Composition and Action of Buffered Solutions

Learning Goal 21
Describe how a buffer of a certain pH is made, and how it works to control pH.

Buffered solutions or buffers are solutions which resist a change in pH when small amounts of acid or base are added.

Buffers contain an acidic species to neutralize OH- ions and a basic species to neutralize H+ ions. However, these two species must be able to coexist in a solution, without completely neutralizing each other. Buffers are therefore made of weak acid-base conjugate pairs, such as HC2H3O2 and C2H3O2-.

If we have a weak acid, HX, and its conjugate base, X-, the following equilibrium occurs:

HX (aq) H+ (aq) + X- (aq)

For this reaction,

Ka = { [H+][X-] / [HX] }

If an amount of H+ ions are added, the above reaction will shift to the left. This will cause the [H+] to decrease, to close to what it was before, and thus the pH will stay fairly constant.

If OH- ions are added they will remove H+ ions to form water, thus increasing the pH. However, the equilibrium reaction will shift to the right as H+ ions are removed. The [H+] will therefore remain fairly constant, as will the pH.

The most effective buffering solutions are those which have similar concentrations of HX and X-, because then the buffer has the capacity to absorb both acid and base, with the same effectiveness in either direction.

When choosing an appropriate conjugate acid-base pair to form a buffer at a specific pH, the most effective buffers have the desired pH within 1.0 of the conjugate acid's pKa.


Ways to make a buffer


Buffering Capacity and pH

The buffering capacity is the amount of acid or base a buffer can accept without the pH changing appreciably. The greater the amounts of the conjugate acid-base pair, the more resistant they are to change in pH.

If we solve the acid-dissociation-constant expression for [H+] we get:

[H+] = Ka { [HX] / [X-] },

assuming that the amount of added acid or base is less than 5% of the conjugate acid / base molarity.

We can use this to determine the pH of a buffer:

We first take the negative log of both sides:

-log [H+] = -log [Ka { [HX] / [X-] }] = -log Ka - log { [HX] / [X-] }

We know -log [H+] = pH and -log Ka = pKa.

pH = pKa - log { [HX] / [X-] } = pKa + log { [X-] / [HX] }

In general:

pH = pKa + log { [base] / [acid] }

This is known as the Henderson-Hasselbalch equation.


Additions of Acids or Bases to Buffers

Learning Goal 22
Calculate the change in pH of a simple buffer solution, whose composition is known, caused by adding small amounts of strong acid or strong base.

EXAMPLE

A 1 L solution containing 0.100 moles of HC2H3O2 and 0.100 moles of C2H3O2 forms a buffer solution, pH = 4.74. What is the pH after 0.020 moles of NaOH are added?

STEP 1: Stoichiometry calculation:

The OH- will react completely with the weak acid HC2H3O2:

HC2H3O2 (aq) + OH- (aq) H2O (l) + C2H3O2- (aq)

HC2H3O2 OH- C2H3O2-
INITIAL
0.100 mol
0.020 mol
0.100 mol
CHANGE
-0.020 mol
-0.020 mol
+0.020 mol
FINAL
0.080 mol
0 mol
0.120 mol

STEP 2: Equilibrium calculation:

The solution contains the weak acid and its conjugate base. We shall therefore use their equilibrium equation, and create an I.C.E. table :

HC2H3O2 (aq) H+ (aq) + C2H3O2- (aq)

HC2H3O2 H+ C2H3O2-
Initial 0.080 M 0 0.120 M
Change -x M +x M +x M
Equilibrium (0.080 - x) M x M (0.120 + x) M

Ka = { [C2H3O2-][H+] / [HC2H3O2] } = { (0.120 + x)(x) / (0.080 - x) } = 1.8 x 10-5

1.8 x 10-5 = { (0.120) x / (0.080) }

x = [H+] = 1.2 x 10-5 M

pH = -log (1.2 x 10-5) = 4.92


Applications of Buffer Calculations

In many systems, such as biological applications, there is extreme sensitivity to minute pH changes. To study these systems, the pH must be controlled by a buffer that is effective within given limits. How does one determine what the buffer concentration is?

Learning Goal 23
Calculate the specific amounts of species necessary to make the buffer effective within given requirements.

EXAMPLE

What must the minimum concentration of HC2H3O3 be in a one litre buffer solution of HC2H3O3-C2H3O3- (pH = 4.74) if the pH changes by less than 0.1 if 0.050 moles of HCl are added?

STEP 1: Stoichiometry calculation:

The amount of HCl added will react completely with the conjugate base, C2H3O3-:

C2H3O3- + H+ HC2H3O3

The pH of the buffered solution, 4.74, matches the pKa of acetic acid, so from the Henderson-Hasselbach equation, we know that there are originally equal concentrations of acid and conjugate base, which we denote as x:

For a complete derivation

HC2H3O3 H+ C2H3O3-
INITIAL
x mol
0.050 mol
x mol
CHANGE
+0.050 mol
-0.050 mol
-0.050 mol
FINAL
(x+0.050) mol
0 mol
(x-0.050) mol

Because there is one litre of solution, [H2H3O3] = (x + 0.050)M and [C2H3O3-] = (x - 0.050)M.

STEP 2: Calculate the equilibrium concentration that will produce the desired pH:

Having added acid, the pH will decrease. However, the tolerance given is 0.1, so the lowest pH that the final solution can have is 4.64.

The hydronium concentration is thus:

[H+] = 10-pH

[H+] = 10-4.64 = 2.29 x 10-5 M

STEP 3: Equilibrium calculation:

I.C.E. table:

HC2H3O3 H+ + C2H3O3-

HC2H3O3 H+ C2H3O3-
INITIAL
(x+0.050)M
0M
(x-0.050)M
CHANGE
-2.29 x 10-5M
+2.29 x 10-5M
+2.29 x 10-5M
EQUILIBRIUM
(x+0.050)M
2.29 x 10-5M
(x-0.050)M

Ka = { [H+][C2H3O2-] } / [HC2H3O2] = { (2.29x10-5)(x-0.050) } / (x+0.050) = 1.8x10-5

x = [HC2H3O2] = 0.417M

Thus, the molarity of both [HC2H3O2] and [C2H3O2-] must be equal to or greater than 0.417M to produce the required effectiveness.