ions can be dissolved per liter of solution. CO 3 7.1 x 10 -5 moles A B half of that C twice as much D one-third as much E one-fourth as much
Slide 15 / 57 If the solubility of Ag 2 CrO 4 is 6.5 x 10 -5 M, this means 2- , can 5 that a maximum of _______chromate ions, CrO 4 be dissolved per liter of solution. 6.5 x 10 -5 moles A twice 6.5 x 10 -5 moles B half 6.5 x 10 -5 moles C one-fourth 6.5 x 10 -5 moles D four times 6.5 x 10 -5 moles E
Slide 16 / 57 If the solubility of Ag 2 CrO 4 is 6.5 x 10 -5 M, this means that a maximum of _______ Ag + ions can be dissolved 6 per liter of solution. 6.5 x 10 -5 moles A twice 6.5 x 10 -5 moles B half 6.5 x 10 -5 moles C one-fourth 6.5 x 10 -5 moles D four times 6.5 x 10 -5 moles E
Slide 17 / 57 Calculating K sp from the Solubility 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. [Pb 2+ ] = 1.0 x 10 -2 mol/L [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 10 -6
Slide 18 / 57 7 For the slightly soluble compound, AB, the molar solubility is 3 x 10 -8 moles per liter. Calculate the K sp for this compound. No calculator. AB <--> A + + B - A 3 x 10 -8 1/2 (3 x 10 -8 ) B (3 x 10 -8 )^1/2 C 2 (3 x 10 -8 ) D (3 x 10 -8 )^2 E AgCl < - - > Ag+ + Cl-
Slide 19 / 57 8 For the slightly soluble compound, XY, the molar solubility is 5 x 10 -5 M. Calculate the K sp for this compound. No calculator. XY <--> X + + Y - A 5 x 10 -5 10 x 10 -5 B 25 x 10 -5 C 5 x 10 -10 D 25 x 10 -10 E BaCO 3 < - - > Ba 2+ + CO 3 2-
Slide 20 / 57 9 For the slightly soluble compound, MN, the molar solubility is 4 x 10 -6 M. Calculate the K sp for this compound. No calculator. MN <--> M + + N - A 4 x 10 -6 16 x 10 -6 B 16 x 10 -12 C 16 x 10 -36 D
Slide 21 / 57 10 For the slightly soluble compound, AB 2 , the molar solubility is 3 x 10 -4 M. Calculate the solubility- product constant for this compound. No calculator. AB 2 <--> A 2+ + 2B - A 9 x 10 -4 9 x 10 -8 B 18 x 10 -8 C 36 x 10 -8 D 108 x 10 -12 E PbCl 2 < - - > Pb 2+ + 2Cl -
Slide 22 / 57 11 For the slightly soluble compound, X 3 Y, the molar solubility is 1 x 10 -4 M. Calculate the solubility product for this compound. No calculator. X 3 Y <-> 3X + + Y 3- A 3 x 10 -4 3 x 10 -8 B 27 x 10 -12 C 27 x 10 -16 D Fe(OH) 3 < - - > Fe 3+ + 3(OH) - Na 3 P <--> 3Na+ + P 3-
Slide 23 / 57 Calculating Solubility from the K sp 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 0.080 M Ca(NO 3 ) 2 solution The solubility product for calcium fluoride, CaF 2 is 3.9 x 10 -11
Slide 24 / 57 Calculating Solubility from the K sp a) pure water CaF 2 < - - > Ca 2+ + 2F - If we assume x as the dissociation then, Ca 2+ ions = x and [F - ] = 2x K sp = [Ca 2+ ] [F - ] 2 = ( x )(2 x ) 2 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 ) 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 Calculating Solubility from the K sp b) a 0.15 M KF solution remember KF is a strong electrolyte, is completely ionized. the major source of F- ions, then [F - ] =0.15M 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 Ksp = 3.9 x 10 -11 = 0.0225 x So x = ______ mol/L Solubility is = ______ x (78 g/mol CaF 2 ) = ______ g/L 1.73 x10 -9 Ca 2+ mol/L x 1mol/L CaF 2 78g/L --------------------- x ------------- 1mol/L Ca 2+ ions 1 mol CaF 2
Slide 26 / 57 Calculating Solubility from the K sp Calculate the solubility of CaF 2 in grams per liter in c) a 0.080 M Ca(NO 3 ) 2 solution [Ca 2+ ] = 0.08M The solubility product for calcium fluoride,CaF 2 is 3.9 x 10 -11 Ca 2+ 2 F - CaF 2 (s) <---> + (aq) (aq) Ksp = [Ca 2+ ] [F-] 2 = (0.080)( x ) 2 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 Solubility is 0.000858 g/L
Slide 27 / 57 Calculating Solubility from the K sp 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 Ca 2+ 2 F - CaF 2 (s) <---> + (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 28 / 57 12 Calculate the concentration of silver ion when the solubility product constant of AgI is 10 -16 . 0.5 (1 x 10 -16 ) A 2 (1 x 10 -16 ) B (1 x 10 -16 ) 2 C (1 x 10 -16 ) D
Slide 29 / 57 The Ksp of a compound of formula AB 3 is 1.8 x 10 -18. 13 The molar solubility of the compound is ----
Slide 30 / 57 14 The Ksp of a compound of formula AB3 is 1.8 x 10-18. The solubility of the compound is ---- The molar mass is 210g/mol
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