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1 Unless otherwise stated, all images in this file have been reproduced from: Blackman, Bottle, Schmid, Mocerino and Wille, Chemistry, 3 rd Edition 2016 (John Wiley & Sons) The University of Sydney Page 1

2 b.socrative.com room CHEM1002 b.socrative.com CHEM1002 The University of Sydney Page 2

3 Acids and bases (3) After lecture 23, you should be able to: Complete its worksheet and practice examples Complete acid/base calculations for strong and weak acids and bases Use ph, pk w, pk a and pk b Define strong and weak acids and bases Answer Review Problems in Blackman Lecture 24: Salts of Acids and Bases Buffer systems Blackman Chapter 11, Sections The University of Sydney Page 3

4 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) The University of Sydney Page 4

5 The Common Ion Effect If you add the salt of an acid to a solution of the same acid then the equilibrium will shift towards neutral. CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq) 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 + ]. The University of Sydney Page 5

6 Buffer System Buffer after addition of H 3 O + Buffer with equal concentrations of conjugate base and acid Buffer after addition of OH - CH 3 COO - CH 3 COOH H 3 O + CH 3 COO - CH 3 COOH OH - CH 3 COO - CH 3 COOH H 2 O + CH 3 COOH H 3 O + + CH 3 COO - CH 3 COOH + OH - H 2 O + CH 3 COO - A solution containing both a weak acid and its salt withstands ph changes when acid or base (limited amounts) are added. The University of Sydney Page 6

7 Buffer systems and ph Change Consider change in ph of pure water (ph = 7) if we add an equal amount of M HCl: [H + ] = M; ph = -log 10 (0.001) = 3 ph goes from 7 to 3! Huge change! What about buffers? The University of Sydney Page 7

8 Henderson - Hasselbalch Equation For a buffer solution, which contains similar concentrations of a conjugate acid/base pair of a weak acid: K a = + [H ][A [HA] - ]» + [H ][initial base] [initial acid] 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: ph = pka + [initial base] log [initial acid] The University of Sydney Page 8

9 Henderson - Hasselbalch Equation Consider a buffer solution with M each of sodium acetate (NaAc) & acetic acid (HAc). What is the ph? pk a of HAc is [initial base] ph = pk a + log = log(0.100/0.100) = 4.76 [initial acid] The University of Sydney Page 9

10 Buffer systems and ph Change Consider a buffer solution with M each of sodium acetate (NaAc) & acetic acid (HAc). What is the ph when M HCl is added? pk a of HAc is H + will react with base (Ac - ) to form HAc H + (aq) + Ac (aq) HAc(aq) [Ac - ] will go down: from M to M [HAc] will go up: from M to M [initial base] ph = pk a + log = log(0.099/0.101) [initial acid] = 4.75 The University of Sydney Page 10

11 Buffer preparation and capacity 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. The University of Sydney Page 11

12 Buffer Preparation and Capacity Most effective buffers have acid/base ratio less than 10 or more than 0.1 ph range is ±1 The University of Sydney Page 12

13 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» In a healthy person, blood ph is never more than 0.2 ph units from its average value. ph < 7.2, acidosis ; ph > 7.6, alkalosis. Death occurs if ph < 6.8 or > 7.8. The University of Sydney Page 13

14 Buffer System in Blood Extracellular buffer (outside cell) H + (aq) + HCO 3 (aq) H 2 CO 3 (aq) H 2 CO 3 (aq) H 2 O(l) + CO 2 (g) 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) + OH (aq) HCO 3 (aq) + H 2 O(l) The University of Sydney Page 14

15 Choosing a buffer 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: pk a1 = 2.14, pk a2 = 7.20, pk a3 = To make up a buffer, we need ph near pk a Must use mixture of H 2 PO 4- and HPO 4 2 The University of Sydney Page 15

16 Choosing a buffer - continued ph = pk + log a2 éhpo ë éhpo ë ù û ù û original amounts Require buffer with a ph of 7.40 é 2- HPO ù ë = log û é - HPO ù ë 2 4 û é 2- HPO ù ë 4 û 0.20 Þ = 10 = 1.58 é - HPO ù ë 2 4 û The required ratio of Na 2 HPO 4 to NaH 2 PO 4 = 1.58:1 The University of Sydney Page 16

17 Learning outcomes: acids and bases (3) 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 the Practice Examples (next slide) and Review Problems and in Blackman The University of Sydney Page 17

18 Practice Examples 1 What is the ph of a M solution of KOBr? The pk a of HOBr is (a) 4.74 b) 4.99 c) 8.25 d) 9.01 e) 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, what 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 The University of Sydney Page 18