Buffer Calculation Examples

It will be noted that many buffer Problems involve a determination of the ratio of the [salt] to [acid], each of which must be expressed in the same units, and the total of the [salt] and [acid] together. The general situation calls, therefore, for two simultaneous equations

[salt]/[acid]=a, [salt]+[acid]=b

It is essential in any buffer problem to identify from the given description the values of pKa, pH, a, and b, which can be expressed in many ways. pKa can be given as Ka, or the equilibrium concentrations of the weak acid. pH can be given as [H+], pOH, or [OH-]. [Salt] or [acid] may be derived by titration, but in all cases the four unknowns must be obtained from the given information or by calculation. The most comprehensive buffer problem asks for the preparation of a volume of buffer of a certain concentration and pH from a weak acid of given pKa, and a strong base. The general approach may be by the following steps.

1.      Since pH and pKa are given, determine the [salt]/[acid] ratio, which is a pure number.

1. The concentration of the buffer is usually expressed as the sum of the salt and acid forms.

If the salt (or the acid) concentration is given, then the corresponding acid (or salt) can be calculated directly from the ratio of [salt] to [acid] in step 1. If the total concentration, [salt] + [acid], is given, then the solution of the two simultaneous equations is required. Note that the unit of concentration of the salt and acid so calculated will be the one used for the [salt] + [acid] concentration. For example, if

[salt]/[acid]=2

[salt] + [acid]=0.3 M

Then, since [salt] = 2[acid]

and by substitution ,

2[acid] + [acid] = 0.3 M

[acid] = 0.3/3 M = 0.1 M

Now, if we substitute this value ,

[salt] = 2 x 0.1 M = 0.2 M

Check these values in (i) and (ii):

[salt[/[acid]=0.2 M/0.1 M = 2

[salt] + [acid] = 0.2 M + 0. 1 M = 0.3 M

3.         From the concentrations of salt and acid the actual amounts of salt and acid can be calculated for the volume required from their molecular weights. In the most general case, where the given materials are pure acid and pure base, the salt will be generated from equivalent amounts of acid      and base; it is necessary to measure an amount of pure acid equivalent to            salt and acid and add to it an amount of base equivalent to the salt.

In the example given in step 2 above, if 1 liter of buffer were required and if the acid were monobasic, then to 0.3 mole of acid would be added 0.2 mole of base and the volume brought to 1 liter. This would give 0.2 M salt and leave 0. 1 M free acid.

An alternative approach to that developed above for the solution of problems involving equilibria is to use the fraction of dissociation, a, which is defined for any system as

a = (moles of HA dissociated)/( total moles of species containing A).

= (moles HA dissociated)/(AT).

For the dissociation of a weak acid

[At] = [HA] + [A-],

then

a = [A-]/[HA] + [A-],

This applies equally to the salt of a weak acid and, in such cases as buffer systems,

[salt]/[acid] = [A-]/[HA] = a/1 – a

The equation becomes pH = pKa + log (a)/(1-a)

Example Question similar to Exp. 2 Pre-lab Q.5

A 10.0 litres of 0.045 M phosphate buffer, pH 7.5 (pKa2 = 7.2).  You are provided with 2.00 M HCl and 1.00 M K2HPO4 solutions.  Calculate volume of KOH and K2HPO4 solutions.

Before doing the calculation, always write down the titration equations for H3PO4 in the presence of KOH and HCl.

The pH of this buffer is a little above the pKa2 of H3PO4, consequently, the two major ionic species present in this buffer are H2PO4­- (conjugate acid) and HPO42- (conjugate base).

