1. SOLUBILITY – The maximum amount of solute that will dissolve in a specific amount of solvent EQUILIBRIA WITH SALTS SATURATED – A solution where the solid and dissolved forms of the solute are in equilibrium, and contains as much dissolved solute as possible Dissolving RateCrystallizing Rate MOLAR SOLUBILITY – The solubility of a solute, measured in mol/L 2D-1 (of 11)

2. SOLUBILITY PRODUCT CONSTANT (K sp ) – The equilibrium constant for the reaction of a solid salt dissolving in water AgCl (s) ⇆ Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ] silver chloride Ca 3 (PO 4 ) 2 (s) ⇆ 3Ca 2+ (aq) + 2PO 4 3- (aq) K sp = [Ca 2+ ] 3 [PO 4 3- ] 2 calcium phosphate 2D-2 (of 11)

3. The solubility of calcium fluoride is 0.017 g/L. Calculate its K sp. x2x CaF 2 (s) ⇆ Ca 2+ (aq) + 2F - (aq) Initial M’s Change in M’s Equilibrium M’s 0 0 + x+ 2x K sp = [Ca 2+ ][F - ] 2 0.017 g CaF 2 ________________ L = 2.18 x 10 -4 M CaF 2 = (x)(2x) 2 = 4x 3 x = molar solubility of CaF 2 x mol CaF 2 ________________ 78.08 g CaF 2 K sp = 4(2.18 x 10 -4 ) 3 = 4.1 x 10 -11 M 3 2D-3 (of 11) CALCULATING THE SOLUBILITY PRODUCT CONSTANT (K sp ) (- x)

4. The solubility of iron (III) hydroxide is 0.013 g/L. Calculate its K sp. Molar mass of iron (III) hydroxide is 106.88 g/mol. x3x Fe(OH) 3 (s) ⇆ Fe 3+ (aq) + 3OH - (aq) Initial M’s Change in M’s Equilibrium M’s 0 0 + x+ 3x K sp = [Fe 3+ ][OH - ] 3 0.013 g Fe(OH) 3 ____________________ L = 1.22 x 10 -4 M Fe(OH) 3 = (x)(3x) 3 = 27x 4 x = molar solubility of Fe(OH) 3 x mol Fe(OH) 3 ______________________ 106.88 g Fe(OH) 3 K sp = 27(1.22 x 10 -4 ) 4 = 5.9 x 10 -15 M 4 2D-4 (of 11)

5. MF 2 (s) ⇆ M 2+ (aq) + 2F - (aq) Which is more soluble, BaF 2 (K sp = 1.4 x 10 -7 ) or MgF 2 (K sp = 6.4 x 10 -9 )? The bigger the K sp, the more soluble the salt … provided the number of ions per formula unit are identical 2E-6 (of 11) PREDICTING RELATIVE SOLUBILITIES x2x Initial M’s Change in M’s Equilibrium M’s 0 0 + x+ 2x K sp = [M 2+ ][F - ] 2 = (x)(2x) 2 = molar solubility of MF 2 = 4x 3 K sp = x ____ 4 3  BaF 2

6. Ag 2 S (s) ⇆ 2Ag + (aq) + S 2- (aq) Which is more soluble, Ag 2 S (K sp = 1.6 x 10 -49 ) or CuS (K sp = 8.5 x 10 -45 )? The number of ions per formula unit are not identical 2E-7 (of 11) 2xx 0 0 + 2x+ x K sp = [Ag + ] 2 [S 2- ] = molar solubility 1.6 x 10 -49 = 4x 3 3.4 x 10 -17 = x CuS (s) ⇆ Cu 2+ (aq) + S 2- (aq) xx 0 0 + x K sp = [Cu 2+ ][S 2- ] = molar solubility 8.5 x 10 -45 = x 2 9.2 x 10 -23 = x Bigger molar solubility, so Ag 2 S is more soluble E I C

