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Buffers and the Uses of Buffer Solutions

Key Concepts

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Effect of a Buffer

In a typical experiment to determine the effectiveness of buffer solutions, a student adds 0.10 mol of H+ (HCl(aq)) to 1.0 L of each of the following:

The pH of the solutions are measured before and after the addition of H+.
The results of the experiment are shown below:

at 25oC water acetic acid acetic acid-
acetate buffer
pH before addition
of 0.10 mol L-1 H+
7.0 2.37 4.74

pH after addition
of 0.10 mol L-1 H+
1.0 1.00 4.66

change in pH as a result
of adding 0.10 mol L-1 H+
6.0 1.37 0.08

Adding acid, H+, decreases the pH in all the sample solutions, but by different amounts:

WHY?

First, let's look at what is happening in the unbuffered acetic acid solution.
(we've included the mathematics behind the logic in the boxes on the right hand side of the page, ignore them if you find the maths distracting)

In 1 L of 1.0 mol L-1 acetic acid, the acetic acid molecules, CH3COOH, are in equilibrium with acetate ions, CH3COO-, and hydrogen ions, H+:

CH3COOH CH3COO- + H+

but because acetic acid is a weak acid (Ka is small) it dissociates only to a very small extent so that in the 1 mol L-1 solution there are:

  • ≈ 1.0 mol L-1 CH3COOH (undissociated acid molecules)
  • very small amounts of CH3COO- and H+ ions
    [CH3COO-] = [H+] = 4.24 × 10-3 mol L-1
    (acetate and hydrogen ion concentration calculations are shown in the box on the right)
CH3COOH CH3COO- + H+
Ka = 1.8 x 10-5
[CH3COOH] = 1.0 mol L-1
Ka = [CH3COO-][H+]
[CH3COOH]
Since [H+] = [CH3COO-]
Ka = [H+][H+]
[CH3COOH]
1.8 × 10-5 = [H+][H+]
[1.0]
1.8 × 10-5 = [H+]2
√1.8 × 10-5 = √[H+]2
[H+] = 4.2 × 10-3 mol L-1

When acid, H+, is added to the acetic acid, the equilibrium position should shift to the left by Le Chatelier's Principle in order to consume the additional H+.

BUT, we added 0.10 mol of H+, and there are only 4.2 × 10-3 moles of CH3COO- that are available to react!

That means there is going to be a lot of H+ in excess !

There will be a significant increase in the concentration of H+, [H+], in solution.

Which means that the pH of the acetic acid will fall
(since pH = -log10[H+])
(pH calcuations are shown in the box on the right)

In the unbuffered acetic acid, the pH falls when acid is added to it because there is an excess of H+, or, in other words, because the acetate ion, CH3COO-, is the limiting reagent.

Unbuffered Acetic Acid
concentration (mol L-1)
before 0.1 mol H+ addedafter 0.1 mol H+ added
H+4.2 × 10-3 ≈ 0.1
CH3COO-4.2 × 10-3 ≈ 0
CH3COOH1.0 ≈ 1.0
pH before H+ added = 2.4
and pH after H+ added = 1

What would happen if you could make an acetic acid solution in which the acetate ion was no longer the limiting reagent, that is, an acetic acid solution in which the acetate ions are in excess?

Wouldn't that mean that if you added acid, H+, to the solution then the extra H+ would react with the CH3COO- in excess to produce CH3COOH ?

Wouldn't the concentration of hydrogen ions in solution then remain about the same ?

So wouldn't the pH of the solution stay about the same ?

Yes! And this solution would be called an acetic acid-acetate buffer solution !

The acetic acid-acetate buffer used in the experiment described above is a mixture that contains:

  • ≈ 1.0 mol L-1 CH3COOH (undissociated acetic acid molecules)
  • ≈ 1.0 mol L-1 CH3COO- (which was added in the form of soluble sodium acetate)
  • and a small amount of H+ (from the partial dissociation of CH3COOH)

When acid, H+, is added to this acetic acid-acetate buffer, the additional H+ can react with the CH3COO- which is now in excess:

H+ + CH3COO- → CH3COOH

The concentration of undissociated CH3COOH molecules increases.

Because acetic acid is a weak acid, it dissociates to a small extent to produce hydrogen ions:

CH3COOH H+ + CH3COO-

So the concentration of hydrogen ions in solution, H+(aq), will increase slightly.

