CHEM 3.6 (5 credits) Demonstrate understanding of equilibrium principals in aqueous systems

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1 CHEM 3.6 (5 credits) Demonstrate understanding of equilibrium principals in aqueous systems sparingly soluble ionic solids acidic and basic solutions concentrations of dissolved species K s calculations common ion effect predicting precipitation/dissolution K a and pk a calculations concentration of species present ph and conductivity titration curves and selection of indicators

2 Do now: Write K a expressions for the following acids: NH 4+ + H 2 O NH 3 + H 3 O + HOCl + H 2 O OCl - + H 3 O + K a = [NH 3 ] [H 3 O + ] [NH 4+ ] K a = [OCl - ] [H 3 O + ] [HOCl] K a (HCOOH) K a = [HCOO - ] [H 3 O + ] [HCOOH]

3 Equilibrium expressions You learnt last year in 2.6 how to write and calculate equilibrium expressions. Quick recap: For the equation Ca(NO 3 ) 2(s) Ca 2+ (aq) + 2 NO 3 - (aq) The equilibrium expression is: K c = [Ca 2+ ][NO 3- ] 2 [Ca(NO 3 ) 2 ] [products] [reactants] Key words Solvent Solute Saturated solution Solubility Solubility product

4 Solubility constant (K s ) K s is a new constant this year the solubility product or solubility constant Just like K c except we don t include solids (as they aren t soluble!) For the equation Ca(NO 3 ) 2(s) Ca 2+ (aq) + 2 NO 3 - (aq) The solubility constant is: K s = [Ca 2+ ][NO 3- ] 2 K s is constant so long as the temperature is constant. The greater the value of K s the better the solubility. We are dealing with sparingly soluble ionic compounds this year K s values will be small

5 Do now: Write K s expressions for the following ionic solids MgS Mg 2+ + S CaF 2 Ca F - K s = [Mg 2+ ] [S ] K s = [Ca 2+ ] [F - ] 2 Ba(OH) 2 Ba(OH) 2 Ba OH - K s = [Ba 2+ ] [OH - ] 2 magnesium oxide MgO Mg 2+ + O K s = [Mg 2+ ] [O ]

6 mol.l -1 and g.l -1 Solubilities of ions in solution will be given to you as either mol.l -1 or g.l -1. We need to be able to convert between these two units g x mol = mol L g L workbook pg 192 Q3 1/molar mass The solubility of Mg(OH) 2 is 1.16 x 10-2 g.l -1, what is its solubility in mol.l -1? 1.98 x 10-4 mol.l -1 Stronium flouride (SrF 2 )has a solubility of g per 100 ml of water. What is its solubility in mol.l -1? M(SrF 2 ) = g.mol g.l x 10-4 mol.l -1 Workbook pg

7 Calculations with K s There are two types of ionic solids depending on their formula. AB - NaCl, CaSO 4 A 2 B/AB 2 - Na 2 SO 4, Ag 2 CO 3 MgCl 2, Cu(NO 3 ) 2 We are expected to be able to calculate K s if given the concentration of ions for a sparingly soluble ionic salt OR calculate the solubility (and/or concentration of ions in solution) of a sparingly soluble ionic salt if given K s. A x B y x A + y B K s = [A] x [B] y In both cases always write the dissolving equation and the K s expression first

8 Calculating K s from solubility The solubility of AgI in pure water is 9.1 x 10-9 mol.l -1, calculate the solubility constant, K s. AgI (s) Ag + (aq) + I - (aq) K s = [Ag + ] [I - ] [Ag + ] = 9.1 x 10-9 mol.l -1 [I - ] = 9.1 x 10-9 mol.l -1 K s = (9.1 x 10-9 ) 2 K s no units! = x The solubility of CaF 2 in pure water is 2.3 x 10-4 mol.l -1, calculate the solubility constant, K s. CaF 2(s) Ca 2+ (aq) + 2 F - (aq) K s = [Ca 2+ ] [F - ] 2 [Ca 2+ ] = 2.3 x 10-4 mol.l -1 [F- ] = 2 x 2.3 x 10-4 mol.l -1 = 4.6 x 10-4 mol.l -1 K s = 2.3 x 10-4 x (4.6 x 10-4 ) 2 = 4.87 x 10-11

