Learning outcomes

1
Learning outcomes

• At the end of this lesson, students should be
  able to
• Derive the Henderson-Hasselbalch equation.
• Calculate the pH of buffer solutions.

2
pH of buffer solution
a) pH of acidic buffer solution

• CH3COOH(aq) CH3COO-(aq) H(aq)
  
• CH3COONa(aq) CH3COO- (aq) Na(aq)
  

3
Calculate the pH of the buffer solution by using
CH3COOH / CH3COONa pair Ka ?CH3COO-??H?
?CH3COOH? ?H? ?CH3COOH? x Ka
?CH3COO-? ?H?
?acid? x Ka conjugate base?
4
Taking -log at both sides of the equation -
log ?H3O? (- log Ka) x -log ?acid? )
?conjugate
base? pH pKa log ?conjugate
base? ?acid ?
5
Example 1
Calculate the pH of a solution prepared by adding
2.00 g of benzoic acid, C6H5COOH 2.00 g sodium
benzoate, C6H5COONa in enough water to make 1.00
L solution (Ka (C6H5COOH) 6.3 x 10-5 M) No of
moles of C6H5COOH 2.00 g/122.1 g mol-1
1.64
x 10-2 moles No of moles of C6H5COONa 2.00 g
/ 144 g mol-1
1.39 x 10-2 moles
6
?C6H5COOH? 1.64 x 10-2 / 1.00 1.64 x
10-2 M ?C6H5COONa? 1.39 x 10-2 / 1.00
1.39 x 10-2 M pH pKa
log ?conjugate base? ?acid? pH
- log 6.3 x 10-5 log 1.39 x 10-2
1.64 x 10-2 4.13
7
Example 2
1.0 litre buffer solution is prepared by mixing
0.10 mole of CH3COOH with 0.10 mole of CH3COONa.
i. Calculate the pH of a the buffer
solution. ii. Calculate the pH of the buffer
solution after the addition of (a) 0.02
mole HCl (b) 0.02 mol NaOH (Assume that the
volume of the solution does not change when HCl
and NaOH is added) Ka for CH3COOH 1.8 x 10-5 M

8
Solution
i) V 1 L CH3COOH 0.10 mol / 1 L
0.10 M CH3COONa 0.10 mol / 1 L 0.10 M
pH pKa log pH pKa log pH
-log(1.8 x 10-5) log pH 4.74
9
ii. (a) pH when 0.02 mol HCl is added
Buffer action CH3COO-(aq) H(aq)
CH3COOH(aq) ninitial 0.10
0.02 0.10 nchange -0.02 -0.02
0.02 nfinal 0.08 0
0.12
10
CH3COO- 0.08 mol / 1 L 0.08 M CH3COOH
0.12 mol / 1 L 0.12 M pH pKa log
pH -log(1.8 x 10-5) log
pH 4.57
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ii.(b) pH when 0.02 mol NaOH is added
Buffer action CH3COOH(aq) OH-(aq)
CH3COO-(aq) H2O(l) ninitial 0.10
0.02 0.10 nchange -0.02
-0.02 0.02 nfinal 0.08
0 0.12
12
CH3COO- 0.12 mol / 1 L 0.12 M CH3COOH
0.08 mol / 1 L 0.08 M pH pKa log
pH -log(1.8 x 10-5) log
pH 4.92
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Example 3

• A buffer solution is prepared by dissolving
• 0.025 mol CH3COONa in 250 cm3 aqueous solution
• of CH3COOH(0.10 mol dm-3). Calculate the pH of
• The buffer solution
• The buffer solution after the addition of 1.0 cm3
  
• H2SO4(0.1 M)
• c) The buffer solution after the addition of 1.0
  cm3
• NaOH(0.1 M)
• Ka(CH3COOH) 1.74 x 10-5 moldm-3

14
Solution
a) CH3COOH 0.1 mol dm-3 CH3COO- 0.025
mol 0.10 mol dm-3 0.250 pH
pKa log CH3COO- CH3COOH
-log (1.74 x 10-5) log (0.10/0.10) 4.76
15
b) After adding 1.0 cm3 H2SO4 (0.1 M) Moles of
H2SO4 MV/1000 0.1 x 1 1.0 x 10-4
mol 1000 In 1 mol of H2SO4, contains 2
mol H Mol of acid added 2( 1x 10-4) 2
x 10-4
16

(b) pH when 1.0 cm3 H2SO4 (0.1 M)
Buffer action CH3COO-(aq)
H(aq) CH3COOH(aq) ninitial
0.025 2 x 10-4 0.025 nchange -
2 x 10-4 - 2 x 10-4 2 x 10-4 nfinal
0.0248 0 0.0252

pH pKa log CH3COO-
CH3COOH - log (1.74 x 10-5) log
(0.0248/0.251) (0.0252/0.251) 4.76
6.95 x 10-3 4.75
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c) pH when addition of 1.0 cm3 NaOH(0.1 M)