Since pH = -log10[H+] the pH will decrease slightly.
The pH actually falls from 4.74 ( ≈ 4.7) to 4.66 ( ≈ 4.7)
([H+] and pH calculations are shown in the box on the right)

Buffered Acetic Acid
Ka = [H+][CH3COO-]
[CH3COOH]
rearrange equation to find [H+]:
[H+] = Ka[CH3COOH]
[CH3COO-]
Ka = 1.8 × 10-5 at 25°C :
[H+] = 1.8 × 10-5[CH3COOH]
[CH3COO-]
Initially, [CH3COOH] = 1.0 mol L-1 and [CH3COO-] = 1.0 mol L-1
[H+] = 1.8 × 10-5[1.0]
[1.0]
[H+] = 1.8 × 10-5 mol L-1
and pH = 4.74
Add 0.1 mol L-1 H+
new [CH3COO-]=1.0-0.1=0.9 M
new [CH3COOH]=1.0+0.1=1.1 M
new [H+] = 1.8 × 10-5[1.1]
[0.9]
new [H+] = 2.2 × 10-5 mol L-1
and pH = 4.66
Adding 0.1 M H+ to 1 M acetic acid-acetate buffer has little effect on the pH of the solution.

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Composition of a Buffer

We have seen above that adding acetate ions, CH3COO-, to acetic acid, CH3COOH, produces a buffer solution, that is, a solution that resists changes to pH because:

  1. Excess acetate ions consume the added H+ to produce more acetic acid molecules:
    CH3COO- + H+ → CH3COOH

    and

  2. acetic acid dissociates only to a very small extent, so very few H+ ions are produced:
    CH3COOH H+ + CH3COO-

The buffer solution resists changes to pH because the both the undissociated acid molecules (CH3COOH) and its conjugate base (CH3COO-) are present in appreciable amounts in the solution.

Can we produce other buffer solutions by adding the conjugate base to the acid?

Acid-Conjugate Base Buffer

Compare the effect of adding the salt of the conjugate base (A-) of acids of different strengths to solutions of the acid (HA):

[HA] = 0.1 mol L-1 Strength of Acid
strong acid weak acid
acid dissociation equation:
fully dissociated
HA → H+ + A-
partially dissociated
HA H+ + A-
before addition of A- [HA] = 0 [HA] bit less than 0.1
after addition of A- [HA] = 0 [HA] increases a bit Le Chatelier's principle

A strong acid cannot be used to make a buffer because the strong acid completely dissociates, there are no undissociated acid molecules in solution.
Adding H+ to a strong acid results in an increase in H+ concentration and therefore a fall in pH because the H+ ions will not be consumed by the conjugate base to produce undissociated acid molecules.

Strong acids unsuitable for use in buffers include:
HCl, HNO3, HBr, HI, HClO4

A weak acid can be used to make a buffer solution because it is only partially dissociated, that is, undissociated acid molecules exist in the solution.
Weak acids are suitable for use in buffers when enough of the salt of the
conjugate base can be dissolved in solution so that the concentration
of the undissociated acid is the same as the concentration its conjugate base.

Buffer solutions can be produced using the following weak acids and the salts of their conjugate bases:

Example of a BufferBuffer Composition
Weak AcidSalt of Conjugate Base
acetic acid-acetate bufferCH3COOHCH3COONa
carbonic acid-bicarbonate bufferH2CO3NaHCO3
dihydrogen phosphate-hydrogen phospahte bufferNaH2PO4Na2HPO4
bicarbonate-carbonate bufferNaHCO3Na2CO3

Note that sodium or potassium salts are often used because the salts of Group 1 metals tend to be soluble at 25oC in aqueous solutions.*

Base - Conjugate Acid Buffer

Buffer solutions can also be produced using a weak base and the salt of its conjugate acid.

A base, B-, accepts a proton from water to produce hydroxide ions, OH- :

B- + H2O(l) HB(aq) + OH-(aq)

The other species produced, HB, is called the conjugate acid.

Hydroxides of Group 1 or Group 2 metals are strong bases, that is, the only species in the aqueous solution are metal cations and hydroxide ions, there are no "metal hydroxide molecules" in solution. Hydroxides of Group 1 and Group 2 metals can not be used to make a buffer.

Weak bases, such as ammonia, can be used to make a buffer solution since a solution of weak base will contain molecules of the unreacted base molecules as well as molecules of its conjugate acid.

Kb for ammonia is 1.8 × 10-5 which is very small, so ammonia is a weak base.
The ammonium ion, NH4+, is the conjugate acid of ammonia, NH3.
The ammonia-ammonium ion buffer solution contains about equal concentrations of unreacted ammonia (NH3) and a soluble salt of its conjugate acid (NH4+).

