1. Obj. finish 17.2, 17.3

2. 1.) The pH range is the range of pH values over which a buffer system works effectively. 2.) It is best to choose an acid with a pK a close to the desired pH.

3. …it is safe to assume that all of the strong acid or base is consumed in the reaction.

4. 1. Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution. 2. Use the Henderson–Hasselbalch equation to determine the new pH of the solution.

5. A buffer is made by adding 0.300 mol HC 2 H 3 O 2 and 0.300 mol NaC 2 H 3 O 2 to 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 mol of NaOH is added.

6. Before the reaction, since mol HC 2 H 3 O 2 = mol C 2 H 3 O 2 − pH = pK a = −log (1.8  10 −5 ) = 4.74

7. Note: use moles!! The 0.020 mol NaOH will react with 0.020 mol of the acetic acid: HC 2 H 3 O 2 (aq) + OH − (aq)  C 2 H 3 O 2 − (aq) + H 2 O(l) HC 2 H 3 O 2 OH − C2H3O2−C2H3O2− Before reaction0.300 mol0.020 mol0.300 mol After reaction0.280 mol0.000 mol0.320 mol

8. Now use the Henderson–Hasselbalch equation to calculate the new pH: pH = 4.74 + log (0.320 M) (0.200 M) pH = 4.74 + 0.06pH pH = 4.80

9. Sample Exercise 17.5 Calculating pH Changes in Buffers Determine (a) the pH of the original buffer described before after the addition of 0.020 mol HCl and (b) the pH of the solution that would result from the addition of 0.020 mol HCl to 1.00 L of pure water Practice Exercise

10. A.) Titration: In this technique a known concentration of base (or acid) is slowly added to a solution of acid (or base).

11. B.) A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.

12. 1.) From the start of the titration to near the equivalence point, the pH goes up slowly.

13. 2.) Just before (and after) the equivalence point, the pH increases rapidly.

14. 3.) At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid.

15. 4.) As more base is added, the increase in pH again levels off.

16. Sample Exercise 17.6 Calculating pH for a Strong Acid-Strong Base Titration Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution: (a) 49.0 mL, (b) 51.0 mL.

17. Sample Exercise 17.6 Calculating pH for a Strong Acid-Strong Base Titration Calculate the pH when the following quantities of 0.100 M HNO 3 have been added to 25.0 mL of 0.100 M KOH solution: (a) 24.9 mL, (b) 25.1 mL. Practice Exercise

18. 1.) Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed. 2.) At the equivalence point the pH is >7. 3.) Phenolphthalein is commonly used as an indicator in these titrations.

19. 4.) At each point below the equivalence point, the pH of the solution during titration is determined from the amounts of the acid and its conjugate base present at that particular time.

20. Sample Exercise 17.7 Calculating pH for a Weak Acid-Strong Base Titration Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH (K a = 1.8 ×10 -5 ).

21. Sample Exercise 17.7 Calculating pH for a Weak Acid-Strong Base Titration (a) Calculate the pH in the solution formed by adding 10.0 mL of 0.050 M NaOH to 40.0 mL of 0.0250 M benzoic acid (C 6 H 5 COOH, Ka = 6.3 × 10 -5 ). (b) Calculate the pH in the solution formed by adding 10.0 mL of 0.100 M HCl to 20.0 mL of 0.100 M NH 3. Practice Exercise

22. Sample Exercise 17.8 Calculating the pH at the Equlvalence Point Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100M CH 3 COOH (K a = 1.8 ×10 -5 ) with 0.100 M NaOH.

23. Sample Exercise 17.8 Calculating the pH at the Equivalence Point Calculate the pH at the equivalence point when (a) 40.0 mL of 0.025 M benzoic acid (C 6 H 5 COOH, K a = 6.3 × 10 -5 ) is titrated with 0.050 M NaOH; (b) 40.0 mL of 0.100 M NH 3 is titrated with 0.100 M HCl. Practice Exercise

24. 5.) With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.

25. 1.) The pH at the equivalence point in these titrations is < 7. 2.) Methyl red is the indicator of choice.

26. 1.) When one titrates a polyprotic acid with a base there is an equivalence point for each dissociation.