Chapter 17 – The common ion effect and buffers

Lewis Acids/Bases

Metal Cations as Acids

Before starting, lets define –log K_a = pK_a

Comparing three weak acids

HF	Ka = 6.8 X 10^-4	3.17
HC2H3O2	1.8 x 10^-5	4.74
HCN	4.9 X 10^-10	9.31

As you can see, the stronger the acid, the smaller pKa. You will have a similar trend with pK_b.

Common Ion Effect

Consider the ionization of acetic acid in water

HC₂H₃O₂ = H⁺ + C₂H₃O₂^-

We calculated before that a 1.0 X 10^-2 M solution would have [H⁺] = 4.2 X 10^-4 M and pH = 3.37

How does Le Chatelier suggest the above equilibrium would shift if NaC₂H₃O₂ was added?

How would the pH change?

Say we added enough NaC₂H₃O₂ to make the solution 1.0 X 10^-2 M in C₂H₃O₂^-. What would the pH be

HC₂H₃O₂ = H⁺ + C₂H₃O₂^-

Initial	1.0 X 10^-2	0 *	1.0 X 10^-2
Change	-x	+x	+x
Equilibrium	1.0 X 10^-2 – x	X	1.0 X 10^-2 + x

K_a = x (1.0 X 10^-2 + x) Assume x is negligible compared to 1.0 X 10^-2 M

(1.0 X 10^-2 – x )

x = 1.8 X 10^-5 = [H⁺] => pH = 4.74

So the addition of acetate, (the conjugate base of acetic acid)

a). Reduced the [H⁺]

b). Shifted the equilibrium to the left

c). Increase the pH, (as we would expect a base to do).

How would you expect the addition of NH₄Cl to change a 1.0 X 10^-2 M NH₃ solution?

a). How would the [OH^-] change?

b.) How would this equilibrium shift?

NH₃ + H₂O = NH₄⁺ + OH^-

c). How would the pH change?

(pH of the .010 M NH₃ solution was 10.63, by adding the ammonium chloride, the pH is 9.26.

The acetate ion is called a common ion of acetic acid and sodium acetate. This ion is common in the formulas of the two different substances. Ammonium is a common ion of ammonia and ammonium chloride.

A solution that is made up of a appreciable concentration of both species of a conjugate pair is called a buffer solution. Buffers resist large changes in pH. Blood is a buffered solution at a pH of 7.4. Death may result if the pH falls below 6.8 or goes over 7.8. Sea water is a buffer with a pH of 8.1-8.3 near the surface. The pH of the buffer depends on the K_a (and conjugate K_b) of the substance in it and the relative concentrations of the conjugate pair. How well it resist a change in pH (called the buffer capacity) depends on the concentrations of the pair. No buffer can resist a change in pH if an enormous amount of acid or base is added.

Consider the HC₂H₃O₂/C₂H₃O₂^- buffer. If you add a strong acid

H⁺ + C₂H₃O₂^- ® HC₂H₃O₂

If you add a strong base

OH^- + HC₂H₃O₂ ® H₂O + C₂H₃O₂^-

So a buffer can neutralize a strong acid or a strong base. We would have had the same thing happen with the NH₃/NH₄⁺ pair.

Consider any weak acid

HA = H⁺ + A^- K_a = [H⁺] [A^-]

[HA]

Lets rearrange it

[H⁺] = K_a [HA] / [A^-] now take the common Log of both sides

Log [H⁺] = log K_a + log [HA] / [A^-]

Now if you multiplied through by –1 you would get

pH = pK_a – log [HA] / [A^-] =

pH = pK_a + log [A^-] / [HA] = pK_a + log base/acid

This is the Henderson-Halsselbalch equation.

Consider if [A^-] = 10 * [HA]

Then pH = pK_a + 1

And if [A^-] = 0.10 * [HA]

Then pH = pK_a - 1

What this shows is the pH of a mixture of a conjugate pair of acid and base will depend more on the pK_a than on the relative amount of the conjugate pair.