1.      The first step involves calculating the proportion and amounts of the two ionic species in the buffer.

The buffer contains a total of 10 litres x 0.045 M = 0.45 mole of phosphate (PO43-).

pH = pKa2 + log [HPO42-]/[ H2PO4­-]

7.5 = 7.2 + log [HPO42-]/[ H2PO4­-] ,   0.3 = log [HPO42-]/[ H2PO4­-]

Antilog of 0.3 = 2 = 2/1 ratio = [HPO42-]/[ H2PO4­-] (NOTE: Always add 1 to the denominator to convert the number to a ratio)

Therefore, mole of HPO42- needs = 2/3 x 0.45 mole = 0.30 mole

Mole of H2PO4­- needs = 1/3 x 0.45 mole = 0.15 mole

Because we want to end up with an [HPO42-]/[ H2PO4­-]  ratio of 2/1, we want to convert ONLY 1/3 of the HPO42-to  H2PO4­-

Mole of H2PO4­- needs = mole of HCl needs = 0.15 mole

Vol of 2.0 M of HCl = 0.15 mole/2.0 M = 0.75 litre (final answer)

Mole of HPO42- needs = mole of HPO42- + mole of H2PO4­- = 0.45 mole

Vol of 1.0 M K2HPO4 = 0.45 mole/1.0 M = 0.45 litre (final answer).

Problem example 1: How many moles of sodium acetate and acetic acid are required

to prepare I liter of a buffer, pH 5.0, which is 0. 1 M in total available acetate

(dissociated and undissociated). Acetic acid has a pK of 4.74.

Solution: From Buffer Equation ,

5.0 = 4.74 + log [salt]/ [acid]

or log [salt]/[acid] = 0.26.

The ratio of the concentration of sodium acetate to acetic acid is the antilog 0.26, or 1.82; there are

1.82 molecules of NaAc for each molecule of HAc. The mole fraction of salt is 1.82/2.82 and the

mole fraction of acid is 1.00/2.82. We are told that the total molar concentration of acetate is 0. 1 M

Therefore, of the 0. 1 mole, the salt must provide its fraction,

1.82 / 2.82 x 0. 1 mole = 0.065 mole.

and the acid must provide its fraction,

1.00 / 2.82 x 0. 1 mole = 0.035 mole,

Problem Example 2: To 100 ml of 0.2 M formic acia, pKa 3.77, is added 10 mmole

of solid NAOH. What will be the final pH?

Solution: The given data are

PKa = 3.77

salt + acid = 20 mmole

The ratio of [salt]/[acid] must be calculated:

10 mmole of NAOH will neutralize 10 mmole of formic acid

Therefore (20 - 10) mmole of acid will be converted to salt:

salt = 10 mmole in 100 ml,

acid = 10 mmole in 100 ml,

Therefore

[salt]/[acid] = 100mM /100 mM = 1.

Substituting in Equation ,

pH = 3.77 + log 1 , pH = 3.77

Problem Example 3: The dissociation constant for lactic acid is 1.38 x 10-4

How much lactic acid must be added to 2 g of NAOH to make 500 ml of buffer at pH 4.00?

Solution:         We are given pH = 4.00

pKa, = -log 1.38 x 10-4.

= - (0. 1399 - 4.00).

= - (- 3.86) (to 2 decimals).

= 3.86.

salt = 2 / 40 x 1000 /500 M.

= 0. 10 M

,

We do not know the acid concentration. When we substitute the known values

in Buffer Equation ,

4.00 = 3.86 + log ( 0.10) / [acid].

or

0.14 = log (0.10)/[acid].

Therefore

0.10 / [acid]= 1.38.

or

[acid] = 0.1/ 1.38 M = 0.0725 M.

Thus 500 ml requires 0.0363 mole of the free acid and to this must be added the

amount to react with the 2 g of NAOH, or 0.05 mole. Therefore

[salt] + [acid] = (0.0363 + 0.05) mole/500 ml.

and the total acid needed is 0.0863 mole or 0.09 mole (2 significant figures).

Problem Example 4 : 0.10 M ammonium hydroxide, K, 1.8 x 10-,5, is added to an equal volume of 0.03 M HCI. What is the resulting pH?

