Unless otherwise stated, all images in this file have been reproduced from: Blackman, Bottle, Schmid, Mocerino and Wille,Chemistry, 2007 (John Wiley) ISBN: 9 78047081 0866
CHEM1002 [Part 2] Dr Michela Simone Weeks 8 – 13 Office Hours: Monday 3-5, Friday 4-5 Room: 412A (or 416) Phone: 93512830 e-mail:email@example.com
Acids & Bases 3 • Lecture 2: • Strong/Weak Acids and Bases • Calculations • Polyprotic Acids • Lecture 3: • Salts of Acids and Bases • Buffer systems • Blackman Chapter 11, Sections 11.3-11.6 Reproduced from ‘The Extraordinary Chemistry of Ordinary Things, C.H. Snyder, Wiley, 2002(Page 245)
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 H2O(aq) + CN–(aq) OH–(aq) + HCN(aq) Does a solution of NH4Cl have pH > 7 or < 7? Salt of NH4OH (weak base) and HCl (strong acid) acid “wins”: pH < 7 and reaction is H2O(aq) + NH4+(aq) NH3(aq) + H3O+(aq) Salts of Weak Acids and Bases
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. CH3COOH(aq) + H2O(l) CH3COO-(aq)+ H3O+(aq) • Addition of CH3COO-: • By Le Chatelier’s principle the equilibrium will shift to the left to remove CH3COO- and therefore decrease [H3O+]. • Addition of CH3COOH: • By Le Chatelier’s principle the equilibrium will shift to the right to remove CH3COOH and therefore increase [H3O+].
A solution containing both a weak acid and its salt withstands pH changes when acid or base (limited amounts) are added. Buffer with equalconcentrations of conjugate base and acid Buffer after addition of H3O+ Buffer after addition of OH- H3O+ OH- CH3COO- CH3COOH CH3COO- CH3COOH CH3COO- CH3COOH Buffer System H2O + CH3COOH H3O+ + CH3COO- CH3COOH + OH- H2O+ CH3COO-
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! Buffer systems and pH Change Huge change!What about buffers?
Buffer systems and pH Change • 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? HAc(aq) H+(aq) + Ac–(aq) Ka = 10-4.7 0 0.1 +10-3 0.1 - 10-3
Buffer systems and pH Change the pH hardly changes from 4.7! Solution is buffered against pH change
Henderson - Hasselbalch Equation • For a buffer solution, which contains similar concentrations of a conjugate acid/base pair of a weak 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:
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 and Capacity • Buffer Preparation • If the pH of a required buffer is pKa of available acid then use equimolar amounts of acid and conjugate base • If the required pH differs from the pKa then use the Henderson-Hasselbalch equation.
Buffer Preparation and Capacity Most effective buffers have acid/base ratio less than 10 and more than 0.1 pH range is ±1
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 – 7.45. 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. Buffers in Natural Systems
“Extracellular” buffer (outside cell) Buffer System in Blood H+(aq) + HCO3–(aq)H2CO3(aq) H2CO3(aq) H2O(l) + CO2 (g) • Removal of CO2 shifts equilibria to right, reducing [H+], i.e., raising the pH • The pH can be reduced by: H2CO3(aq) + OH–(aq) HCO3–(aq) + H2O(l)
Example In the H3PO4 / NaH2PO4 / Na2HPO4 / Na3PO4 system, how could you make up a buffer with a pH of 7.40? DATA: Ka1 = 7.2 x 10-3, Ka2 = 6.3 x 10-8, Ka3 = 4.2 x 10-13 • To make up a buffer, we need pH near pKa pKa1 = 2.14, pKa2 = 7.20, pKa3 = 12.38 must use mixture of H2PO4- and HPO4- • Could go through whole procedure... or simply use Henderson-Hasselbalch equation ...
original amounts Example (Continued) • Require buffer with a pH of 7.40 • the required ratio of Na2HPO4 to NaH2PO4 = 1.58:1
Practice Examples • 1 What is the pH of a 0.045 M solution of KOBr? • The pKa of HOBr is 8.63. • (a) 4.74 b) 4.99 c) 8.25 d) 9.01 e) 10.64 • A buffered solution is 0.0500 M CH3COOH and 0.0400 M NaCH3CO2. If 0.0100 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, pKa = 4.76 • (a) 4.76 (b) 4.46 (c) 4.66 (d) 4.86 (e) 4.54
Summary: Acids & 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 Review Problems 11.36-11.37 and 11.96-106 in Blackman • Next lecture: • Titrations