When a base, OH-, is added to the ammonia-ammonium ion buffer:

  1. ammonium ion which is in excess reacts with hydroxide ions to produce ammonia:
    NH4+ + OH- → NH3 + H2O
  2. ammonia molecules react with water to a very small extent to produce a small amount of hydroxide ions:
    NH3 + H2O NH4+ + OH-

The increased NH3 concentration produces a very small increase in OH- concentration.
Remember that pOH = -log10[OH-] and for aqueous solutions at 25° pH = 14 - pOH
so, since the [OH-] increases slightly, pOH decreases slightly, therefore, pH increases, but only very slightly.

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Buffer Example: Carbonic Acid - Hydrogen Carbonate Ion Buffer

A buffer solution can be made using carbonic acid and hydrogen carbonate ions (also known as bicarbonate ions) because:

  1. carbonic acid, H2CO3, is a weak acid, and,
  2. hydrogen carbonate ion (or bicarbonate ion), HCO3-, is the conjugate base of carbonic acid

The carbonic acid - hydrogen carbonate ion buffer system is used to regulate pH in chemical systems exposed to carbon dioxide gas.

Cell respiration in your body produces carbon dioxide gas, CO2(g).
CO2(g) dissolves in water to form carbonic acid, H2CO3.
Carbonic acid dissociates to form hydrogen ions, H+.
If your blood wasn't buffered, the pH of your blood would steadily decrease as it became more and more acidic!

Similarly, the water in swimming pools is exposed to carbon dioxide gas in the atmosphere.
Carbon dioxide gas dissolves in the water to form carbonic acid.
The carbonic acid dissociates to form hydrogen ions.
So without buffering the pH of the pool water would decrease as it became more acidic.

In an aqueous system that is not buffered and exposed to carbon dioxide gas:

The carbonic acid - hydrogen carbonate ion buffer can act to reduce the effect of dissolving CO2(g) on the pH of the solution.
The carbonic acid - hydrogen carbonate ion buffer is produced by mixing together the weak acid, carbonic acid (H2CO3), and a salt of its conjugate base such as sodium hydrogen carbonate (NaHCO3).
Enough sodium hydrogen carbonate (NaHCO3) is added so that the concentration of hydrogen carbonate ions (HCO3-) in solution is about the same as the concentration of the undissociated carbonic acid molecules in solution (H2CO3):
In the buffer solution: [HCO3-] = [H2CO3]
By adding the conjugate base (HCO3-) in appreciable quantities, we introduce a way in which the addition of significant amounts of H+ can be consumed so that the pH of the solution can remain about the same.

In an aqueous system that is buffered by carbonic acid and hydrogen carbonate ions and exposed to carbon dioxide gas:

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Effective pH of a Buffer

A buffer is most effective when the concentration of the weak acid is the same as the concentration of its conjugate base, or, when the concentration of the weak base is the same as the concentration of its conjugate acid.

Consider a weak acid, HA, dissolving in water. The following equilibrium is established:

HA H+ + A-

and the equilibrium expression, Ka, for this system is:

Ka = [H+][A-]
[HA]

the expression can be rearranged to find the concentration of H+ as shown below:

Ka × [HA]
[A-]
= [H+]

Remember that pH = -log10[H+]
Therefore the pH of the solution can be calculated:

-log10[H+] = -log10( Ka × [HA]
[A-]
)

pH = -log10( Ka × [HA]
[A-]
)

Remember that Ka is a constant for a particular acid at a particular temperature
Rearrange the equation to separate out the constant, Ka:

pH = -log10Ka -log10( [HA]
[A-]
)

A buffer is most effective when enough of the salt of the conjugate base has been added to the solution of the weak acid so that [HA] = [A-], so let y = [HA] = [A-]

pH = -log10Ka -log10( y
y
)
pH = -log10Ka -log10 (1)
pH = -log10Ka  

This equation can then be used to determine the pH at which a buffer system will be most effective:

buffer Ka
(25°C)
-log10Ka = effective
pH
[weak acid] = [conjugate base]
acetic acid (ethanoic acid)
and acetate ion (ethanoate ion)
[ CH3COOH ] = [ CH3COO- ] 1.8 × 10-5 -log101.8 × 10-5 = 4.7
carbonic acid and
hydrogen carbonate ion
(bicarbonate ion)
[ H2CO3 ] = [ HCO3- ] 4.2 × 10-7 -log104.2 × 10-7 = 6.4
dihydrogen phosphate ion
and hydrogen phosphate ion
[ H2PO4- ] = [ HPO42- ] 6.3 × 10-8 -log106.3 × 10-8 = 7.2
hydrogen carbonate ion
and carbonate ion
[ HCO3- ] = [ CO32- ] 5.6 × 10-11 -log105.6 × 10-11 = 10.3

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* Also, the ions of Group 1 metals (alkali metal ions) do not hydrolyse, that is, do not react with water.