1. Additional Aspects of Aqueous Equilibria

2. Roundtable problems P.757: 3, 6, 12, 14, 18, 24, 30, 38, 44, 50, 54, 56, 58, 64, 68, 70, 72, 103.

3. The Common Ion Effect HC 2 H 3 O 2 in the presence of NaC 2 H 3 O 2 HC 2 H 3 O 2  H + + C 2 H 3 O 2 - NaC 2 H 3 O 2 is a strong electrolyte Addition of C 2 H 3 O 2 - shifts equilibrium to left decreasing [H + ] and pH increases.

4. Common Ion Effect Shift in equilibrium that occurs when an ion that is a component of the equilibrium system is added. Calculate the pH of a 0.100 M acetic acid solution when 8.20 g of sodium acetate is added.

5. Common Ions Generated by Acid/Base Reactions Reaction that mixes weak acid with strong base: HC 2 H 3 O 2 + NaOH  C 2 H 3 O 2 - + H 2 O Acid/Base combinations may generate common ions Reactions with strong acids and strong bases; strong acids and weak bases; weak acids and strong bases proceed to completion.

6. pH Of an Acid/Base Mixture If solution contains only a conjugate acid/base pair, consider equilibrium only. If solution of strong acid and weak base or weak acid and strong base, consider stoichiometry first and then equilibrium. Calculate the pH if 0.60 L of 0.10 M NH 4 Cl is mixed with 0.40 L of 0.10 M NaOH.

7. Determine the pH of a solution consisting of 500 mL of 0.015 M NaOH and 500 mL of 0.030M HC 7 H 5 O 2 (benzoic acid).

8. Buffered Solutions Resist changes in pH because they contain an acidic species to offset OH - additions and a basic species to offset H + additions. Generally weak acid / weak base conjugate pair since a strong acid / strong base pair would neutralize each other.

10. Buffer: HX + MX representing a weak acid and its salt. HX  H + + X - K a = [H + ][X - ] [HX] H + = K a [HX] [X - ]

11. [H + ] depends on K a and ratio of [Acid]/[Base] Addition of base, OH - reacts with acid HX: HX + OH -  H 2 O + X - [HX]  [X - ]  Addition of acid, H + reacts with base X - H + + X -  HX [HX]  [X - ]  Most effective when [HX] and [X - ] are about the same.

13. Buffering Capacity and pH Buffering capacity is the amount of acid or base the buffer can neutralize before pH begins to change appreciably. Capacity depends on K a of acid and relative concentrations of acid and base in the buffer.

14.  Buffering capacity depends on amount of acid and base from which the buffer is made.  The greater the amounts of the conjugate acid/ base pair, the more resistant the ratio and thus the pH is to changes.

15. Henderson-Hasselbalch Equation: pH = pK a – log [HX] [X] pH = pK a + log [base] [acid]

16. What is the pH of a buffer consisting of 0.12 M lactic acid (HC 3 H 5 O 3 ), K a = 1.4 x 10 -4, And 0.10 M sodium lactate? How many moles of NH 4 Cl must be added to 1.0 L of a 0.10 M NH 3 solution to form a buffer whose pH = 9.00?

17. Addition of Acids or Bases to Buffers Addition of strong acids or strong bases proceed to completion. 0.100 mol HC 2 H 3 O 2 + 0.100 mol NaC 2 H 3 O 2 produce a buffer with a pH = 4.74 Calculate the pH after 0.020 mol NaOH is added. Calculate the pH after 0.020 mol HCl is added.

18. Titration Curves Graph pH versus volume of acid or base. Equivalence point = point where equivalent quantities of acid and base are present. Point of inflection = middle point of the vertical part of the curve.

20. Titration Curves Examine the curves at the left. What is common to both curves? What is different about the curves?

23. Titration Curves When a strong base is added to a strong acid there is a gradual rise in pH. Equivalence point = pH of salt. Before equivalence point, pH depends on how much acid is not neutralized. Calculate the pH when 49.00 mL of 0.100 M NaOH is added to 50.00 mL of 0.100 M HCl.

24. Titration of a weak acid or weak base A weak acid solution has a higher initial pH. pH rises more rapidly early on in titration; more slowly at equivalence point pH at equivalence point is not 7. Calculate the pH when 15.0 mL of 0.100 M HCl and 25.0 mL of 0.100 M NH 3 are mixed.

25. Solubility Equilibria Equilibrium occurs between a solid and a saturated solution. Heterogeneous equilibrium. Solubility product constant, K sp, equals the product of ions involved in the equilibrium.

26. BaSO 4  Ba 2+ (aq) + SO 4 2- (aq) K sp = [Ba 2+ ] [SO 4 2- ] Calculate K sp if [Ba 2+ ] = 1.04 x 10 -5 M

27. Common Ion Effect Common ions shift equilibrium to left reducing solubility. Calculate the molar solubility of CaF 2 upon the addition of 0.010 M Ca(NO 3 ) 2. K sp = 3.9 x 10 -11

28. Use of reaction quotient, Q, to determine direction of reaction: Q < K sp solute dissolves until Q = K sp Q = K sp at equilibrium Q > K sp precipitation occurs until Q = K sp

29. Will a precipitate form when 0.100 L of 3.0 x 10 -3 Pb(NO 3 ) 2 is added to 0.400 L of 5.0 x 10 -3 M Na 2 SO 4 ?

30. Solubility and pH Solubility of a substance whose anion is basic is affected by pH. Solubility of slightly soluble salts with basic anions increases as [H + ] increases, pH . Anions of strong acids are unaffected.

31. CaF 2  Ca 2+ (aq) + 2F - (aq) Addition of H + causes F - to react to form HF. According to Le Chatelier’s Principle, removal of F - by addition of H + shifts equilibrium to right increasing solubility of salt. Which salt is more soluble in acidic solution? Ni(OH) 2 CaCO 3 BaSO 4 AgCl

32. Amphoterism Metal hydroxides that dissolve in strongly acidic and in strongly basic media. Interpreted in terms of the behavior of water molecules that surround the metal. Al 3+ acts as a Lewis acid  Al(H 2 O) 6 3+ As NaOH is added a water molecule is lost and an OH - is added: Al(H 2 O) 6 3+ + OH -  Al(H 2 O) 5 OH 2+ + H 2 O

33. Qualitative Analysis Used to determine the presence or absence of a particular metal ion. Separate ions into broad groups; perform a separation with each group and then perform individual ion tests. Order is important.

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