Acids and Bases 6
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IB Topic 18.4: Buffers
Text Reference: Higher Level Chemistry p. 299 - 305
IB Assessment Statements
18.2.1
Describe the composition of a buffer solution and
explain its action.
18.2.2
Solve problems involving the composition and pH of
a specified buffer system.
Only examples involving the transfer of one proton will be assessed.
Examples should include ammonia solution/ammonium chloride and
ethanoic acid/sodium ethanoate.
Students should state when approximations are used in equilibrium calculations.
The use of quadratic equations will not be assessed.
Starter Activity
pH Tug of War
1. Set up three sets of 3 test tubes.
2. Add the following to each test tube in the set, to a depth of 2-3 cm:
A: tap water
B: pH 7 buffer
C: your choice - pH 4 or pH 10 buffer
3. Add several drops of universal indicator to each test tube and mix.
Note the colours.
4. Add the following to each set (A, B, C) and note any changes that occur:
test tube 1: nothing
test tube 2: 2-3 drops 0.1 M HCl
test tube 3: 2-3 drops 0.1 M NaOH
Starter Activity
pH Tug of War Results
A
solution
original
colour
after adding
HCl
after adding
NaOH
Q: How is a buffer different from water?
B
C
What is a Buffer Solution?
Buffers are solutions that resist changes in pH
when small amounts of acid or base are added.
Application 1:
The pH of blood must be maintained
between 7.35 to 7.45.
Changes of only 0.2 - 0.4 pH units is fatal.
HCO3– ⇄ H+ + CO32–
Application 2:
Several chemical equilibria (buffer
systems) help maintain the pH of
seawater at around 8.0.
This helps to maintain corals.
Buffer Composition
Buffer solutions are made to
control the pH at a desired value.
http://students.cis.uab.edu/ash13y/ph-buffer-solution-470674.jpg
Buffer solutions are prepared by mixing two substances together:
1. a weak acid + its salt (containing the conjugate base)
2. a weak base + its salt (containing the conjugate acid)
Equations for Buffer Systems
Acidic Buffers (pH < 7)
weak acid mixed with its salt
1: ethanoic acid & sodium ethanoate
CH3COOH & NaCH3COO
The chemical equation for this buffer system is simply
the equation for the weak acid:
CH3COOH ⇄ H+ + CH3COO–
NOT
CH3COOH + NaCH3COO ⇄ ???
2: citric acid & potasssium citrate
3: benzoic acid & sodium benzoate
Equations for Buffer Systems
Basic Buffers (pH > 7)
weak base mixed with its salt
1: ammonia & ammonium chloride
NH3 & NH4Cl
The chemical equation for this buffer system is simply
the equation for the weak base:
NH3 + H2O ⇄ NH4+ + OH–
NOT
NH3 + NH4Cl ⇄ ???
2: carbonate ions & sodium bicarbonate
Equations for Buffer Systems
Equations for Buffers
1. Write the equation for the buffer prepared by mixing benzoic acid
(C6H5COOH) with sodium benzoate (NaC6H5COO).
C6H5COOH ⇄ H+ + C6H5COO–
2. Write the equation for the buffer prepared by mixing methylamine
(CH3NH2) with its salt, methylammonium chloride (CH3NH3Cl).
CH3NH2 + H2O ⇄ CH3NH3+ + OH–
Calculating the pH of a Buffer
Acidic Buffers
HA ⇄ H+ + A–
The [H+ ] depends on the concentration of the acid and its salt (i.e. A–):
+
–
[H ] [A ]
Ka =
[HA]
[HA]
[H ] = K a x
–
[A ]
+
Assume the equilibrium concentration
of HA is the initial concentration.
Assume the equilibrium concentration of
A– is the initial concentration of the salt.
Calculating the pH of a Buffer
Acidic Buffers
HA ⇄ H+ + A–
Nifty trick: Take the -log10 of both sides of that last equation:
[HA]
[H ] = K a x
–
[A ]
⎛
[HA] ⎞
+
-log10 [H ] = -log10 ⎜ K a x
– ⎟
[A ] ⎠
⎝
+
Only pKa values
are given in the
Data Booklet!