9 Calculating K s from solubility The solubility of Ag 2 O in pure water is mol.l -1, calculate the solubility constant, K s. Ag 2 O (s) 2 Ag + (aq) + O (aq) K s = [Ag + ] 2 [O ] [O ] = mol.l -1 [Ag + ] = 2 x mol.l -1 = mol.l -1 K s = x = 5.04 x 10-9 The solubility of BaSO 4 in pure water is 1.05 x 10-5 mol.l -1, calculate the solubility constant, K s. BaSO 4(s) Ba 2+ (aq) + SO 4 (aq) K s = [Ba 2+ ] [SO 4 ] [SO 4 ] = 1.05 x 10-5 mol.l -1 [Ba 2+ ] = 1.05 x 10-5 mol.l -1 K s = (1.05 x 10-5 mol.l -1 ) 2 = 1.10 x 10-10

10 Do now: (a) Write the equation for PbI 2 dissolving in water (b) Write an expression for the solubility constant for PbI 2 (c) If the solubility of PbI 2 in water is mol.l -1 what is the concentration of Pb 2+ and I - in solution? (d) What is K s for PbI 2?

11 2013 Exam Q 2

12 2013 Exam Q2

13 2013 Exam Q 2 For achieved: K s correct in (a) Working correct OR correct answer but minimal working in (b) (i) Correct ratio of [Ag + ] and [CrO 4 ] in (b) (i) Use of 4s 3 in (b) (ii) For merit: Correct calculation of concentration of Ag 2 CrO 4 in (b) (i) For excellence: Correct calculation of concentration of Ag + ions and CrO 4 ions in (b) (i) AND K s in (b) (ii)

14 Calculating solubility from K s The solubility constant of PbSO 4 is 1.6 x Calculate the solubility of PbSO 4 in water and the concentrations of the ions in solution. PbSO 4(s) Pb 2+ (aq) + SO 4 (aq) K s = [Pb 2+ ] [SO 4 ] If the solubility of PbSO 4 is x then we can substitute into the K s equation x 2 = 1.6 x 10-8 x = x = 2.82 x 10-4 The solubility of PbSO 4 is 2.82 x 10-4 mol.l -1 The concentration of Pb 2+ ions in solution is 2.82 x 10-4 mol.l -1 The concentration of SO 4 ions in solution is 2.82 x 10-4 mol.l -1

15 Calculating solubility from K s The solubility constant of Ag 2 CrO 4 is 9.0 x Calculate the solubility of Ag 2 CrO 4 in water and the concentrations of the ions in solution. Ag 2 CrO 4(s) 2 Ag + (aq) + CrO 4 (aq) K s = [Ag + ] 2 [CrO 4 ] If the solubility of Ag 2 CrO 4 is x then we (2x) can substitute into the K s equation 2 x = K s 4x 3 = 9.0 x x = x = 1.31 x 10-4 The solubility of Ag 2 CrO 4 is 1.31 x 10-4 mol.l -1 The concentration of Ag + ions in solution is 2.62 x 10-4 mol.l -1 The concentration of CrO 4 ions in solution is 1.31 x 10-4 mol.l -1

16 2012 Exam

17 2014 Exam

18 Do now: (a) Write the equation for PbBr 2 dissolving in water (b) Write an expression for the solubility product of PbBr 2 dissolving in water (c) If the solubility of PbBr 2 in water at room temperature is 2.65 x 10-2 mol.l -1 what is the concentration of Pb 2+ and Br - in solution? (d) How would you calculate the solubility of PbBr 2 if K s = 7.62 x 10-6 at 0 C?

19 Do now: a) Write the equation for CuCO 3 dissolving and the K s expression. b) If x = solubility, substitute in to the K s expression to solve for x CuCO 3(s) Cu 2+ (aq) + CO 3 (aq) K s = [Cu 2+ ][CO 3 ] K s = x 2 c) Write the equation for Ag 2 CO 3 dissolving and the K s expression. d) If x = solubility, substitute into the K s expression to solve for x Ag 2 CO 3(s) 2 Ag + (aq) + CO 3 (aq) K s = [Ag + ] 2 [CO 3 ] K s = 4x 3 AB A 2 B e) Write the equation for CuI 2 dissolving and the K s expression. f) If x = solubility, substitute into the K s expression to solve for x CuI 2(s) Cu 2+ (aq) + 2 I - (aq) K s = [Cu 2+ ][I - ] 2 K s = 4x 3 AB 2