Buffers

4/13/2011

Definitions

• Weak Acids: all proton donors that are in equilibrium

• Conjugate Bases: the ions that are left over after a weak acid loses it H+ ions (protons)

• Buffer: – a solution containing a mix of weak acid and the

salt of its conjugate base– A solution that can absorb added acids and bases

w/o a large pH change

The weak acid rxn

• HA H+ + A-

• In a pure acid, the [H+] and [A-] occur in equal amounts

• In any solution, addition of H+ will decrease A- and vice versa

• The conjugate base of HA is A- • Notice that if [A-] goes up, the H+ will go down.

The solution will become more basic.

Salt of a conjugate base

• A “salt” is defined as the product of an acid and a base

• A salt of a conjugate base is the salt produced when a weak acid is neutralized

Examples: Weak acid Salt of Conjugate Base HF NaF HNO2 LiNO2

CH3COOH KCH3CO3

H2SO3 NaHSO3

H2CO3 LiHCO3

To identify the salt:1. Take any weak acid2. Remove the H+

3. Replace with any other positive ion

NaAH-

Weak AcidConjugate BaseSalt of Conjugate Base

Non-Buffered Solutions

• If a strong acid is added to pure water, the pH changes radically

• Original pH = 7• Imagine 1 mL of 1

Molar HCl (1 x 10-3 moles) added to 1 Liter of water

1 x 10-3 HCl

[H+]= 1x10-3 Cl- (spectator)

pH 7 pH 3

Buffered Solutions

• Add roughly equal amounts of weak acid and conjugate base salt

• An equilibrium is set up

• And then altered

HA NaA

HAH+

A-

Na+

A-

HA H+ + A-H+ + A-

(The addition of the A- lowers the H+)

Buffers in ActionThe Buffer solution now has both an acid and a base in it.

Addition of an acid will add more H+, which will be absorbed (neutralized) by the base.

If a base is added, it will react with the H+ , shifting the equilibrium to the right, reducing HA and increasing A-

HAH+

A-

Na+

A-

HA H+ + A-

Acid

H+

Base

OH-

H2O

Buffer CalculationsImagine 1 liter of a buffer with 0.5 moles HA and 0.5 moles NaAPretend (for now) that Ka = 1 x 10-7

Ka = [H+] [A-] = (X )[0.5] [HA] [0.5]So…with equal moles of weak acid and conjugate base, H+ = Ka = 1 x 10-7

And pH = 7

Like before, add 1 mL of 1 molar HCl.(you are adding .001 moles acid)

The equilibrium will change…And so will the pH…

Ka = [H+] [A-] = (X )[0.499] [HA] [0.501]

The H+ has increased and = 1.004 x 10-7

And pH = 6.998

HA H+ + A-H+ + A-

Original 0.5 mol 1 x 10-7 0.5 mol

Add Acid + 0.001 mols

Increase HA + 0.001 - 0.001Decrease A-

New Values 0.501 mol X 0.499 mol

pH 7 pH 6.998

Practice calculations

• What would be the pH of a solution of 0.2 M HF and 0.2 M NaF if the Ka = 1.6 x 10-5?

• What would be the pH if the above solution was diluted by ½ (so each solute is 0.1 M)?

• The Ka for Carbonic acid (H2CO3) is 4.5 x 10-7 (only the first H+ is ionzed). If the pH of a solution is 7.4, and the concentration of HCO3

1- is 0.22 M, what is the concentration of H2CO3?

Buffers in the real world

• Buffers are important in the biological world.– (imagine what could happen to a fish tank if the

water was not buffered)• Your blood is strongly buffered to control the

amount of CO2 in the blood.– CO2(aq) + H2O(l) H2CO3(aq)H+

(aq) + HCO3-(aq)

– Your blood measure pH instead of CO2(aq) – If your pH gets too low, your breathing rate goes

up

Summary of Buffers

• They are made of weak acids and conjugate bases, and are in equilibrium with H+

• Because they contain both base and acid, they can absorb added bases and acids with only a very small change in pH

• Calculations of acidity can be made if the Ka, acid and base concentrations are known.