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

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e CHEM1002 [Part 2]

Dr Michela SimoneWeeks 8 – 13

Office Hours: Monday 3-5, Friday 4-5Room: 412A (or 416)Phone: 93512830e-mail: michela.simone@sydney.edu.au

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e

Lecture 3:

• Salts of Acids and Bases• Buffer systems• Blackman Chapter 11,

Sections 11.3-11.6

Acids & Bases

Reproduced from ‘The Extraordinary Chemistry of Ordinary Things, C.H.

Snyder, Wiley, 2002(Page 245)

Lecture 4:

• Titrations• Blackman Chapter 11, Section 11.7

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e• Strong acids completed dissociates in solution:

HA + H2O A-(aq) + H3O+(aq)

pH = -log10([strong acid]initial)

• If strong base is added, it reacts with the H3O+ so [H3O+(aq)] is reduced:

H3O+(aq) + OH-(aq) H2O(l)

pH = -log10([strong acid]remaining)

• Equivalence point: when the amount of added base = initial amount of acid:

[H+(aq)] = 10-7.0 M

pH = 7.00

• After the equivalence point, any added base increases [OH-(aq)]: pH = 14.00 - pOH = 14.00 - (-log10([excess base]))

Strong Acid/Strong Base Titration

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e

• Initial pH ispH = 14 - pOH

= 14 -(-log10([strong base]initial)

• At the equivalence point, [H+(aq)] = 10-7.0 MpH = 7.00 (at 25 °C)

• After the equivalence point,pH = -log10([excess acid])

Strong Base/Strong Acid Titration

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e Strong Base/Strong Acid Titration

Add HCl(g) in 0.10 mol amounts to 1.0 L of 0.5 M NaOH(aq)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.80

2

4

6

8

10

12

14

16

pH

amount of added acid

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x Weak Acid/Strong Base Titration

Add NaOH(s) in 0.10 mol amounts to 1.0 L of 0.5 M acetic acidInitial pH: initially the solution contains just acetic acid and its pH must be calculated using the procedure outlined in slide 7 of lecture 2. pKa = 4.76.

CH3COOH H2O H3O+ CH3COO–

initial (I) 0.5 large 0 0

change (C) negligible

equilibrium (E) large

Ka =

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x Weak Acid/Strong Base Titration

Add NaOH(s) in 0.10 mol amounts to 1.0 L of 0.50 M acetic acid

cf lecture 3

Acidic region: in the acidic region, OH- has reacted with some of the acetic acid to make acetate. The solution then contains both acetic acid and its conjugate base and the Henderson-Hasselbalch equation can be used to calculate the pH.

CH3COOH(aq) + OH-(aq) CH3COO-(aq) + H2O(aq)

a[ ]

pH = p log[ ]

baseK

acid

10

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x Weak Acid/Strong Base Titration

Add NaOH(s) in 0.10 mol amounts to 1.0 L of 0.50 M acetic acid

cf lecture 2

Equivalence point: all of the acetic acid has reacted with OH- to form CH3COO-. At this point, the solution contains a weak base so is basic. The pOH must be calculated.

CH3COO- H2O CH3COOH OH-

initial (I) 0.5 large 0 0

change (C) negligible

equilibrium (E) large

b ap 14.00 p 9.24 K K

Kb =

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x Weak Acid/Strong Base Titration

Add NaOH(s) in 0.10 mol amounts to 1.0 L of 0.50 M acetic acidAlkaline region: after the equivalence point, the major source of OH- is that from the excess NaOH. Very little is contributed from the acetate ion.After the equivalence point, the calculations are then exactly the same as for the strong acid/strong base titration.

When 0.60 mol of NaOH(s) is added, 0.50 mol react with the CH3COOH leaving a 0.10 M solution of OH-:

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e Weak Acid/Strong Base Titration

• Initial pH is higher than for strong acid / strong base A weak acid is present initially

• In acidic region, weak acid and conjugate base present Buffering region with slow change in pH

At the 1/2 equivalence point, pH = pKa

• At equivalence point, conjugate base present pH > 7

• In alkaline region, excess strong base present Curve is the same as for strong acid/strong base titration

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e Weak Acid/Strong Base Titrations

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e Weak Base/Strong Acid Titration

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e Titrations

• Equivalence Point:

When number of moles of added base = original number of moles of acid

Strong acid/strong base pH = 7

Weak acid/strong base pH > 7

Strong acid/weak base pH < 7

• End Point:

When a colour change in the indicator is observed

Choose an indicator that changes colour close to the equivalence point

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e Indicators

• The pH at which acid base depends on the pKa of the indicator

weak acid base– each form has a different colour

pH 3.2 pH 4.4 pH 4.8 pH 5.4

methyl orange methyl purple bromothymol blue phenolphthalein

pH 6.0 pH 7.6 pH 8.2 pH 10.0

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e

1.Which one of the following combinations does the titration curve to the right represent?

A.Addition of a strong base to a weak acidB.Addition of a weak base to a strong acidC.Addition of a weak acid to a strong baseD.Addition of a strong acid to a strong baseE.Addition of a strong acid to a weak base

2. What is the value of the pKa that can be obtained from this titration curve?

A.11.3B.10.0C.9.3D.5.3E.1.8

Practice Examples

pH

8

10

12

6

4

2

10 30 504020

Amount of solution added (mL)

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e

Learning Outcomes - you should now be able to:

• Complete the worksheet• Understand strong acid/strong base, strong

base/strong acid, weak acid/strong base and weak base/strong acid titrations

• Be able to extract pKa from the half equivalence point

• Be able to work at the pH at the equivalence point• Answer Review Problems 11.38-11.44 in Blackman

Summary: Acids & Bases 4

Next lecture:

• Periodic Trends