biopharmaceutics and pharmacokinetics - minia

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Physical Pharmacy

BuffersFaculty of Pharmacy, Minia University

By: Kareem Ebeid, Ph.D.17/3/2020Lecture# 1

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Buffers• Buffers are compounds or mixtures of compounds that, by their presence in

solution, resist changes in pH upon the addition of small quantities of acid or alkali.• If a small amount of a strong acid or base is added to water or a solution of

sodium chloride, the pH is altered considerably; such systems have no buffer action.

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THE BUFFER EQUATIONCommon Ion Effect and the Buffer Equation for a Weak Acid and Its Salt

• The pH of a buffer solution and the change in pH upon the addition of an acid or base can be calculated by use of the buffer equation.

• This expression is developed by considering the effect of a salt on the ionization of a weak acid when the salt and the acid have an ion in common.

• For example, when acetic acid is added in waterHAc + H2O ⇄ H3O+ + Ac−

Ka = [H3O+][Ac−][HAc] = 1.75 × 10−5

Ka is acid dissociation constant

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• When sodium acetate is added to acetic acid, the equilibrium is shifted to the left towards the formation of more HAc, and this is accompanied by reduction in H3O+ concentration, which subsequently results in elevation of the pH of the solution.

HAc + H2O ⇄ H3O+ + Ac−

Ac− Ac− Ac−

HAc + H2O H3O+ + Ac−

• Thus we can say that, the ionization of acetic acid, is repressed upon the addition of the common ion, Ac−. This is an example of the common ion effect.

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• The pH of the final solution is obtained by rearranging the equilibrium expression for acetic acid: HAc + H2O ⇄ H3O+ + Ac−

Ka = [H3O+][Ac−][HAc]

[H3O+] =Ka[HAc][Ac−]

• If the acid is weak and ionizes only slightly, the expression [HAc] may be considered to represent the total concentration of acid, and it is written simply as [Acid].

• In the slightly ionized acidic solution, the acetate concentration [Ac−] can be considered as having come entirely from the salt, sodium acetate. Because 1 mole of sodium acetate yields 1 mole of acetate ion, [Ac−] is equal to the total salt concentration and is replaced by the term [Salt].

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[H3O+] =Ka[HAc][Ac−]

[H3O+] =Ka[Acid][Salt]

• pH = -Log [H3O+]

• Thus in order to calculate the pH, you need to: × -Log

-Log [H3O+] =−Log (Ka[Acid][Salt] )

pH = pKa – Log [Acid][Salt]

pH = pKa + Log [Salt][Acid] 6

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pH = pKa + Log [Salt][Acid]

The buffer equation or the Henderson–Hasselbalch equation, for a weak acid and its salt.

• The term pKa, the negative logarithm of Ka, is called the dissociation exponent.

• The buffer equation is important in the preparation of buffered pharmaceutical solutions; it is satisfactory for calculations within the pH range of 4 to 10.

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• What is the pH of weak acetic acid solution, pKa = 4.76?

HAc + H2O ⇄ H3O+ + Ac−

If you just have weak acid, you can assume that [Ac-] = [H3O+], and [HAc] = C (total weak acid concentration)

Ka = [H3O+][H3O+][HAc]

[H3O+]2 =Ka[HAc]

[H3O+] =2 Ka[HAc]

pH = 12 pKa - 1

2 Log C

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• What is the pH of 0.1 M acetic acid solution, pKa = 4.76?

pH = 12 pKa - 1

2 Log C

pH = 2.38+0.50 = 2.88• What is the pH after enough sodium acetate has been added to make the solution 0.1 M

with respect to this salt?


pH = 4.76 + Log [0.1][0.1] = 4.76

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• What is the molar ratio, [Salt]/[Acid], required to prepare an acetate buffer of pH 5.0


5 = 4.76 + Log [Salt][Acid]

Log [Salt][Acid] = 0.24

[Salt][Acid]= antiLog (0.24) = 1.74

This means that I need to each 1 mole of acid, I need to add 1.74 moles of salt.Thus mole fraction of salt = 1.74/(1+1.74) = 0.635 = 63.5% (mole percent)


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Buffer solutions are not ordinarily prepared from weak bases and their salts because of the volatility and instability of the bases and because of the dependence of their pH on pKw, which is often affected by temperature changes.

