1. Acid-Base Equilibria Chapter 16

2. Revision Acids and bases change the colours of certain indicators. Acids and bases neutralize each other. Acids and bases react to form salts. Acids react with certain metals to release hydrogen.

3. Bronsted-Lowry Acids and Bases These two chemists pointed out that acids and bases can be seen as proton transfer reactions. According to the Bronsted-Lowry concept: An acid is the species donating a proton in a proton-transfer reaction A base is the species accepting the proton in a proton-transfer reaction.

4. If we consider the reaction: HCl (g) + NH 3 (g) NH 4 Cl (s) We see that we can view it as a proton transfer. In any reversible acid-base reaction, both forward and reverse reactions involve proton transfers.

5. Consider the reaction of NH 3 with H 2 O: NH 3 (aq) + H 2 O (aq) NH 4 + (aq) + OH - (aq) Note: NH 3 and NH 4 + are a conjugate acid-base pair. H 2 O and OH - are also a conjugate acid-base pair. A conjugate acid-base pair consists of two species in an acid-base reaction, one acid one base, that differ by the loss or gain of a proton.

6. Example Identify the acid and base species in the following equation: CO 3 2- (aq) + H 2 O(l) HCO 3 - (aq) + OH - Acid species : Base species :

7. An amphiprotic species is a species that can act as either an acid or a base (it can lose or gain a proton), depending on the other reactant. H 2 O + CH 3 O -  OH - + CH 3 OH H 2 O + HBr  H 3 O + + Br - acid base acid base Consider water:

8. Relative strengths of acids and bases The strongest acids have the weakest conjugate bases and the strongest bases have the weakest conjugate acids.

9. Water undergoes self-ionisation  autoprotolysis, since H 2 O acts as an acid and a base. H 2 O + H 2 O  H 3 O + + OH - The extent of autoprotolysis is very small. AUTOPROTOLYSIS K c = [H 3 O + ] [OH - ] [H 2 O] 2 The equilibrium constant expression for this reaction is: The concentration of water is essentially constant. K c =[H 3 O + ]. [OH - ][H 2 O] 2 Therefore: constant= K w

10. We call the equilibrium value of the ion product [H 3 O + ][OH - ] the ion-product constant of water. K w = [H 3 O + ] [OH - ] = 1.0 x 10 -14 at 25°C Using K w you can calculate concentrations of H 3 O + and OH - in pure water. If you add an acid or a base to water the concentrations of H 3 O + and OH - will no longer be equal. But K w will still hold. [H 3 O + ] [OH - ] = 1.0 x 10 -14 But [H 3 O + ] = [OH - ] in pure water  [H 3 O + ] = [OH - ] = 1.0 x 10 -7 M

11. The pH Scale Because concentration values may be very small, it is often more convenient to express acidity in terms of pH. Definition of pH pH is defined as the negative logarithm of the molar hydronium-ion concentration. pH = -log [H 3 O + ] often written as : pH = -log [H + ]

13. For a solution with a hydronium-ion concentration of 1.0 x 10 -3 M, the pH is: Note that the number of places after the decimal point in the pH equals the number of significant figures reported in the hydronium-ion concentration.

14. Calculating the pH from the hydronium-ion concentrationExample Calculate the pH of typical adult blood, which has a hydronium-ion concentration of 4.0 x 10 -8 M.

15. Calculating the hydronium-ion concentration from the pH.Example The pH of natural rain is 5.60. Calculate its hydronium- ion concentration.

16. Other “p” scales – pOH In the same manner that we defined pH we can also define pOH: pOH = -log [OH - ] also remember that: K w = [H 3 O + ]. [OH - ] = 1.0 x 10 -14 therefore we can show that: pH + pOH = 14.00 -log ([H + ] [OH - ]) = -log (1x10 -14 ) -log [H + ] + -log [OH - ] = -log (1x10 -14 ) pH + pOH = 14

17. Calculating concentrations of H 3 O + and OH - in solutions of a strong acid or base. Calculate the concentrations of hydronium ion and hydroxide ion at 25°C in 0.10 M HCl.Example

18. Weak Acids An acid reacts with water to produce hydronium ion and the conjugate base ion. This process is called acid ionization or acid dissociation. Eg. Acetic Acid: CH 3 COOH + H 2 O H 3 O + + CH 3 COO -

19. In general for an acid, HA, we can write: HA(aq) + H 2 O H 3 O + + A - (aq) and the corresponding equilibrium constant expression would be: K c = [H 3 O + ]. [A - ] [HA]. [H 2 O] For a dilute solution, [H 2 O] would be nearly constant, hence: K a = [H 2 O] K c = [H 3 O + ]. [A - ] [HA]

20. Determining K a from the solution pH.Example Sore-throat medications sometimes contain the weak acid phenol, HC 6 H 5 O. A 0.10 M solution of phenol has a pH of 5.43 at 25°C. What is the acid-dissociation constant, K a, for this acid at 25°C? What is its degree of ionization?