Solution: For the NH4+ = NH3 + H+ reaction, (i)

Kb= 1.8 X 10- 5,

Therefore

Ka = 1.0 X 10-14 / 1.8 x 10-5

4

= 5.5 x 10-10

From Equation , pKa = -log 5.5 x 10-10 = 9.26: [acid] = 0.015 M, since 2 the acid is the [NH4+] ion produced by reacting a volume of 0.03 M HCI with an equal volume of ammonia and so diluting the acid to one-half its mormality, or 0.015 M

In this situation the [salt] is the ammonia concentration, as can be seen

from (i):

[NH3] = 0.10 –0.03 / 2 M = 0.035 M

Therefore,

pH = 9.26 + log 0.035/0.015 = 9.26 + log 2.333,

= 9.26 + 0.37 (to 2 decimal places).

pH = 9.63

Buffer Systems

The hydrogen ion concentration of intracellular and extracellular fluids must be maintained within very narrow limits. A pH change in blood plasma of 0.2 to 0.4 may result in serious damage to an organism or even death. A constant pH ensures that acidic and basic biomolecules are in the correct ionic state for proper functioning. This is especially critical for enzymes that are sensitive to pH changes. Metabolic reactions generate high concentrations of organic acids that would change fluid pH values if buffering agents were not present. Consider the following experiment. If 1.0 mL of 10 M HCI is added to 1.0 L of 0. 15 M NaCl solution at pH 7.0, the pH would tumble to pH 2.0. If 1.0 mL of 10 M HCI is added to I L of blood plasma, the pH would fall only from pH 7.4 to pH 7.2. There is nothing magical about blood. Its ability to maintain a constant pH is due to a heterogeneous mixture of biomolecules that can act to neutralize added acids and bases. Blood and other biological fluids contain buffering systems: reagents that resist changes in pH when H+ or OH- are added. Chemically, buffer systems contain acid-base conjugate pairs. Many nonprescription products containing bases are available to neutralize stomach acid. Therefore, solution is buffered or protected against pH changes caused by added acids or bases. In general, an acid-base conjugate pair is most effective as a buffer system at a pH equal to its pKa. The effective buffering range for an acid-base conjugate pair can be estimated:

effective buffering range (pH) = pKa+/-1

Acetic acid-acetate is an effective buffer in the pH range of 3.76 to 5.76. Phosphoric acid is an effective buffer in three pH ranges: (1) pH = 1.14-3.14, (2) pH 6.20-8.20, and (3) pH = 11.4-13.4.

The major buffer system of blood and other extracellular fluids is the carbonic acid-bicarbonate conjugate pair:

A base added to blood would be neutralized by the following reaction:

H2CO3 + OH ==== HCO- + H+

The addition of an acidic substance to blood also results in neutralization:

HCO- + H+ ==== H2CO3

These reactions illustrate how blood is protected against pH changes. The actual pH of blood (pH = 7.4) is at the upper limit of the buffering range of carbonic acid-bicarbonate (I = 5.4-7.4) and may not seem as efficient as desired. This inefficiency is remedied by a reserve supply of gaseous C02 in the lungs, which can replenish H2CO3 in the blood by the following series of equilibrium reactions:

C02(g) === C02(aq)

Lungs Blood

C02(aq) + H20(l) ==== H2CO3(aq)

H2CO3(aq) ==== HCO3- (aq) + H+

The reactions also work in reverse. A major product of metabolism, H+ is removed from cells by the blood plasma. It is neutralized by reaction with HCO-3 and leads to eventual release Of C02(g), which is exhaled from the lungs.

The carbonic acid-bicarbonate conjugate pair is the most important buffer system in extracellular fluids, such as blood. However, it is not the only one. There is a diverse array of amino acids, peptides, and proteins with ionizable functional groups (-COOH and -NH+) that assist in buffering. A major protein constituent of blood, hemoglobin serves many purposes, including (1) transport of oxygen from lungs to peripheral parts of the body, (2) transport of the metabolic product C02 from peripheral tissue to the lungs for exhalation, and (3) buffering of blood by neutralizing H+and OH-.

Another important buffer system in intracellular solutions is the H2PO4- &HPO4-- (dihydrogen phosphate-monohydrogen phosphate) conjugate pair. This solution is an effective buffer in the pH range of 6.2 to 8.2. Because it is a natural substance, phosphoric acid and its salts are often used as buffers when working in the biochemical laboratory. Such buffers may be prepared by using the HendersonHasselbalch equation.