1. C H E M I S T R Y Chapter 15 Applications of Aqueous Equilibria

2. Neutralization Reaction General Formula Acid + Base  Water + Salt

3. Neutralization Reactions Assume complete dissociation: Strong Acid-Strong Base After neutralization: pH = 7 H 2 O(l) +NaOH(aq)HCl(aq)+NaCl(aq) 2H 2 O(l)H 3 O + (aq) + OH – (aq) (net ionic equation)

4. Strong acid-Strong base neutralization When the number moles of acid and base are mixed together [H 3 O + ] = [ - OH] = 1.0 x 10 -7 M Reaction proceeds far to the right

5. Neutralization Reactions H 2 O(l) + CH 3 CO 2 – (aq)CH 3 CO 2 H(aq) + OH – (aq) Assume complete dissociation: Weak Acid - Strong Base After neutralization: pH > 7 (net ionic equation) H 2 O(l) + CH 3 CO 2 H(aq) + NaOH(aq) NaCH 3 CO 2 (aq)

6. Weak acid-strong base neutralization Neutralization of any weak acid by a strong base goes 100% to completion - OH has a great infinity for protons

7. Neutralization Reactions Weak Acid - Strong Base H 2 O(l) +CH 3 CO 2 H(aq) +NaOH(aq)NaCH 3 CO 2 (aq)

8. Neutralization Reactions Assume complete dissociation: Strong Acid - Weak Base After neutralization: pH < 7 NH 4 Cl(aq)HCl(aq)+ NH 3 (aq) H 2 O(l) +H 3 O + (aq) + NH 3 (aq)NH 4 + (aq) (net ionic equation)

9. Strong acid-weak base neutralization Neutralization of any weak base by a strong acid goes 100% to completion H 3 O + has a great infinity for protons

10. Neutralization Reactions Strong Acid - Weak Base NH 4 Cl(aq)HCl(aq)+ NH 3 (aq)

11. Neutralization Reactions NH 4 + (aq) + CH 3 CO 2 – (aq)CH 3 CO 2 H(aq) + NH 3 (aq) Weak Acid-Weak Base After neutralization: pH = ? NH 4 CH 3 CO 2 (aq)CH 3 CO 2 H(aq) + NH 3 (aq) (net ionic equation) Assume complete dissociation:

12. Weak acid-weak base neutralization Less tendency to proceed to completion than neutralization involving strong acids and strong bases

13. The Common-Ion Effect H 3 O 1+ (aq) + CH 3 CO 2 1- (aq)CH 3 CO 2 H(aq) + H 2 O(l) Common-Ion Effect: The shift in the position of an equilibrium on addition of a substance that provides an ion in common with one of the ions already involved in the equilibrium. Example of Le Chatelier’s principle The addition of acetate ion,CH 3 CO 2 1-, to a solution of acetic acid suppresses the dissociation of the acid. The equilibrium shifts to the left.

14. The Common-Ion Effect

15. H 3 O 1+ (aq) + CH 3 CO 2 1- (aq)CH 3 CO 2 H(aq) + H 2 O(l) E.g Adding HCl and NaOH to a solution of acetic acid would shift the equilibrium to which direction? Le Châtelier’s Principle

16. Example The pH of a 0.10M of acetic is 2.87 at 25 o C. Determine the pH at the same temperature of a solution by adding 0.050 mole of sodium acetate to 1.0 L of 0.10 M acetic acid. (Assume that the addition of sodium acetate does not change the volume of the solution) K a = 1.8 x 10 -5

17. Example Dtermine the pH at 25oC of a solution prepared by dissolving 0.35 mol of ammonium chloride in 1.0 L of 0.25 M aqueous ammonia solution. No volume changed. K b = 1.8 x 10 -5

18. Buffer Solutions © 2012 Pearson Education, Inc. Chapter 15/18 Buffer Solution: A solution which contains a weak acid and its conjugate base and resists drastic changes in pH CH 3 CO 2 H + CH 3 CO 2 – HF + F – NH 4 + + NH 3 H 2 PO 4 – + HPO 4 2– Weak acid + Conjugate base For example:

19. Buffer Solutions Conjugate base (NaCH 3 CO 2 ) Weak acid H 3 O + (aq) + CH 3 CO 2 – (aq)CH 3 CO 2 H(aq) + H 2 O(l) H 2 O(l) + CH 3 CO 2 H(aq)CH 3 CO 2 – (aq) + H 3 O + (aq) 100% Addition of H 3 O 1+ to a buffer: Addition of OH 1– to a buffer: H 2 O(l) + CH 3 CO 2 – (aq)CH 3 CO 2 H(aq) + OH – (aq) 100%

20. Buffer Solutions Add a small amount of base ( - OH) to a buffer solution Acid component of solution neutralizes the added base Add a small amount of acid (H 3 O + ) to a buffer solution Base component of solution neutralizes the added acid The addition of – OH or H 3 O + to a buffer solution will change the pH of the solution, but not as drastically as the addition of – OH or H 3 O + to a non-buffered solution

21. Buffer Solutions

22. Buffer capacity A measure of the amount of acid or base that a buffer solution can absorb without a significant change in pH Depends on how much weak acid and conjugated base is present For equal volume of solution, the more concentration the solution, the greater the buffer capacity For solution with the same concentration, increasing the volume increases the buffer capacity

23. Example pH of human blood (pH = 7.4) controlled by conjugated acid-base pairs (H 2 CO 3 /HCO 3 - ). Write a neutralization for each condition With addition of H 3 O + With addition of - OH

24. Example Calculate the pH of the buffer that results from mixing 60.0mL of 0.250M HCHO 2 and 15.0 mL of 0.500M NaCHO 2 Ka = 1.8 x 10 -4

25. Example Calculate the pH of 0.100L of a buffer solution that is 0.25M in HF and 0.50 M in NaF, K a = 6.3 x 10 -4 What is the change in pH on addition of 0.002 mol HCl What is the change in pH on addition of 0.010 moles KOH Calculate the pH after addition of 0.080 moles HBr

26. Example calculate the pH of a 50.0 ml buffer solution that is 0.50 M in NH 3 and 0.20 M NH 4 Cl. For ammonia, pK b = 4.75 Calculate the pH after addition of 150.0 mg HCl

27. The Henderson-Hasselbalch Equation Conjugate baseWeak acid H 3 O 1+ (aq) + Base(aq)Acid(aq) + H 2 O(l) [Acid] [Base] [H 3 O 1+ ] = K a K a = [H 3 O 1+ ][Base] [Acid] H 3 O 1+ (aq) + CH 3 CO 2 1- (aq)CH 3 CO 2 H(aq) + H 2 O(l)

28. The Henderson-Hasselbalch Equation [Base] [Acid] pH = pK a + log [Acid] [Base] [H 3 O 1+ ] = K a -log([H 3 O 1+ ]) = -log(K a ) - log [Acid] [Base]

29. Example Calculate the pH of a buffer solution that is 0.50 M in benzoic acid (HC 7 H 5 O 2 ) and 0.150 M in sodium benzoate (NaC 7 H 5 O 2 ). Ka = 6.5 x 10 -5 How would you prepare a NaHCO 3 -Na 2 CO 3 buffer solution that has pH = 10.40 Ka 2 = 5.6 x 10 -11