B + H2O ⇄ BH+ + OH-

Kb = [BH+][OH−][B]

[OH−] = [Kb][B][BH+] =Kb

[Base][Salt]

[OH−] [H3O+] = KW

[OH−] = KW[H3O+]

KW[H3O+] =Kb

[Base][Salt]

THE BUFFER EQUATIONCommon Ion Effect and the Buffer Equation for a Weak Base and Its Salt

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KW[H3O+] =Kb

[Base][Salt]

[H3O+] =KWKb

[Salt][Base]

pH = pKW – pKb + Log [Base][Salt]


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• What is the pH of a solution containing 0.10 mole of ephedrine and 0.01 mole of ephedrine hydrochloride per liter of solution? Since the pKb of ephedrine is 4.64

pH = pKW – pKb + Log [Base][Salt]

pH = 14.00−4.64 + Log 0.10.01

pH = 9.36+Log 10 = 10.36


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• It is important to recognize that solutions of drugs that are weak electrolytes also manifest buffer action. Salicylic acid solution in a soft glass bottle is influenced by the alkalinity of the glass. It might be thought at first that the reaction would result in an appreciable increase in pH; however, the sodium ions of the soft glass combine with the salicylate ions to form sodium salicylate. Thus, there arises a solution of salicylic acid and sodium salicylate—a buffer solution that resists the change in pH.

Drugs as Buffers

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• Similarly, a solution of ephedrine base manifests a natural buffer protection against reductions in pH. Should hydrochloric acid be added to the solution, ephedrine hydrochloride is formed, and the buffer system of ephedrine plus ephedrine hydrochloride will resist large changes in pH until the ephedrine is depleted by reaction with the acid. Therefore, a drug in solution may often act as its own buffer over a definite pH range.

• Such buffer action, however, is often too weak to counteract pH changes brought about by the carbon dioxide of the air and the alkalinity of the bottle.

• Additional buffers are therefore frequently added to drug solutions to maintain the system within a certain pH range

Drugs as Buffers

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• Thus far it has been stated that a buffer counteracts the change in pH of a solution upon the addition of a strong acid, a strong base, or other agents that tend to alter the hydrogen ion concentration.

• The magnitude of the resistance of a buffer to pH changes is referred to as the buffer capacity, β. It is also known as buffer efficiency, buffer index, and buffer value.

• Koppel and Spiro and Van Slyke introduced the concept of buffer capacity and defined it as the ratio of the increment of strong base (or acid) to the small change in pH brought about by this addition.

β = ∆B∆pH

∆B is the small increment in gram equivalents (g Eq)/liter of strong base added to the buffer solution to produce a pH change of ∆pH.

the buffer capacity of a solution has a value of 1 when the addition of 1 g Eq of strong base (or acid) to 1 liter of the buffer solution results in a change of 1 pH unit.

BUFFER CAPACITY

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.β = 2.3 C Ka [H3O+](Ka + [H3O+])2

Where C is the total buffer concentration, that is, the sum of the molar concentrations of the acid and the salt.This equation permits one to compute the buffer capacity at any hydrogen ion concentration—for example, at the point where no acid or base has been added to the buffer.

BUFFER CAPACITY

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.β = 2.3 C Ka [H3O+](Ka + [H3O+])2

• At a hydrogen ion concentration of 1.75 × 10−5 (pH = 4.76), what is the capacity of a buffer containing 0.10 mole each of acetic acid and sodium acetate per liter of solution? The total concentration, C=[Acid]+[Salt], is 0.20 mole/liter, and the dissociation constant is 1.75 × 10−5

β = 2.3 x 0.2 1.75 × 10−5 [1.75 × 10−5 ](1.75 × 10−5+ [1.75 × 10−5])2 = 0.115

BUFFER CAPACITY

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BUFFER CAPACITY

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.β = 2.3 C Ka [H3O+](Ka + [H3O+])2

• The buffer capacity is affected not only by the [Salt]/[Acid] ratio but also by the total concentrations of acid and salt

• when 0.01 mole of base is added to a 0.1 molar acetate buffer, the pH increases from 4.76 to 4.85, for a ∆ pH of 0.09.

• If the concentration of acetic acid and sodium acetate is raised to 1 M, the pH of the original buffer solution remains at about 4.76, but now, upon the addition of 0.01 mole of base, it becomes 4.77, for a ∆ pH of only 0.01.

• Therefore, an increase in the concentration of the buffer components results in a greater buffer capacity or efficiency

BUFFER CAPACITY: influence of C on β

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.β = 2.3 C Ka [H3O+](Ka + [H3O+])2

• The maximum buffer capacity occurs where pH = pKa, or, in equivalent terms, where [H3O+] = Ka.

• Substituting [H3O+] for Ka in both the numerator and the denominator of the equation=

βmax = 2.3 C [H3O+]2

(2 [H3O+])2

βmax = 2.3034 C

βmax = 2.3034 C

βmax = 0.576C

Maximum BUFFER CAPACITY

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.β = 2.3 C Ka [H3O+](Ka + [H3O+])2

• What is the maximum buffer capacity of an acetate buffer with a total concentration of 0.020 mole/liter?

βmax = 0.576×0.020 = 0.01152

Maximum BUFFER CAPACITY

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Thank you

biopharmaceutics and pharmacokinetics - minia

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