1. Aqueous Solutions and Solubility Equilibria Chapter 9

2. 9.1 The Acid-Base Properties of Salt Solutions Salt: any ionic compound that is formed in a neutralization reaction from the anion of an acid and the cation of a base. E.g. HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l)

3. The acid/base property of a salt results from reactions between water and the dissociated ions of the salt. – Ion that react with water produce a solution with excess H 3 0 + or OH

4. Salts that form Neutral Solutions Salts of strong bases and strong acids dissolve in water and form neutral solutions. – The conjugate bases of strong acids are very ___________. – The conjugate acids of strong bases are very ___________. E.g. Neither ion will react with water: too weak!

5. Salts the Dissolve and form Acidic and Basic Solutions Salts of weak bases and strong acids dissolve in water and form acidic solutions. Salts of strong bases and weak acids dissolve in water and form basic solutions. (LOOKING AT THE CONJUGATE ACIDS/BASES!)

6. Salts of Weak Acids and Bases Ions in a salt from weak acids and bases BOTH react with water. Acidity/basicity depends on relative strength of the ions. Determine which ion is stronger by comparing the Ka and Kb associated with the cation and anion. If Ka>Kb, acidic. If Kb>Ka, basic.

10. CLASSWORK/HOMEWORK Read through SP on pg. 423. Make notes. Do PPs 1-3.

11. Calculating pH at Equivalence Equivalence point: just enough acid and base have been mixed for complete reaction to occur, with no excess of either reactant. Acid-base indicator: weak, monoprotic acid. It is in equilibrium between undissociated acid (one colour) and conjugate base (different colour) End-point: indicator changes colour.

15. PPs 8 & 9

16. SR (pg. 429) #2, 6, 7.

17. 9.2 – Solubility Equilibria

18. Solubility as an Equilibrium Process Change in enthalpy, entropy, and temperature determine whether or not a change is favoured. Same is important to determine how much salt will dissolve.

20. Change is favoured when G is negative. When salt dissolves, entropy is increased.

21. Ex// Barium sulfate crystals in water: As ions enter solution, rate of reverse change, recrystallisation, increases. Eventually, rate of recrystallisation becomes equal to the rate of dissolving.

22. Solubility Equilibria Equilibrium exists between the solid ionic compound and its dissociated ions in solution.

23. Solubility Product Constant

24. Solubility product constant: Ksp!!!

29. PPs 13, 15

30. Using the Solubility Product Constant Use the value of Ksp for a compound to determine the concentration of its ions in a saturated solution. Similar to finding equilibrium amounts using Kc for homogeneous equilibria.

32. WORK ON PPs 17, 18, 19,

33. The Common Ion Effect When ionic compound added to a solution that already contains one of its ions. Adding a common ion to a solution increases the concentration of that ion in solution – EQUILIBRIUM SHIFTS AAWY FROM THE ION. – Can form precipitates.

35. The Effect of a Common Ion on Solubility

36. Work on PPs 21-23

37. 9.3 Predicting the Formation of a Precipitate The Ion Product: Qsp: expression that is identical to the solubility product constant, but values are not necessarily at equilibrium. (Where have we seen this before?)

38. E.g. -Adding magnesium sulfate to water. -Initially, all magnesium sulfate dissolves. -Saturation: no more salt will dissolve. -More solid will form.

39. -Calculate Qsp by substituting the conc. Of each ion into the expression.

40. Using the Ion Product Expression If Qsp > Ksp, compound will form an ionic compound. How do we know which ionic compounds are soluble?

42. E.g.

44. PPs 29, 30, 31

46. PPs. 33, 34, 35

47. Test Review

48. Buffers and the Common Ion Effect A buffer consists of a ________________________ and a ________________________ or vise versa. Need to consider initial concentration of reactants since homogeneous.