Chapter 17

Buffers

Buffered solutions A solution that resists a change in pH. Buffers are:

– A solution that contains a weak acid-weak base conjugate pair.

– Often prepared by mixing a weak acid, or a weak base, with a salt of that acid or base.

Buffered solutions

The solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base.

We can make a buffer of any pH by varying the concentrations of these solutions.

What is the pH of a buffer that is 0.12M in lactic acid, HC3H5O3 , and 0.10M in sodium lactate? For lactic acid,

K = 1.4 x 10 -4

I 0.12 M 0 0.10 M

C -x +x +x

E 0.12 – x +x 0.10 + x

HC3H5O3 ↔ H+ + C3H5O3-1

[ H +1] [C3H5O3-1]

Ka = [HC3H5O3]

1.4 x 10-4 = (x) (0.10 – x)

(0.12 – x)

Because of the small Ka and the presence of the common ion, it is expected that x will be small relative to 0.12 or 0.10M.

x = 1.7 x 10 -4 M therefore [H+1] = 1.7 x 10 -4 M

pH = -log (1.7 x 10 -4 M) = 3.77

Use Henderson-Hasselbach Instead

[base]

pH = pKa + log [acid]

[ 0.10]

pH = -log (1.4 x 10-4) + log [0.12]

pH = 3.85 + (-0.08) = 3.77

Practice ProblemCalculate the pH of a buffer composed of 0.12M benzoic acid (HC7H5O2)and 0.20M sodium benzoate. Ka = 6.3 x 10-5

ans: 4.42

Remember the Aqueous Equilibrium Constants are located in Appendix D of your textbook.

Buffered Solutions Buffers resist changes in pH because

they contain both an acidic species to neutralize OH-1 ions and a basic one to neutralize H+1 ions.

It is important that the acidic and basic species of the buffer do not consume each other through a neutralization reaction.

Buffered Solutions HA H+ + A-

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

Buffers most effectively resist a change in pH in either direction when the concentrations of HA and A- are about the same. When [HA] equals [A-] then [H+] equals Ka.

Scientists usually try to select a buffer whose acid form has a pKa close to the desired pH.

General equation Ka = [H+] [A-]

[HA] so [H+] = Ka [HA]

[A-] The [H+] depends on the ratio [HA]/[A-] Take the negative log of both sides pH = -log(Ka [HA]/[A-])

pH = -log(Ka)-log([HA]/[A-])

pH = pKa + log([A-]/[HA])

This is called the Henderson-Hasselbach equation

pH = pKa + log([A-]/[HA])

pH = pKa + log(base/acid)

Try an Acid Calculate the pH of the following mixture: Prob. #1:

0.75 M lactic acid (HC3H5O3) and 0.25 M

sodium lactate (Ka = 1.4 x 10-4)

Answer: pH = -log(1.4x10-4) + log[0.25/0.75] pH = 3.38

Now Try a Base Calculate the pH of the following mixture: 0.25 M NH3 and 0.40 M NH4Cl (Kb = 1.8 x 10-5) Answer: NH3 NH4

+1 + OH-1 (NH3 is B, NH4

+1 is HB+1) pOH = pKb + log[ HB+1 / B ] pOH = -log(1.8x10-5) + log[0.4/0.25] pOH – 4.94 now convert to pH

Buffering Capacity

This is the amount of acid or base the buffer can neutralize before the pH begins to change to an appreciable degree.

Buffering Capacity The buffering capacity, that is, the

effectiveness of the buffer solution, depends on the amount of acid and conjugate base from which the buffer is made.

The larger the amount the greater the buffering capacity.

In general, a buffer system can be represented as salt-acid or conjugate base-acid.

**Buffer capacity** The pH of a buffered solution is

determined by the ratio [A-]/[HA]. As long as this doesn’t change much

the pH won’t change much. The more concentrated these two are

the more H+ and OH- the solution will be able to absorb.

Larger concentrations means bigger buffer capacity.

Buffer Capacity

Calculate the change in pH that occurs when 0.010 mol of HCl(g) is added to 1.0L of each of the following:

5.00 M HAc and 5.00 M NaAc 0.050 M HAc and 0.050 M NaAc Ka= 1.8x10-5

Buffer capacity

The best buffers have a ratio [A-]/[HA] = 1

This is most resistant to change True when [A-] = [HA] Make pH = pKa (since log1=0)

Addition of Strong Acids or Bases to Buffers

See pg. 648 B&L for explanation and examples.

Adding a strong acid or base Do the stoichiometry first. A strong base will take protons from the

weak acid reducing [HA]0

A strong acid will add its proton to the anion of the salt reducing [A-]0

Then do the equilibrium problem.

B & L pg 649

Try the Buffer worksheet from the “Chang” chemistry text.

Prove they’re buffers

What would the pH be if 0.020 mol of HCl is added to 1.0 L of both of the preceding solutions.

What would the pH be if 0.050 mol of solid NaOH is added to each of the proceeding.

Thanks to Mr. Green