Buffer Solutions and the Common Ion effect

Buffer solutions and common ion effect

A buffer solution resists (or buffers) a change in its pH. That is, we can add a small amount of an acid or base to a buffer solution and the pH will change very little.

How to calculate pH of buffer solution containing both acid and conjugate base? Dissociation constant definition 1.1 can be rearranged into or

(note that due to sign change [A-] was moved to nominator).
This is so called Henderson-Hasselbalch equation (or buffer equation). It can be used for pH calculation of solution containing pair of acid and conjugate base - like HA/A-, HA-/A2- or B+/BOH. For solutions of weak bases sometimes it s more convenient to use equation in the form
15.3

Two common types of buffer solutions are :

(1) a weak acid together with a salt of the same acid with a strong base. These are called Acid buffers e.g. CH3COOH + CH3COONa.

(2) a weak base and its salt with a strong acid. These are called Basic buffers. e.g.NH4OH + NH4Cl.

Let us illustrate buffer action by taking example of a common buffer system consisting of solution of acetic acid and sodium acetate (CH3COOH/CH3COONa).
CH3COOH --- H+ + CH3COO–
CH3COONa ---- Na+ + CH3COO-

since the salt is completely ionised, it provides the common ions CH3COO– in excess. The common ion effect suppresses the ionisation of acetic acid. This reduces the concentration of H+ ions which means that pH of the solution is raised. Thus, a 0.1 M acetic acid solution has a pH of 2.87 but a solution of 0.1 M acetic acid and 0.1 M sodium acetate has a pH of 4.74. Thus 4.74 is the pH of the buffer. On addition of 0.01 mole NaOH the pH changes from 4.74 to 4.83, while on the addition of 0.01 mole HCl the pH changes from 4.74 to 4.66. Obviously the buffer solution maintains fairly constant pH and the changes in pH could be described as marginal.

Buffer Action:

A buffer solution containing equimolar amounts (0.10 M) of acetic acid and sodium acetate has pH 4.74. Now we proceed to discuss how the addition of a small amount of HCl or NaOH to the buffer solution affects its pH. The pH of the buffer is governed by the equilibrium

CH3COOH --- CH3COO– + H+

The buffer solution has a large excess of CH3COO– ions produced by complete ionisation of sodium acetate,

CH3COONa ---CH3COO- + Na+

(1) Addition of HCl.

Upon the addition of HCl, the increase of H+ ions is counteracted by association with the excess of acetate ions to form unionised CH3COOH. Thus the added H+ ions are neutralized and the pH of the buffer solution remains virtually unchanged. However owing to the increased concentration of CH3COOH, the equilibrium shifts slightly to the right to increase H+ ions. This explains the marginal increase of pH of the buffer solution on addition of HCl.

(2) Addition of NaOH.

When NaOH is added to the buffer solution, the additional OH– ions combine with H+ ions of the buffer to form water molecules. As a result the equilibrium (1) shifts to the right to produce more and more H+ ions till practically all the excess OH– ions are neutralised and the original buffer pH restored. However, a new equilibrium system is set up in which [CH3COOH] is lower than it was in the original buffer. Consequently [H+] is also slightly less and pH slightly higher than the buffer pH values Operation of a Basic buffer as NH4OH/NH4Cl can also be explained on the same lines as of an acid buffer upon addition of HCl the H+ ions combine with OH– ions of the buffer to form water molecules. The equilibrium,

NH4OH--- NH4+ + OH- is shifted to the right till all the additional H+ ions are neutralised and the original buffer pH restored. When NaOH is added to the buffer solution, OH– ions associate with excess of NH4+ ions to form unassociated NH4OH. Thus the pH of the buffer is maintained approximately constant.

Problems

What is the pH of a solution containing 0.02 M HA and 0.01 M A-? pKa of HA = 5.0.

Solution

pH = pKa + log [salt] / [acid]

= 5 + log 0.01/0.02 = 4.7

Problem
a) A solution was prepared by dissolving 0.02 moles of acetic acid (HOAc; pKa = 4.8) in water to give 1 liter of solution. What is the pH?
b) To this solution was then added 0.008 moles of concentrated sodium hydroxide (NaOH). What is the new pH? (In this problem, you may ignore changes in volume due to the addition of NaOH).
c) An additional 0.012 moles of NaOH is then added. What is the pH?

Solution
a) This is a class 1 problem:

HOAc

H+
+
OAc- pKa = 4.8
Ka = 1.6 x10-5 M
Initial Conc
0.02 M

~0

0

Change
-X

+X

+X

Net
0.02-X

X

Ka = = = 1.6 x10-5
[HOAc]initial > 100 x Ka, so... X can be ignored in the denominator.
X2/0.02 = 1.6 x10-5 ; X2 = 3.2 x10-7
X = 5.6 x10-4 M = [H+] pH = -log(5.6 x10-4) = 3.25
b) NaOH is a strong base. It will react nearly quantitatively with HOAc to produce OAc-.
HOAc + OH- OAc- + H2O
0.008 moles HOAc + 0.008 moles OH- 0.008 moles OAc-
Amount of HOAc remaining = 0.02 - 0.008 = 0.012 moles. Thus,
[HOAc] = 0.012 M and [OAc-] = 0.008 M
Since both [HOAc] and [OAc-] are present, this is a class 3 problem. pH = pKa + log
= 4.8 + log = 4.8 + log(0.67)
= 4.8 - 0.17 pH = 4.63
c) After the addition of another 0.012 moles of NaOH, all of the HOAc has been converted to NaOAc. thus, [OAc-] = 0.02 M. Since only the "base form" of the HOAc is now present, this is a class 2 problem:

OAc-
+
H2O

HOAc
+
OH-

Keq =
Initial Conc
0.02 M

[H2O]

~0

~0

Change
-X

-X

+X

+X

Net
0.02-X

[H2O]

X

X

So,
Kw/Ka = = =
Neglecting the denominator gives,
X2/0.02 = 6.25 x10-10
X2 = 12.5 x10-12
X = 3.5 x10-6 = [OH-]
[H+] = 10-14/(3.5 x10-16) = 2.9 x10-9 pH = 8.54