Unless otherwise stated, all images in this file have been reproduced from: Blackman, Bottle, Schmid, Mocerino and Wille, Chemistry, 2007 (John Wiley) ISBN:

CHEM1002 [Part 2] A/Prof Adam Bridgeman (Series 1) Dr Feike Dijkstra (Series 2) Weeks 8 – 13 Office Hours: Monday 2-3, Friday 1-2 Room: 543a

Slide 3/18 e Lecture 2: Strong/Weak Acids and Bases Calculations Polyprotic Acids Lecture 3: Salts of Acids and Bases Buffer systems Blackman Chapter 11, Sections Acids & Bases 3 Reproduced from ‘The Extraordinary Chemistry of Ordinary Things, C.H. Snyder, Wiley, 2002 (Page 245)

Slide 4/18 e Salts of Weak Acids and Bases Is a solution of NaCN acidic or basic?  NaCN is the salt of NaOH (strong base) and HCN (weak acid). The base “wins”: pH > 7  Overall reaction is H 2 O(aq) + CN – (aq) OH – (aq) + HCN(aq) Does a solution of NH 4 Cl have pH > 7 or < 7?  Salt of NH 4 OH (weak base) and HCl (strong acid)  acid “wins”: pH < 7 and reaction is H 2 O(aq) + NH 4 + (aq) NH 3 (aq) + H 3 O + (aq)

Slide 5/18 e The Common Ion Effect CH 3 COOH(aq) + H 2 O(l) CH 3 COO  (aq)  + H 3 O + (aq) If you add the salt of an acid to a solution of the same acid then the equilibrium will shift towards neutral. Addition of CH 3 COO - :  By Le Chatelier’s principle the equilibrium will shift to the left to remove CH 3 COO- and therefore decrease [H 3 O + ]. Addition of CH 3 COOH :  By Le Chatelier’s principle the equilibrium will shift to the right to remove CH 3 COOH and therefore increase [H 3 O + ].

Slide 6/18 e Buffer System A solution containing both a weak acid and its salt withstands pH changes when acid or base (limited amounts) are added. CH 3 COOH + OH -  H 2 O + CH 3 COO - H 2 O + CH 3 COOH  H 3 O + + CH 3 COO - Buffer after addition of H 3 O + H3O+H3O+ CH 3 COO - CH 3 COOH Buffer with equal concentrations of conjugate base and acid CH 3 COO - CH 3 COOH Buffer after addition of OH - OH - CH 3 COO - CH 3 COOH

Slide 7/18 e Buffer systems and pH Change Consider change in pH of pure water (pH = 7) if we add an equal amount of 10 –3 M HCl:  [H + ] = 1 x 10 –3 M  pH goes from 7 to 3! Huge change! What about buffers?

Slide 8/18 x Consider a buffer solution with 0.1 M each of sodium acetate (NaAc) & acetic acid (HAc):  What is the pH when 10 –3 M HCl is added? Buffer systems and pH Change Initial: Neutralization (I): Change (C): Equilibrium (E): -x-x +x+x +x+x x x x cf slide 6 HAc(aq) H + (aq) + Ac – (aq)K a =

Slide 9/18 x x = 1.02  K a = <<  pH = – log x = 4.69 Buffer systems and pH Change the pH hardly changes from 4.7! Solution is buffered against pH change

Slide 10/18 e Henderson - Hasselbalch Equation The dissociation of HA or protonation of A - does not lead to a significant change in the concentrations of these species. Taking logs and rearranging gives: For a buffer solution, which contains similar concentrations of a conjugate acid/base pair of a weak acid:

Slide 11/18 e Buffer Capacity  Buffer capacity is related to the amount of strong acid or base that can be added without causing significant pH change.  Depends on amount of acid & conjugate base in solution:  highest when [HA] and [A – ] are large.  highest when [HA]  [A – ] Buffer Preparation  If the pH of a required buffer is pK a of available acid then use equimolar amounts of acid and conjugate base  If the required pH differs from the pK a then use the Henderson-Hasselbalch equation. Buffer Preparation and Capacity

Slide 12/18 e Buffer Preparation and Capacity Most effective buffers have acid/base ratio less than 10 and more than 0.1  pH range is ±1

Slide 13/18 e Buffers in Natural Systems Biological systems, e.g. blood, contain buffers: pH control essential because biochemical reactions are very sensitive to pH. Human blood is slightly basic, pH  7.39 – In a healthy person, blood pH is never more than 0.2 pH units from its average value. pH 7.6, “alkalosis”. Death occurs if pH 7.8.

Slide 14/18 e Buffer System in Blood “Extracellular” buffer (outside cell) H 2 CO 3 (aq) + OH – (aq) HCO 3 – (aq) + H 2 O(l)  Removal of CO 2 shifts equilibria to right, reducing [H + ], i.e., raising the pH  The pH can be reduced by: H 2 CO 3 (aq) H 2 O(l) + CO 2 (g) H + (aq) + HCO 3 – (aq) H 2 CO 3 (aq)

Slide 15/18 x In the H 3 PO 4 / NaH 2 PO 4 / Na 2 HPO 4 / Na 3 PO 4 system, how could you make up a buffer with a pH of 7.40? DATA: K a1 = 7.2 x 10 -3, K a2 = 6.3 x 10 -8, K a3 = 4.2 x Example To make up a buffer, we need pH near pK a pK a1 = 2.14, pK a2 = 7.20, pK a3 =  must use mixture of H 2 PO 4 - and HPO 4 - Could go through whole procedure... or simply use Henderson-Hasselbalch equation...

Slide 16/18 x Example (Continued) original amounts  the required ratio of Na 2 HPO 4 to NaH 2 PO 4 = 1.58:1 Require buffer with a pH of 7.40

Slide 17/18 x Practice Examples 1What is the pH of a M solution of KOBr? The pK a of HOBr is (a) 4.74b) 4.99c) 8.25d) 9.01e) A buffered solution is M CH 3 COOH and M NaCH 3 CO 2. If mol of gaseous HCl is added to 1.00 L of the buffered solution, wahat is the final pH of the solution? For acetic acid, pK a = 4.76 (a) 4.76(b) 4.46(c) 4.66(d) 4.86(e) 4.54

Slide 18/18 e Learning Outcomes - you should now be able to: Complete the worksheet Understand acid and base equilibria Identify conjugate acid/base pairs Perform calculations with strong acids/bases Use the buffer concept and construct buffers Apply the Henderson-Hasselbalch equation Answer Review Problems and in Blackman Summary: Acids & Bases 3 Next lecture: Titrations