© Silvia Kolchens
Pima Community College
In this module you will learn:
Example: Blood buffer system
The normal pH of human blood is 7.4. Some people suffer from alkalosis when experiencing severe anxiety. Alkalosis is a condition in which the pH of the blood is too high. The situation can easily be improved when the affected person breathes into a paper bag. The opposite condition - a blood pH lower than 7.4 is called acidosis.
How does blood then maintain its normal pH?
The pH of the blood is maintained through a conjugated acid/base pair which includes carbonic acid and the bicarbonate anion. When CO2 is produced through metabolism, some of the CO2 is soluble in blood -or its aqueous component, water. The rest is removed from the body through exhalation. Dissolution of CO2 in water forms carbonic acid (H2CO3), a weak acid, which then dissociates into hydronium ions and bicarbonate ions and the following equilibrium is established:
CO2(g) + H2O(l) ó H2CO3(aq)
H2CO3(aq) + H2O(l) ó H3O+(aq) + HCO3-(aq)
An excess of carbon dioxide will shift the equilibrium to the right and produces hydronium ions, which lowers the pH, and cure the condition of alkalosis.
In case the pH of blood drops below the normal range, bicarbonate ions reacts to form carbonic acid, which then decomposes into water and carbondioxide, which is exhaled.
What is a buffer and how does it work? - The concept
A buffer must contain an acid component and a basic component. Typically a buffer may either consist of an acid and its conjugate base or of a base and its conjugate acid. In either case it contains a "common" ion, i.e. an ion that is common to the acid/base equilibrium:
CH3COOH + H2O ó CH3COO- + H3O+
Acid conjugated base
NH3 + H2O ó NH4+ + + OH-
Base conjugated acid
In the discussion that follows below, we will discuss the acetic acid/acetate buffer system.
CH3COOH + H2O ó CH3COO- + H3O+
When a small amount of strong acid is added to the buffer solution, it will react with the conjugated base (CH3COO-), and the equilibrium will shift to the left, increasing concentration of CH3COOH. Consequently, the pH of the buffer solution will decrease, but only slightly.
Likewise, if a small amount of a strong base is added, it will shift the equilibrium to the right, and the pH of the buffer solution will slightly increase.
How can we calculate the pH of a buffer solution?
Example: A buffer solution consists of 1.0 M CH3COOH and 1.5 M NaCH3COO
CH3COOH + H2O ó
CH3COO- + H3O+
We assume that x<< initial concentrations and simplify so that:
1.0-x ~ 1.0 [CH3COOH] and 1.5+x ~ 1.5 for [CH3COO-]
The Ka value for acetic acid is 1.8x10-5 (table on p. 799). Now we can solve for x and obtain:
Solve for x:
Now we can calculate the pH:
pH = - log [H3O+] = -log 1.20x10-5 = 4.93
What is the pH of the buffer solution after some base has been added?
Example: 0.4 g NaOH are added to 1 L buffer solution
First, we need to calculate the molar concentration of NaOH in the buffer solution:
Then we can set up an iCe table to determine the new concentrations
of acetic acid and acetate:
The equilibrium for this reaction is on the right. Now we can use the equilibrium concentration to set up a new iCe table and calculate the pH of the solution:
We make the assumption that x<< initial concentration and at equilibrium we obtain
CH3COOH = 1.01
Now we can calculate the pH for the buffer solution just as we did above:
Solve for x=[H3O+]:
pH = 4.93
Indeed, the pH did not change. It will change somewhat if the concentration of NaOH is increased. Compare this value to the pH of an unbuffered solution of 0.01 M NaOH:
[OH-] of 0.01 M NaOH = 10-2
pOH = 2
pH = 12!!!
Exercise: HF/F- buffer system
a) Calculate the pH of a buffer solution containing 0.5 M HF and 0.45 M F-:
b) What is the pH of the solution after 0.40 g naOH are added to 1 L of buffer?
The Henderson Hasselbach equation:
The Henderson-Hasselbach equation allows us to calculate the pH of a buffer directly without the use of iCe tables. It is derived from the expression for the acid dissociation constant of weak acid:
Since we are interested in finding the pH of the solution, we take the negative log on both sides of the equation and obtain:
A general expression for the Henderson Hasselbach equation is
Using this equation, we can calculate the pH of the acetic acid/acetate buffer solution we discussed above. Look up the Ka value for acetic acid: Ka = 1.8x10-5.
pKa = - log Ka = 4.74
1. Determine the pH of a buffer solution composed of 0.1 M NH3 and 0.1 M NH4Cl
Buffer Capacity and Buffer range:
The buffer capacity is the ability to resist change in pH. The more concentrated the components of a buffer, the greater the buffer capacity
Let's consider the following cases:
(1) The concentrations of acid and its conjugated base are about equal:
[base] ~ [acid] : pH = pKa
(2) The base concentration is about 10 times as high as the acid concentration:
pH = pKa + 1
(3) The acid concentration is about 10 times as high as the base concentration:
pH = pKa -1
As a rule, not more than about 5% (mols) of a strong acid or a strpong base should be added to a buffer in order to maintain a constant or nearly constant pH
The buffer range: is determined by the pKa value of the acid.
A buffer works best when its pH is close to the pKa value of the weak acid thatís used in the buffer system (or pOH is close to pKb value of the weak base.)
© copyright Silvia Kolchens, Pima Community College 2000