A buffer solution is a solution which resists change in pH when small additional amounts of strong acid or base are added to it. (like a shock absorber)
buffer solutions
A buffer can resist pH changes if the pH is at or near a weak acid pK value: pK= -log K pKa=-log Ka pKb=-log Kb
PH change of a non Buffer:
Water at 25°C
PH =7
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Addition of 0.1 cm3 1.0 MHCl
Addition of 0.1 cm3 1.0 NaOH
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PH = 4
PH = 2.9 PH change of a Buffer:
Addition of 0.2 cm3 1.0 M HCl
Addition of 0.2 cm3 1.0 M NaOH
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Buffer at 25°CPH = 3
PH = 3.1
The Buffer resists against a big change in PH:
There are two types of Buffers: Acidic Buffers and Basic Buffers
1 Acidic Buffers
Are made from a mixture of A weak acid HA
Its conjugate base,
A-E.g. sodium ethanoate in ethanoic acid. CH3COOH(aq) + NaCH3COO(aq)
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The acid is partially dissociated: CH3COOH(aq) D H+
(aq) + CH3COO-(aq)
The salt dissociates completely since it is ionic: NaCH3COO(aq) Na→ +
(aq) + CH3COO-(aq)
The mixture contains relative high concentrations of both CH3COOH(aq)
and CH3COO
-(aq) that is an Acid and it’s conjugate base
Response to added acid and base
CH3COO
-(aq) + H+(aq) D CH3COOH(aq)
Adding H+
(aq)
Adding OH
-(aq)
CH3COOH(aq) +OH
-(aq) D CH3COO-(aq) + H2O (l)
After adding the H+
(aq) or OH-(aq) an equilibrium will form (the PH
has changed slightly)
2 Basic Buffers
Are made from a mixture of
A weak base
A-Its conjugate acid, HA
E.g. NH3(aq) with NH4Cl(aq) weak base with salt of a weak base with strong acid
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=======
The Base is partially dissociated: NH3 (aq) + H2O D NH4+
(aq) + OH-(aq)
The salt dissociates completely since it is ionic: NH4Cl (aq) D NH4+
(aq) + Cl-(aq)
The mixture contains relative high concentrations of both NH3 (aq) and
NH4+
(aq) that is a base and it’s conjugate Acid.
NH4+ to react with OH
-
Adding H+ (aq)
Adding OH
-(aq)
Response to added acid and base
After adding the H+
(aq) or OH-(aq) an equilibrium will form (the PH
has changed slightly) NH3 (aq) + H+ D NH
4+(aq)
NH4+
(aq)+ OH- D NH3 (aq) + H2O
Determining the PH of a buffer solutions.
Since there is an acid dissociation taking place the pH of a buffer solution must depend on the Ka of the acid present AND
© Tom Selegue Pima Community College 2005
The equilibria that exist in the buffer are:
HA(aq) D H+
(aq) + A-(aq) dissociation weak acid partly
MA(aq) D M+
(aq) + A-(aq) dissociation salt completely
Ka = [H+
(aq) ][A-(aq)] [HA(aq)]
Equilibrium for an acid
we assume that [HA(aq)]equil = [HA(aq)]initial And [A-(aq)]equil = [MA(aq)]uinitial
pH = pK
a+ log [A
-(aq)]
[HA
(aq)]
pH = pK
a+ log [A
-(aq)]
[HA
(aq)]
Taking the negative log of both sides of the equation [H+(aq) ]= Ka x [HA] [A
-(aq)]
You need to memorize this equation it is not in the data booklet !
pH = pK
a+ log [salt]
[Acid]
pH = pK
a+ log [salt]
[Acid]
for a base
[OH- (aq) ]= Kb x [Base]
[salt(aq)] And
pOH = pK
[base]
b+ log [salt]
pOH = pK
b+ log [salt]
[base]
Example
A buffer solution can be made containing 0.600 mol dm-3 propanoic acid and 0.800 mol dm-3 sodium propanoate.
Ka, for propanoic acid is 1.3 x 10-5 mol dm-3
[H+
(aq) ] = Ka x [HA] gives us
[A
-(aq)]
[H+
(aq) ] = 1.3 x 10-5 x 0.600 mol dm-3 = 9.75 x 10-6 mol dm-3
0.800
And pH = -log(9.75 x 10-6) = -(-5.01) = 5.01
Execises
a) 0.0500 mol dm-3 methanoic acid and 0.100 mol dm-3 sodium methanoate. Ka =1.6 x 10-4 mol dm-3
b) 0.0100 mol dm-3 benzoic acid and 0.0400 mol dm-3 sodium benzoate Ka = 6.3 x 10-5 mol dm-3.
Ans = 4.8
c) 0.100 mol dm-3 ethanoic acid and 0.200 mol dm-3 sodium ethanoate (pKa ethanoic acid = 4.8)
Ans = 5.1
Calculate the pH of a solution containing:
d) 0.02 mol dm-3 ethanoic acid and 0.05 mol dm-3 sodium ethanoate. Ka = 1.7 x 10-5 mol dm-3
Ans = 5.17
e) 0.200mol dm-3 citric acid and 0.100 mol dm-3 sodium citrate. Ka = 7.24 x 10-4 mol dm-3
Ans = 2.8
f) Calculate the pH of a buffer made by dissolving 18.5g of propanoic acid, C2H5COOH and 12.0g of sodium propanoate,
C2H5COONa, in water and then making up the volume to 250cm3.