[HA]
pH = pK a - log10 –
[A ]
Calculating the pH of a Buffer
Acidic Buffers
Example: Calculate the pH of a buffer prepared by mixing 0.050 mol of solid
sodium acetate with 0.50 dm3 of 0.10 mol dm–3 ethanoic acid.
CH3COOH ⇄ H+ + CH3COO–
[CH3COOH] = 0.10 mol
dm-3
[CH3COO–] = [NaCH3COO]
moles
=
volume
0.050 mol
=
0.50 dm3
= 0.10 mol dm−3
[HA]
pH = pK a - log10 –
[A ]
0.10
= 4.76 - log10
0.10
= 4.76 - log10 (1)
= 4.76 - 0
= 4.76
Calculating the pH of a Buffer
Acidic Buffers
TIP!
[HA]
pH = pK a - log10 –
[A ]
0.10
= 4.76 - log10
0.10
= 4.76 - log10 (1)
= 4.76 - 0
= 4.76
Whenever you have
“equimolar” amounts of
the weak acid and the salt
in an acidic buffer:
pH = pKa
Calculating the pH of a Buffer
Basic Buffers
B + H2O ⇄ BH+ + OH–
The [H+ ] depends on the concentration of the base and its salt (i.e. BH+):
+
–
[BH ] [OH ]
Kb =
[B]
[B]
−
[OH ] = K b x
+
[BH ]
[B]
pOH = pK b - log10
+
[BH ]
Calculating the pH of a Buffer
Basic Buffers
Calculate the pH of a buffer prepared by mixing 100 cm3 of 0.40 mol dm-3
ammonia with 100 cm3 of 0.60 mol dm–3 ammonium chloride.
NH3 + H2O ⇄ NH4+ + OH–
Find [NH3] and [NH4+] in the buffer
mixture using C1V1 = C2V2
[NH3]buffer = C1V1 / Vtotal
= (0.40 mol dm-3)(100 cm3) / 200 cm3
= 0.20 mol dm-3
[NH4]+buffer = [NH4Cl]buffer
= C1V1 / Vtotal
= (0.60 mol dm-3)(100 cm3) / 200 cm3
= 0.30 mol dm-3
pOH = pK b - log10
[NH3 ]
[NH4 + ]
= 4.76 - log10 (0.20)/(0.30)
= 4.76 - (-0.17)
= 4.93
∴ pH = 14.00 - 4.93
= 9.07
= 9.1
Preparing Buffer Solutions
Method 1: Mix the weak/acid base with its salt.
Example 1: 50 ml of 0.25 M ethanoic acid and
100 mL of 0.25 M sodium ethanoate
Example 2: 50 ml of 0.25 M ethanoic acid and
5.0 g of solid sodium ethanoate
Example 3: 200 ml of 0.40 M ammonia and
5.0 g of solid ammonium chloride
Preparing Buffer Solutions
Method 2A: Add a strong base to a weak acid to reach the
half-neutralization point.
The neutralization reaction produces the salt (A–).
buffer reaction:
HA ⇄
acid
neutralization reaction:
HA
HA
HA
HA
HA
HA
salt
acid + salt = buffer
HA
HA
HA
–
+A
HA + NaOH ⇄ H2O + NaA
acid
HA
+
H
add enough NaOH to
neutralize half the HA
A–
HA
HA
HA
HA
HA
A–
A–
A–
HA
HA
A–
HA
A–
Preparing Buffer Solutions
Method 2A: Add a strong base to a weak acid to reach the
half-neutralization point.
buffer reaction:
CH3COOH ⇄ H+ + CH3COO–
Add 25 ml of 0.10 mol dm–3 NaOH to 50 mL of 0.10 mol dm–3 CH3COOH.