7. Find the molar solubility of calcium sulfate if its K sp = 2.4 x 10 -5. xx CaSO 4 (s) ⇆ Ca 2+ (aq) + SO 4 2- (aq) Initial M’s Change in M’s Equilibrium M’s 0 0 + x K sp = [Ca 2+ ][SO 4 2- ] 4.90 x 10 -3 mol CaSO 4 ____________________________ L = 0.67 g/L 2.4 x 10 -5 = x 2 x = 4.90 x 10 -3 M x 136.15 g CaSO 4 ____________________ mol CaSO 4 = molar solubility of CaSO 4 Find the solubility of calcium sulfate in g/L. 2D-5 (of 11) CALCULATING THE SOLUBILITY OF A COMPOUND (- x)

8. Find the molar solubility of lead (II) iodide if its K sp = 8.3 x 10 -9. x2x PbI 2 (s) ⇆ Pb 2+ (aq) + 2I - (aq) Initial M’s Change in M’s Equilibrium M’s 0 0 + x+ 2x K sp = [Pb 2+ ][I - ] 2 8.3 x 10 -9 = (x)(2x) 2 x = 1.28 x 10 -3 M = 4x 3 2D-8 (of 11) = molar solubility of PbI 2 (- x)

9. Find the molar solubility of lead (II) iodide in a 0.10 M sodium iodide solution. x0.10 + 2x PbI 2 (s) ⇆ Pb 2+ (aq) + 2I - (aq) Initial M’s Change in M’s Equilibrium M’s 0 0.10 + x+ 2x K sp = [Pb 2+ ][I - ] 2 8.3 x 10 -9 = (x)(0.10 + 2x) 2 x = 8.3 x 10 -7 M= (x)(0.10) 2 The NaI contains an ion involved in the PbI 2 equilibrium reaction 2D-9 (of 11) = molar solubility The I - shifts the equilibrium to the left, making PbI 2 less soluble in this solution than in pure water (8.3 x 10 -7 M < 1.3 x 10 -3 M)

10. 2D-10 (of 11) If a solution is 0.10 M in sodium hydroxide and 0.15 M in barium nitrate, determine if a precipitate of barium hydroxide will form. Ba(OH) 2 (s) ⇆ Ba 2+ (aq) + 2OH - (aq) Initial M’s Change in M’s Equilibrium M’s 0.15 0.10 ?? K sp = [Ba 2+ ][OH - ] 2 Q < K sp = 1.5 x 10 -3  forward reaction spontaneous NaOH completely dissociates  0.10 M Na + and 0.10 M OH - Ba(NO 3 ) 2 completely dissociates  0.15 M Ba 2+ and 0.30 M NO 3 - Look up the K sp : 5.0 x 10 -3 Q = [Ba 2+ ][OH - ] 2 = (0.15 M)(0.10 M) 2  no precipitate forms PREDICTING PRECIPITATION

11. If a solution is 0.010 M in nickel (II) nitrate and 0.025 M in sodium carbonate, determine if a precipitate of nickel (II) carbonate (K sp = 1.4 x 10 -7 ) will form. NiCO 3 (s) ⇆ Ni 2+ (aq) + CO 3 2- (aq) Initial M’s Change in M’s Equilibrium M’s 0.010 0.025 ?? K sp = [Ni 2+ ][CO 3 2- ] Q > K sp = 2.5 x 10 -4  reverse reaction spontaneous Ni(NO 3 ) 2 completely dissociates  0.010 M Ni 2+ and 0.020 M NO 3 - Na 2 CO 3 completely dissociates  0.050 M Na + and 0.025 M CO 3 2- Q = [Ni 2+ ][CO 3 2- ] = (0.010 M)(0.025 M)  a precipitate forms 2D-11 (of 11)

