1. Ch. 17 – Other Aspects of Equilibrium The concept of equilibrium may be used to describe the solubility of salts and the buffering action of a solution. A buffer is a solution that resists changes in pH. Buffers are found in living organisms to pharmaceuticals. Some salts are very soluble while some are not. Both of these concepts revolve around the central ideal of a common-ion.

2. The Common-Ion Effect Imagine the you made the following solution and you allowed it to reach equilibrium – CH 3 COOH (aq)  H + (aq) + CH 3 COO - (aq) Now imagine you added sodium acetate, NaCH 3 COO, to this system at equilibrium. What would happen to the position of the equilibrium?

3. Buffers A buffer is a solution that resists changes in pH when either an acid or a base is added. Buffers consist of either a weak acids with one of its salts, or a weak base with one of its salts.

4. Buffers For example, if a solution is make by dissolving carbonic acid (weak acid) and sodium bicarbonate (salt of the acid) we get the following; H 2 CO 3(aq) + H 2 O (l)  H 3 O + (aq) + HCO 3 -1 (aq)

5. Buffers H 2 CO 3(aq) + H 2 O (l)  H 3 O + (aq) + HCO 3 -1 (aq) If we add a base to this buffered solution, the H 3 O + will scoop it up. H 3 O + (aq) + OH (aq) -  2H 2 O (l)

6. Buffers H 2 CO 3(aq) + H 2 O (l)  H 3 O + (aq) + HCO 3 -1 (aq) If we add an acid to this buffered solution, the HCO 3 -1 will scoop it up forming a weak acid. H + (aq) + HCO 3 -1 (aq)  H 2 CO 3(aq)

7. Buffers Write the chemical reaction for the phosphoric acid – dihydrogen phosphate buffer reaction.

8. Buffers What is the pH of a buffer that is 0.12 M in lactic acid (HC 3 H 5 O 3 ), and 0.10 M in sodium lactate, NaC 3 H 5 O 3 ? K a for lactic acid = 1.4 x 10 -4.

9. Buffers Henderson-Hasselbalch equation – pH = pKa + log [base] [acid]

10. Buffers Calculate the pH of a buffer composed of 0.12 M benzoic acid (HC 7 H 5 O 2 ) and 0.20 M sodium benzoate (NaC 7 H 5 O 2 ). K a of benzoic acid = 6.3 x 10 -5.

11. Buffers Calculate the pH of a buffered solution that was made by mixing 65.0 mL of 0.20 M NaHCO 3 with 75.0 mL of 0.15 M Na 2 CO 3.

12. Buffers How many moles of NH 4 Cl must be added to 2.0 L of a 0.10 M NH 3 to form a buffer whose pH is 9.00? K b of NH 3 = 1.8 x 10 -5.

13. Buffers Buffer Capacity – The amount of acid or base that can be added before the pH begins to change considerably. The optimal pH of any buffer is when the pH = pK a. It is acceptable to make buffers that have a pH = pK a +/- 1. What is the optimal pH buffered by a solution that contains CH 3 COOH and NaCH 3 COO (pK a for CH 3 COOH = 1.8 x 10 -5 )?

14. Addition of Strong Acids and Strong Bases to Buffers A buffer is made by adding 0.300 mol of CH 3 COOH and 0.300 mole of NaCH 3 COO in enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 moles of NaOH is added. K a of acetic acid = 1.8 x 10 -5.

15. Addition of Strong Acids and Strong Bases to Buffers For giggles, what would the the pH if 0.020 moles of NaOH is added to pure water ( a non-buffered solution)?

16. Titrations A titration is the process by which a known concentration of a base is added to an acid (or vice versa) until the equivalence point is reached. The equivalence point is when an equal number of moles of the base and acid has reacted.

17. Titrations An indicator is used to signal when the equivalence point has been reached. Indicators are selected depending on the pH range of the equivalence point.

18. Titrations Calculate the pH when the following quantities of 0.100 M NaOH is added to 50.0 mL of a 0.100 M HCl solution. A.) 49.0 mL B.) 51.0 mL

19. Titrations Thing become more complicated if a weak acid or a base it titrated with a strong acid or base. Why? Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of a 0.100 M solution of CH 3 COOH (K a = 1.8 x 10 -5 ).

20. Titrations Calculate the pH of a the solution formed when 10.0 mL of 0.050 M NaOH is added to 40.0 mL of 0.0250 M benzoic acid (C 6 H 5 COOH), K a = 6.3 x 10 -5.

21. Titrations Titration Curves of Weak Acid / Base with a Strong Acid / Base.

22. Titrations Titration Curves of Polyprotic Acids with a Strong Base.

23. Solubility Equilibria Imagine a saturated solution of CuS; o There is a dynamic equilibrium at the interphase where dissolved Cu +2 and S -2 ions come in contact with undissolved CuS. CuS (s)  Cu +2 (aq) + S -2 (aq) K sp = [Cu +2 ] [S -2 ]

24. Solubility Equilibria Imagine a saturated solution of CuSO 4 ; CuS (s)  Cu +2 (aq) + S -2 (aq) K sp = [Cu +2 ] [S -2 ] = 6.0 x 10 -37 Since the K sp of CuS is very small, we can conclude that CuS is not very soluble at all.

25. Solubility Equilibria Write the Solubility Product expression for silver sulfate. Based in its K sp value, is silver sulfate soluble in water? K sp Ag 2 SO 4 = 1.5 x 10 -5.

26. Solubility Equilibria A saturated solution of silver chromate is made and the concentration of silver ion is determined to be [Ag +1 ] = 1.3 x 10 -4. Calculate the value of K sp for silver chromate.