Moles of NaOH added 0.1mol L-1 x 1.0 L 1.0
x 10-4 mol
Buffer action CH3COOH(aq) OH-(aq)
CH3COO-(aq) H2O ninitial 0.025 1
x 10-4 0.025 nchange - 1 x 10-4 -
1 x 10-4 1 x 10-4 nfinal 0.0249
0 0.0251
18
pH pKa log CH3COO-
CH3COOH - log 1.74 x 10-4 log (
0.0251/0.251) (0.0249/0.251)
4.76
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b) pH of basic buffer solution

• NH4Cl(aq) NH4(aq)
  Cl-(aq)
• NH3(aq) H2O(l) NH4(aq)
  OH-(aq)

• Calculating pH of the basic buffer solution
• Kb ?NH4? ?OH??
• ?NH3?
• ?OH?? ?NH3? x Kb
• ?NH4?

20
Taking the ? log of both sides of the
equation ?log?OH?? ( - log Kb ) x ? log
?NH3? ?NH4? pOH pKb
log ?NH4? ?NH3?
pOH pKb log ?conjugate acid?
?base? pH 14 ? pOH
21
Example 1
Calculate the pH of the solution prepared
by mixing 500 mL 0.1 M hydrazinium
chloride, N2H5Cl with 500 mL 0.2 M hydrazine,
N2H4 (Kb 1.7 x 10-7) Solution No. of moles of
N2H5Cl 500 x 0.1 1000
0.05 moles
?N2H5Cl? 0.05 (0. 50
0.50) 0.05 M
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No of moles of N2H4 500 x 0.2
1000 0.1 moles
?N2H4? 0.1 (0.50
0.50) 0.1 M
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Using the equation, pOH pKb
log ?N2H5Cl? ?N2H4?
pOH -log (1.7 x 10-7 ) log
0.05 0.1
6.469 pH 14 -
pOH 14 - 6.469 7.531
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Example 2

• A buffer solution is prepared by mixing 400 mL
  of 1.50 M NH4Cl solution with 600 mL of 0.10 M
  NH3.
• i. Calculate the pH of a the buffer solution.
• Calculate the pH of the buffer solution after the
  addition of
• (a) 0.15 M NaOH (b) 0.011 M H
• (Assume that the volume of the solution does not
  change when HCl and NaOH is added)
• Kb for NH3 1.8 x 10-5 M

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Solution
i. V 400 ml 600 ml 1000 ml
1 L
mol NH4Cl 1.50 mol L-1 x 0.4 L 0.6
mol mol NH3 0.10 mol L-1 x 0.6 L
0.06 mol NH4Cl 0.6
M NH3 0.06 M
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pOH pKb log _at_ pOH pKb
log -log(1.8 x 10-5) log
5.74 pH 14 5.74 8.26
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ii. (a) pH when 0.15 M NaOH is added
Buffer action NH4(aq)
OH-(aq) NH3(aq) H2O(l)
initial 0.60 M 0.15 M 0.06
M change -0.15 M -0.15 M
0.15 M final 0.45 M 0
0.21 M
pOH pKb log
-log(1.8 x 10-5) log

5.08 pH 14 5.08 8.92
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ii.(b) pH when 0.011 M HCl is added
Buffer action NH3(aq) H(aq)
NH4(aq) initial
0.06 M 0.011 M 0.6 M
change -0.011 M -0.011 M
0.011 M final 0.049 M
0 0.611 M
pOH pKb log
-log(1.8 x 10-5) log
5.84 pH 14 5.84
8.16
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Exercise

• How many moles of NH4Cl must be added to 2.0 L of
  0.10 M NH4 to form a buffer whose pH is 9.00 ? (
  Assume that the addition of NH4Cl does not change
  the volume of the solution ) (0.36 mol)
• Calculate the concentration of sodium benzoate
  that must be present in a 0.20 M solution of
  benzoic acid ( C7H6O2 ) to produce a pH of 4.00.
  (0.13 M)
• Calculate the pH of a buffer composed of 0.12 M
  benzoic acid and 0.20 M sodium benzoate. Ka 6.3
  x 10-5. ( 4.43 )
• What is the pH of a buffer that is 0.12 M in
  lactic acid ( HC3H5O3) and 0.10 M in
  sodium lactate? Ka 1.4 x 10-4 (3.77)

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Exercise
A solution is prepared by mixing 100 cm3 aqueous
NH3 (0.1 moldm-3) with 100 cm3 NH4Cl (1.0 mol
dm-3). Calculate the pH of the solution before
and after the addition of 1 cm3 HCl (0.1 mol
dm-3). Kb (NH3) 1.74 x 10-5 mol dm-3

• Answer
• 8.24
• 8.23