13. 0.100x BaF 2 (s) ⇆ Ba 2+ (aq) + 2F - (aq) Initial M’s Change in M’s Equilibrium M’s = [F - ] eq K sp = [Ba 2+ ] [F - ] 2 Calculate the concentration of fluoride ions needed to produce a precipitate in a 0.100 M barium nitrate solution. The K sp of barium fluoride is 1.4 x 10 -7. Ba(NO 3 ) 2 completely dissociates  0.100 M Ba 2+ and 0.200 M NO 3 - 1.4 x 10 -5 M= x 2E-1 (of 11) PREDICTING ONE EQUILIBRIUM ION CONCENTRATION 1.4 x 10 -7 = (0.100) x 2 Equilibrium is when the precipitate just starts to form  want the [F - ] eq

14. x0.200 Sr(OH) 2 (s) ⇆ Sr 2+ (aq) + 2OH - (aq) Initial M’s Change in M’s Equilibrium M’s 0.0080 M x 100 = 8.0% ___________ 0.100 M K sp = [Sr 2+ ][OH - ] 2 Crystals of NaOH are added to a solution that is 0.100 M Sr(NO 3 ) 2 until the OH - concentration is 0.200 M. Calculate the percentage of Sr 2+ ions left in the solution at this point. The K sp of Sr(OH) 2 is 3.2 x 10 -4. 2E-2 (of 11) 3.2 x 10 -4 = x(0.200) 2 0.0080 M= x The answer is not 0% because the reaction does not go to completion, it reaches an equilibrium  solve for [Sr 2+ ] eq

15. AgOH (s) ⇆ Ag + (aq) + OH - (aq) K sp for AgOH is 2.0 x 10 -8 and K sp for Mn(OH) 2 is 2.0 x 10 -13. 2E-3 (of 11) PREDICTIONS INVOLVING TWO PRECIPITATES 0.100x K sp = [Ag + ][OH - ] = [OH - ] eq to ppt 2.0 x 10 -8 = (0.100)x 2.0 x 10 -7 = x Mn(OH) 2 (s) ⇆ Mn 2+ (aq) + 2OH - (aq) 0.200x K sp = [Mn 2+ ][OH - ] 2 = [OH - ] eq to ppt 2.0 x 10 -13 = (0.200)(x) 2 1.0 x 10 -6 = x AgOH will precipitate first because it requires a smaller [OH - ] eq E I C Solve for [OH-] eq needed to precipitate each (a)If hydroxide ions are added to a solution 0.100 M Ag + and 0.200 M Mn 2+, what will precipitate first?

16. (b)What will be the concentration of Ag + when the Mn(OH) 2 starts to precipitate? 2E-4 (of 11) [OH - ] eq needed to precipitate Mn(OH) 2 was 1.0 x 10 -6 M x1.0 x 10 -6 AgOH (s) ⇆ Ag + (aq) + OH - (aq) Initial M’s Change in M’s Equilibrium M’s K sp = [Ag + ][OH - ] 2.0 x 10 -8 = x(1.0 x 10 -6 ) 0.020 M= x K sp for AgOH is 2.0 x 10 -8 and K sp for Mn(OH) 2 is 2.0 x 10 -13. PREDICTIONS INVOLVING TWO PRECIPITATES = [Ag + ]

17. Whenever a solution has its “pH adjusted”, this means an external source of acid or base has been added If a solution is saturated with H 2 S (0.10 M) and the pH adjusted to 1.00 with HCl, determine the molar solubility of FeS. K sp for FeS = 3.7 x 10 -19 K a1 for H 2 S= 1.0 x 10 -7 K a2 for H 2 S= 1.0 x 10 -15 Use the K a1,2 to determine [S 2- ] K a1,2 = K a1 K a2 = (1.0 x 10 -7 )(1.0 x 10 -15 )= 1.0 x 10 -22 2E-5 (of 11) EFFECT OF pH ON PRECIPITATION In this case, adjusting the pH controls the [S 2- ]

