1. Solubility Equilibrium Chapter 7

2. The Solubility Equilibrium Remember from SPH3U: Solubility is the amount of solute that dissolves in a given amount of solvent (usually water) at a particular temperature.  Measured in g/100 mL or mol/L (molar concentration)  Very soluble - > 10 g/100 mL Soluble – 1 – 10 g/100 mL Slightly soluble – 0.1 – 1 g/100 mL Insoluble - < 0.1 g/100 mL

3. The Solubility Equilibrium Produced solubility curves  Most solids increased in solubility as the temperature increased. While, most gases decreased in solubility as the temperature increased.  Terminology – Saturated, Unsaturated, Supersaturated

4. The Solubility Process Solubility (gm/100 ml) Temperature ( o C) Saturated solution Unsaturated solution Supersaturated solution

5. The Solubility Equilibrium Dissociation of ionic compounds into ions  Ionic equations, net ionic equations, spectator ions  Electrolytes, non-electrolytes Solubility charts  Predict precipitates in ionic solution combinations (see periodic table given)

6. Solubility Product Constant – K sp

7. The Solubility Product Constant A solution equilibrium can be treated as a chemical equilibrium. The solubility equilibrium must be saturated and have excess solute (salt), therefore they occur with low solubility compounds. The balanced chemical reaction (dissociation equation) can be used to generate an equilibrium expression. Remember, solids have a constant concentration and are not used in the equilibrium law expression. The equilibrium constant is simplified to the product of the ion concentration raised to their proper exponents. The equilibrium constant (K eq ) can be considered a solubility product constant (K sp ) in an solubility equilibrium.

8. The Solubility Product Constant A solution equilibrium can be treated as a chemical equilibrium. PbI 2 (s)  Pb 2+ (aq) + 2 I 1- (aq) (saturated sol’n) PbI 2 is a solid so its concentration does not alter and is therefore removed from the equilibrium expression.

9. The Solubility Product Constant A general application of the solubility equilibria results in the following K sp expression. BC (s)  b B + (aq) + c C - (aq) Where BC (s) is a slightly soluble salt, and B + (aq) and C - (aq) are dissociated aqueous ions. K sp values are reported in reference tables such as those in the appendix. (Pg 802)

10. The Solubility Product Constant usage 1. Calculating solubility constants from solubility data. (i.e.- solubility of 0.0003 g/100 mL @ 25 o C) 2. Predicting whether a precipitate will form in a solubility equilibrium. 1. Trial ion product, Q– the reaction quotient applied to the ion concentrations. 2. Q > K sp (supersaturated) precipitate will form Q = K sp (saturated) precipitate will not form Q < K sp (unsaturated) precipitate will not form

11. Calculating solubility constants from solubility data Determine the solubility of AgCl (s) at SATP given that its K sp is 1.8X 10 -10. Strategy 1. Balanced chemical equation for the system. 2. Equilibrium Law expression. 3. Initial, Change in & Equilibrium Concentration table. 4. Sub-n-solve.

12. Calculating solubility constants from solubility data Determine the solubility of AgCl (s) at SATP given that its K sp is 1.8X 10 -10. Strategy 1. Balanced chemical equation for the system. AgCl (s)  Ag +1 (aq) + Cl -1 (aq)

13. Calculating solubility constants from solubility data Determine the solubility of AgCl (s) at SATP given that its K sp is 1.8X 10 -10. 2. Equilibrium Law expression. K sp = [Ag +1 ] [Cl -1 ] = 1.8 x 10 -10

14. Calculating solubility constants from solubility data Determine the solubility of AgCl (s) at SATP given that its K sp is 1.8X 10 -10. 3. Initial, Change in & Equilibrium Concentration table. AgCl (s)  Ag+1(aq)+ Cl+1(aq) [Initial][constant]00 [change in][constant]+x [Equilibrium][constant]0+x = x

15. Calculating solubility constants from solubility data Determine the solubility of AgCl (s) at SATP given that its K sp is 1.8X 10 -10. 4. Sub-n-solve.