Preparing the buffer:
initial moles CH3COOH
= C xV
= 0.10 x 0.050
= 0.0050 mol
moles NaOH added
= C xV
= 0.10 x 0.025
= 0.0025 mol
Resultant buffer:
moles CH3COOH left
= 0.0050 - 0.0025
= 0.0025 mol
moles CH3COO– formed
= 0.0025 mol
pH = pK a - log10
[HA]
0.0025/0.075
=
4.76
log
= 4.76 - log10 (1) = 4.76
10
–
0.0025/0.075
[A ]
Preparing Buffer Solutions
Method 2A: Add a strong base to a weak acid to reach the
half-neutralization point.
buffer reaction:
CH3COOH ⇄ H+ + CH3COO–
Add 50 ml of 0.050 mol dm–3 NaOH to 50 mL of 0.10 mol dm–3 CH3COOH.
Preparing the buffer:
initial moles CH3COOH
= C xV
= 0.10 x 0.050
= 0.0050 mol
moles NaOH added
= C xV
= 0.050 x 0.050
= 0.0025 mol
Resultant buffer:
moles CH3COOH left
= 0.0050 - 0.0025
= 0.0025 mol
moles CH3COO– formed
= 0.0025 mol
pH = pK a - log10
[HA]
0.0025/0.075
=
4.76
log
= 4.76 - log10 (1) = 4.76
10
–
0.0025/0.075
[A ]
Preparing Buffer Solutions
Method 2B: Add a strong acid to a weak base to reach the
half-neutralization point.
The neutralization reaction produces the salt (BH+).
buffer reaction:
neutralization reaction:
NH3 + H2O ⇄ NH4+ + OH–
NH3 + HCl ⇄ NH4+ + Cl–
Add 25 ml of 0.10 mol dm–3 HCl to 50 mL of 0.10 mol dm–3 NH3.
moles HCl = C x V
moles NH3 = C x V
= 0.10 x 0.025
= 0.10 x 0.050
= 0.0025 mol
= 0.0050 mol
Add 50 ml of 0.050 mol dm–3 HCl to 50 mL of 0.10 mol dm–3 NH3.
moles HCl = C x V
= 0.050 x 0.050
= 0.0025 mol
moles NH3 = C x V
= 0.10 x 0.050
= 0.0050 mol
How Buffers Resist pH Changes
Consider this general acidic buffer system made by mixing HA with its salt, NaA:
HA ⇄ H+ + A–
HA ⇄
high concentration
(from HA)
http://collective.chem.cmu.edu/buffers/images/Pict3buffer.gif
–
H+ + A
high concentration
(from NaA)
HIGH concentrations of BOTH
of these species helps maintain
[H+]
How Buffers Resist pH Changes
Consider how the pH of this system is calculated:
HA ⇄ H+ + A–
+
–
[H ] [A ]
Ka =
[HA]
[HA]
+
[H ] = K a x
–
[A ]
This ratio determines the [H+]
As long as this ratio remains the
same (or very close to it), [H+]
will remain the same.
How Buffers Resist pH Changes
What happens when small amounts of acid (H+)
or base (OH–) are added to an acidic buffer?
HA
add H+:
⇄
H+
–
+A
the large amount of A– can react to use up the added H+
A– + H+ → HA
∴ there is little change in the concentration of A– and HA
Look at this example just showing number of molecules:
HA
A–
[HA]
ratio
[A – ]
before
10,000
10,000
1
after adding “10” H+
10,010
9,990
1.0020
[HA]
pH = pK a - log10 –
[A ]
How Buffers Resist pH Changes
What happens when small amounts of acid (H+)
or base (OH–) are added to an acidic buffer?
HA
⇄
H+
–
+A
add OH–: the large amount of HA can react to use up the added OH–
HA + OH– → H2O + A–
there is little change in the concentration of HA and A–
Look at this example just showing number of molecules:
HA
A–
[HA]
ratio
[A – ]
before
10,000
10,000
1
after adding “10” OH–
9990
10010
0.9980
[HA]
pH = pK a - log10 –
[A ]