18. 0.10x H 2 S (aq) + 2H 2 O (l) ⇆ 2H 3 O + (aq) + S 2- (aq) Initial M’s Change in M’s Equilibrium M’s 0.10 = [S 2- ] eq K a1,2 = [H 3 O + ] 2 [S 2- ] ______________ [H 2 S] If a solution is saturated with H 2 S (0.10 M) and the pH adjusted to 1.00 with HCl, determine the molar solubility of FeS. [H 3 O + ] eq = antilog (-1.00)= 0.10 M 1.0 x 10 -21 M= x 2E-6 (of 11) EFFECT OF pH ON PRECIPITATION 1.0 x 10 -22 = (0.10) 2 x ____________ (0.10)

19. If a solution is saturated with H 2 S (0.10 M) and the pH adjusted to 1.00 with HCl, determine the molar solubility of FeS. FeS (s) ⇆ Fe 2+ (aq) + S 2- (aq) Initial M’s Change in M’s Equilibrium M’s x1.0 x 10 -21 K sp = [Fe 2+ ][S 2- ] 370 M= x  FeS is very soluble at a pH = 1.00 3.7 x 10 -19 = (x)(1.0 x 10 -21 M) 2E-7 (of 11) EFFECT OF pH ON PRECIPITATION = molar solubility of FeS

20. If a solution is saturated with H 2 S (0.10 M) and the pH adjusted to 10.00 with HCl, determine the molar solubility of FeS. 2E-8 (of 11) 1.0 x 10 -10 x H 2 S (aq) + 2H 2 O (l) ⇆ 2H 3 O + (aq) + S 2- (aq) Initial M’s Change in M’s Equilibrium M’s 0.10 = [S 2- ] eq K a1,2 = [H 3 O + ] 2 [S 2- ] ______________ [H 2 S] [H 3 O + ] eq = antilog (-10.00)= 1.0 x 10 -10 M 1.0 x 10 -3 M = x 1.0 x 10 -22 = (1.0 x 10 -10 ) 2 x _________________ (0.10)

21. If a solution is saturated with H 2 S (0.10 M) and the pH adjusted to 10.00 with HCl, determine the molar solubility of FeS. FeS (s) ⇆ Fe 2+ (aq) + S 2- (aq) Initial M’s Change in M’s Equilibrium M’s x 1.0 x 10 -3 K sp = [Fe 2+ ][S 2- ] 2E-9 (of 11) 3.7 x 10 -16 M= x  FeS is very insoluble at a pH = 10.00 3.7 x 10 -19 = (x)(1.0 x 10 -3 M) = molar solubility of FeS

22. Calculate the pH that FeS precipitates in a solution 0.10 M Fe 2+ and saturated with H 2 S (0.10 M). Use the K sp to determine the [S 2- ] eq needed to precipitate FeS 0.10x FeS (s) ⇆ Fe 2+ (aq) + S 2- (aq) Initial M’s Change in M’s Equilibrium M’s 3.7 x 10 -19 = (0.10)x ∴ [S 2- ] eq = 3.7 x 10 -18 M to start to precipitate FeS K sp = [Fe 2+ ][S 2- ] 3.7 x 10 -18 M = x 2E-10 (of 11) Use the K a1,2 to determine the pH

23. x 3.7 x 10 -18 H 2 S (aq) + 2H 2 O (l) ⇆ 2H 3 O + (aq) + S 2- (aq) Initial M’s Change in M’s Equilibrium M’s 0.10 1.0 x 10 -22 = (x) 2 (3.7 x 10 -18 ) __________________ (0.10) pH = -log (1.64 x 10 -3 ) K a1,2 = [H 3 O + ] 2 [S 2- ] ______________ [H 2 S] 1.64 x 10 -3 = x Calculate the pH that FeS precipitates in a solution 0.10 M Fe 2+ and saturated with H 2 S (0.10 M). = 2.78 2E-11 (of 11)