1 mole C2H5COOH weighs 74g. 1 mole C2H5COONa weighs 96g. [C2H5COOH] = 18.5 x 4
74 = 1.00mol dm-3. [C2H5COONa] = 12.0 x4
96 = 0.500 mol dm-3 Ka = 10-pKa
Ka = 10-4.87 = 1.35 x 10-5 mol dm-3 Working out
pH = 4.57
pH = pK
a+ log [salt]
[Acid]
pH = pK
a+ log [salt]
[Acid]
Calculate the pH and pOH of a solution that is 0.17M Na2H2PO4(aq) and 0.25M Na3PO4 pKa= 12.68
pH=pKa+ log[A-]/[HA]
Number of moles acid = 50 x 0.300
1000 = 0.0150 (in a total volume of 150 cm3)
[CH3COOH] = 0.0150 x 1000
150 = 0.100 mol dm-3
Number of moles CH3COONa = 100 x 0.600
1000 = 0.0600 (in a total volume of 150 cm3)
[CH3COONa] = 0.0600 x 1000
15 = 0.400 mol dmK -3
a = [CH3COO-] [H+] [CH3COOH]
1.74 x 10-5 = 0.400 x [H+] 0.100 [H+] =1.74 x 10-5 x 0.100
0.400 = 4.35 x 10-5 pH=5.36
Or make use of the
Henderson-Hasselbalch equation
pH = pK
a+ log [salt]
[Acid]
pH = pK
a+ log [salt]
[Acid]
A buffer solution was made by mixing 50.0cm3 of 0.300 mol dm-3 ethanoic acid with 100 cm3 0f 0.600 mol dm-3 sodium ethanoate.
Consider a solution that contains 1.0M of the weak acid HF and 1.0M of the salt NaF
Ka (HF)= 7.2 x 10-4
HF(aq) + H2O(l) H3O+(aq) + F-(aq)
NaF(s) Na+(aq) + F-(aq)
For the acid:
For the salt:
Keq= 0.97
The acid is dissociating into a solution that already contains 1.0M F- from
the dissociation of NaF
According to LeChatlier’s principle, this makes the acid equilibrium shift toward undissociated acid
HF is less acidic in the NaF solution!
This is called the
common ion effect
The role of a natural buffer in the control of blood pH.
• To remain healthy human blood has to be maintained at a
constant pH 7.4 (7.35-7.45).
• If the blood becomes too acidic and the pH drops ( as in the
medical condition acidosis), we have to breathe rapidly to expel more carbon dioxide.
• The main mechanism for maintaining pH is the buffering
action of several acid/base pairs e.g carbonic acid/hydrogencarbonate and H2PO4
-(aq) and HPO42-(aq), together with the buffering action of plasma proteins and
haemoglobin.
The carbonic acid/hydrogencarbonate ion buffer is the most important buffer system in the blood plasma.
Carbonic acid is a weak acid.
Hydrogencarbonate ions HCO3- are the conjugate base.
H2CO3(aq) DH+
(aq) + HCO3-(aq)
Any increases in [H+] ions in the blood are removed by the conjugate base.
To make the best buffer for your job
pH = pK
a+ log (1) = pK
a1) Pick the pH at which you need to buffer
3) Prepare your buffer using a weak acid that has a pK
aclose
to the pH at which you’re buffering
[A
-]
[HA]
= 1
2) Optimum buffering will occur when
4) To get the exact pH, adjust the ratio slightly to
raise or lower the pH
[A
-]
[HA]
5) Determine how tightly you need to control pH and choose
Note when the buffersolution contains an equal amount
of acids and salt The Henderson Hasselbalch equation = 0
pH = pK
a+ log [salt]
[Acid]
pH = pK
a+ log [salt]
[Acid]
Example
Prepare a solution that will buffer at a pH of 3.20
1) Choose HF as the weak acid, pK
a= 3.17
2) Adjust the ratio to get pH = 3.20
[F
-]
[HF]
3.20 = 3.17 + 0.03
log
[F
-]
= 0.03
[HF]
= pK
a[F
-]
[HF]
+ log
pH
Using
= 1.07
[F
-]
[HF]
3) Decide the buffer capacity you need for your job to choose
How do I make a pH 4.0 buffer? Buffer Table
Formic Acid Ka =1.8 X 10-4 pK
a =3.74
Barbituric Acid Ka = 9.8 X 10-5 pK
a = 4.01
Butanoic Acid Ka 1.52 X 10-5 pK
a = 4.82
Choose a pKa near the desired pH
Answer
pH = pKa + log (base/acid)
4.0 = 3.74 + log (base/acid)
0.26 = log (base/acid)
Factors that influence Buffers.
Buffer Capacity
As long as ratio remains virtually constant, the pH will be virtually constant This is true as long as concentrations of buffering materials (HA/A-) or (B/
BH+) are large compared with H+ or OH- added.
1 The more concentrated the components of the buffer the greater the buffer capacity
2 The PH is distinct from the buffer capacity
3 A buffer has the highest capacity when the concentrations of the components are
Dilution: Does not alter the PH but decreases the buffer capacity