16. Calculating solubility constants from solubility data Determine the solubility of AgCl (s) at SATP given that its K sp is 1.8X 10 -10. 4. Check assumptions

17. The Solubility Product Constant usage

18. Example Calculate the molar solubility of PbSO 4 in 0.100 M Na 2 SO 4 at SATP. K sp = 1.96 x 10 -8 Split this problem into two steps. Initially, NaSO 4(aq) has already dissociated in water to a concentration of 0.100 mol/L Na 2 SO 4 (s)  2Na + (aq) + SO 4 2- (aq) The [SO 4 2- (aq) ] = [Na 2 SO 4 (s) ] = 0.100M PbSO 4 is then added to this solution. PbSO 4  Pb 2+ (aq) + SO 4 2- (aq)

19. Because there are already SO 4 2- ions present in the solution, the PbSO 4 cannot completely dissociate in this solution. We cannot use the assumption that the concentration of ions will be equal to the concentration of PbSO 4. Let molar solubility of PbSO 4 be x. Ksp = [Pb 2+ ] [SO 4 2- ] = 1.96   PbSO 4(s)  Pb 2+ (aq) + SO 4 2- (aq) Initial--00.100 Change--x+ x Equil.--x0.100 + x

20. Ksp = [Pb 2+ ] [SO 4 2- ] (x)(  x   x 2 + 0.100 x -    Using the quadratic equation, x = 2.00 x 10 -7 M

21. Molar solubility of PbSO 4 in 0.100 M Na 2 SO 4 = 2.0    Molar solubility of PbSO 4 in pure water = 1.40    Note the solubility is very much reduced in 0.100 M Na 2 SO 4 because of the common ion effect.

22. Predicting Precipitate

23. Using Q To do so, the concentration of ions in solution can be used to calculate a trial ion product and then compare it to the listed value of K sp  Given a reaction AB (s)  A + (aq) + bB - (aq) Trial Ion Product = [A + ][B - ] b

24. Determining if a precipitate will form Strategy 1. Balanced chemical equation for the system. 2. Identify the concentrations to be considered. 3. Use the equilibrium law expression to generate the “Trial Ion Product (Q sp )” expression. 4. Use given concentration and the expression to solve for Q sp.

25. Determining if a precipitate will form Strategy 5. Compare the Q sp value to the known K sp value to determine if the situation is at equilibrium.  Q sp = K sp  the solution is saturated & at equilibrium so no precipitate will form.  Q sp < K sp  the solution is unsaturated so no precipitate will form.  Q sp > K sp  the solution is “supersaturated” so a precipitate will form.

27. Determining if a precipitate will form If 50 mL of 0.200 mol/L NiCl 2 (aq) and 50 mL of 0.0500 mol/L Na 2 CO 3 (aq) combined at 25 o C will a precipitate form? K sp for NiCO 3 (s) is 8.2 x 10 -4. NiCl 2 (aq) + Na 2 CO 3 (aq) → NiCO 3 (s) + 2 NaCl (aq) (Use the solubility rules to identify which of the products is most likely to form a solid.) NiCO 3 (s) ⇌ Ni 2+ (aq) + 2 Cl - (aq)

28. Determining if a precipitate will form If 50 mL of 0.200 mol/L NiCl 2 (aq) and 50 mL of 0.0500 mol/L Na 2 CO 3 (aq) combined at 25 o C will a precipitate form? K sp for NiCO 3 (s) is 8.2 x 10 -4. NiCl 2 (aq) → Ni 2+ (aq) + 2 Cl - (aq) [Ni 2+ ]= 0.200 mol/L(50 mL)/(100 mL) = 0.100 mol/L Na 2 CO 3 (aq) → 2 Na + (aq) + CO 3 2- (aq) [CO 3 2- ]= 0.0500 mol/L(50 mL)/(100 mL) = 0.0250 mol/L

29. Determining if a precipitate will form If 50 mL of 0.200 mol/L NiCl 2 (aq) and 50 mL of 0.0500 mol/L Na 2 CO 3 (aq) combined at 25 o C will a precipitate form? K sp for NiCO 3 (s) is 8.2 x 10 -4. NiCO 3 (s) ⇌ Ni 2+ (aq) + CO 3 2- (aq)

30. Determining if a precipitate will form If 50 mL of 0.200 mol/L NiCl 2 (aq) and 50 mL of 0.0500 mol/L Na 2 CO 3 (aq) combined at 25 o C will a precipitate form? K sp for NiCO 3 (s) is 8.2 x 10 -4. Q sp > K sp (0.0025 > 8.2 x 10 -4 ) In this case the solution is either supersaturated or a precipitate will be formed.