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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]. A/Prof Adam Bridgeman (Series 1) Dr Feike Dijkstra (Series 2) Weeks 8 – 13

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Unless otherwise stated all images in this file have been reproduced from

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
CHEM1002 reproduced from:[Part 2]

A/Prof Adam Bridgeman (Series 1)

Dr FeikeDijkstra (Series 2)

Weeks 8 – 13

Office Hours: Monday 2-3, Friday 1-2

Room: 543a

e-mail:adam.bridgeman@sydney.edu.au

e-mail:feike.dijkstra@sydney.edu.au


Unless otherwise stated all images in this file have been reproduced from

Acids & Bases 3 reproduced from:

  • 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)


Salts of weak acids and bases

Is a solution of reproduced from: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
The Common Ion Effect reproduced from:

  • 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+].


Buffer system

A solution containing reproduced from: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-


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!

Buffer systems and pH Change

Huge change!What about buffers?


Buffer systems and ph change1
Buffer systems and pH Change equal amount of 10

  • 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

+x

-x

+x

x

0.1+10-3-x

0.1- 10-3+x

cf slide 6


Unless otherwise stated all images in this file have been reproduced from

x equal amount of 10 = 1.02  Ka = 0.000020 << 0.001

pH = – log x = 4.69

Buffer systems and pH Change

the pH hardly changes from 4.7!

Solution is buffered against pH change


Henderson hasselbalch equation
Henderson - Hasselbalch Equation equal amount of 10

  • 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:


Unless otherwise stated all images in this file have been reproduced from

Buffer Capacity equal amount of 10

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.


Unless otherwise stated all images in this file have been reproduced from

Buffer Preparation and Capacity equal amount of 10

Most effective buffers have acid/base ratio less than 10 and more than 0.1  pH range is ±1


Buffers in natural systems

Biological systems, equal amount of 10e.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


Buffer system in blood

“Extracellular” buffer (outside cell) equal amount of 10

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
Example equal amount of 10

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 ...


Unless otherwise stated all images in this file have been reproduced from

original equal amount of 10

amounts

Example (Continued)

  • Require buffer with a pH of 7.40

  • the required ratio of Na2HPO4 to NaH2PO4 = 1.58:1


Unless otherwise stated all images in this file have been reproduced from

Practice Examples equal amount of 10

  • 1What is the pH of a 0.045 M solution of KOBr?

  • The pKa of HOBr is 8.63.

  • (a) 4.74b) 4.99c) 8.25d) 9.01e) 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


  • Unless otherwise stated all images in this file have been reproduced from

    Summary: Acids & Bases 3 equal amount of 10

    • 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


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