21. HA + H 2 O H 3 O + + A - Initial  Eqlbm pH = 5.43 Solution

22. Degree of ionisation (fraction of ions that were ionised): HA + H 2 O H 3 O + + A - x of the original 0.10 M was ionised. Degree of ionisation = Solution

23. Calculating Concentrations of Species in a Weak Acid Solution Using K a. (Approximation Method) Example (See discussion p609-) Para-hydroxybenzoic acid is used to make certain dyes. What are the concentrations of this acid (i.e.hydrogen ion) and para-hydroxybenzoate anion in a 0.200 M aqueous solution at 25°C? What is the pH of the solution and the degree of ionisation of the acid? The K a of this acid is 2.6 x 10 -5.

24. Solution HA + H 2 O H 3 O + + A - Initial  Eqlbm Now we must solve this quadratic. One possible way of simplifying things is via an assumption!

25. Because K a is small (2.6 x 10 -5 ), Hence: [H 3 O + ] = [A - ] = [HA] = = Degree of ionisation

26. Polyprotic Acids So far we have only dealt with acids releasing only one H 3 O + ion. Some acids, have two or more such protons; these acids are called polyprotic acids. Remember the strong acid H 2 SO 4

27. For a weak diprotic acid like carbonic acid, H 2 CO 3, there are two simultaneous equilibria to consider. H 2 CO 3 + H 2 O H 3 O + + HCO 3 - HCO 3 - + H 2 O H 3 O + + CO 3 2- Each equilibrium has an associated acid dissociation constant. For the loss of the first proton: and for the loss of the second proton: K a1 = [H 3 O + ] [HCO 3 - ] [H 2 CO 3 ] = 4.3 x 10 -7 K a2 = [H 3 O + ] [CO 3 2- ] [HCO 3 - ] = 4.8 x 10 -11

28. In general, the second ionisation constant, K a2, of a polyprotic acid is much smaller then the first ionisation constant, K a1. In the case of a triprotic acid, the third ionisation constant, K a3, is much smaller that the second one, K a2.

29. Weak Bases Equilibria involving weak bases are treated similarly to those for weak acids. In general, a weak base B with the base ionization: B(aq) + H 2 O HB + + OH - has a base-ionization constant, K b, equal to: K b = [HB + ] [OH - ] [B]

30. Calculating concentrations of species in a weak base solution using K b.Example Aniline, C 6 H 5 NH 2, is used in the manufacturing of some perfumes. What is the pH of a 0.035 M solution of aniline at 25°C? The K b = 4.2 x 10 -10 at 25°C.

31. B + H 2 O HB + + OH - Initial  Eqlbm Now

32. Remember that forHA(aq) + H 2 O H 3 O + + A - (aq) we have: K a = [H 3 O + ] [A - ] [HA] A - (aq) + H 2 O HA(aq) + OH - and for we can write : K b = [HA] [OH - ] [A - ] Multiplying: Relationship between K a and K b [H 3 O + ] [A - ] [HA] [HA] [OH - ] [A - ] K a.K b = x  K a.K b = K w = [H 3 O + ] [OH - ]

33. Obtaining K a from K b or K b from K aExample Obtain the K b for the F - ion, the ion added to public water supplies to protect teeth. For HF, K a = 6.8 x 10 -4. K b.K a = K w

34. Buffers A buffer is a solution characterised by the ability to resist changes in pH when limited amounts of acid or base are added to it. Buffers contain either a weak acid and its conjugate base or a weak base and its conjugate acid. Thus, a buffer solution contains both an acid species and a base species in equilibrium.

35. Calculate the pH of a buffer from given volumes.Example What is the pH of a buffer made by mixing 1.00 L of 0.020 M benzoic acid, HC 7 H 5 O 2, with 3.00 L of 0.060 M sodium benzoate, NaC 7 H 5 O 2 ? The K a for benzoic acid is 6.3 x 10 -5.

36. Calculating the pH of a buffer when a strong acid or strong base is added. Calculate the pH change that will result from the addition of 5.0 mL of 0.10 M HCl to 50.0 mL of a buffer containing 0.10 M NH 3 and 0.10 M NH 4 +. How much would the pH of 50.0 mL of water change if the same amount of acid were added.?Example