Physical and chemical properties of drug molecules

2 Physical and chemical properties of drug molecules

Introduction

The physical properties of organic molecules, such as pKa and partition coefficient, are dealt with extensively in pharmacy courses ^1,² but do not feature greatly in analytical chemistry courses. It is often surprising that analytical chemists cannot distinguish between, for instance, basic, weakly basic, acidic, weakly acidic and neutral nitrogen functions. The physical properties of drug molecules, along with simple chemical derivatisation and degradation reactions, play an important part in the design of analytical methods. Drug molecules can be complex, containing multiple functional groups that in combination produce the overall properties of the molecule. This chapter will serve as a starting point for understanding the chemical and physico-chemical behaviour of drug molecules, which influence the development of analytical methods. The latter part of the chapter focuses on some typical drugs that are representative of a class of drug molecules and lists their physical properties and the properties of their functional groups in so far as they are known.

Calculation of pH value of aqueous solutions of strong and weak acids and bases

Strong acids and bases

If an acid is introduced into an aqueous solution the [H⁺] increases.

If the pH of an aqueous solution is known, the [H⁺] is given by the expression 10^−pH, e.g. [H⁺] in pH 4 solution = 10⁻⁴ M = 0.0001 M. Since [H⁺][OH⁻] = 10⁻¹⁴ for water, the concentration of [OH⁻] in this solution is 10⁻¹⁰ M.

A strong acid is completely ionised in water and [H⁺] is equal to its molarity, e.g. 0.1 M HCl contains 0.1 M H⁺ (10⁻¹ H⁺) and has a pH of −log 0.1 = 1. For a solution of a strong base such as 0.1 M NaOH, [OH⁻] = 0.1 M and [0.1][H⁺] = 10⁻¹⁴; therefore, [H⁺] = 10⁻¹³ M and the pH of the solution = 13. Although the pH range is regarded as being between 0 and 14, it does extend above and below these values, e.g. 10 M HCl, in theory, has a pH of −1.

Calculate the pH of the following solutions:

(ii) 0.05 M NaOH

(iii) 0.05 M H₂SO₄.

Answers: (i) 1.3; (ii) 12.7; (iii) 1.0, since 0.05 M H₂SO₄ contains 0.1 M H⁺

Weak acids and bases

Weak acids are not completely ionised in aqueous solution and are in equilibrium with the undissociated acid, as is the case for water, which is a very weak acid. The dissociation constant Ka is given by the expression below:

For instance, in a 0.1 M solution of acetic acid (Ka = 1.75 × 10⁻⁵), the equilibrium can be written as follows:

The pH can be calculated as follows:

Since the dissociation of the acetic acid does not greatly change the concentration of the un-ionised acid, the above expression can be approximated to:

In comparison, the pH of 0.1 M HCl is 1.

The calculation of the pH of a weak base can be considered in the same way. For instance, in a 0.1 M solution of ammonia (Kb = 1.8 × 10⁻⁵), the equilibrium can be written as follows:

If the concentration/activity of water is regarded as being 1, then the equilibrium constant is given by the following expression:

The concentration of NH₃ can be regarded as being unchanged by a small amount of ionisation and the expression can be written as:

In comparison, the pH of 0.1 M NaOH is 13.

Calculate the pH of the following solutions:

(i) 0.1 M formic acid (Ka = 1.77 × 10⁻⁴)

(ii) 0.05 M phenol (Ka = 1.3 × 10⁻¹⁰)

(iii) 0.15 M ethylamine (Kb = 5.6 × 10⁻⁴)

Answers: (i) 2.4; (ii) 5.6; (iii) 11.9

Henderson–Hasselbalch equation

Can be rearranged substituting pH for −log [H⁺] and pKa for −log Ka to give:

For example, when acetic acid (pKa 4.76) is in solution at pH 4.76, the Henderson–Hasselbalch equation can be written as follows:

From this relationship for acetic acid it is possible to determine the degree of ionisation of acetic acid at a given pH.

Thus, when the pH = 4.76, then:

Determine the percentage of ionisation of acetic acid at (i) pH 3.76 and (ii) pH 5.76.

Answers: (i) 9.09%; (ii) 90.0%

Acetic acid is 50% ionised at pH 4.76. In the case of a weak acid, it is the protonated form of the acid that is un-ionised, and, as the pH falls, the acid becomes less ionised.

For a base, the Henderson–Hasselbalch equation is written as follows:

For example, when ammonia (pKa 9.25) is in a solution at pH 9.25, the Henderson–Hasselbalch equation can be written as follows:

Ammonia is 50% ionised at pH 9.25. In this case, it is the protonated form of the base that is ionised, and, as the pH falls, the base becomes more ionised.

Calculate the percentage of ionisation of ammonia at (i) pH 8.25 and (ii) pH 10.25.

Answers: (i) 90.9%; (ii) 9.09%

An alternative way of writing the expression giving the percentage of ionisation for an acid or base of a particular pKa value at a particular pH value is:

Ionisation of drug molecules

The ionisation of drug molecules is important with regard to their absorption into the circulation and their distribution to different tissues within the body. The pKa value of a drug is also important with regard to its formulation into a medicine and to the design of analytical methods for its determination.

Calculation example 2.1

Calculate the percentage of ionisation of the drugs shown in Figure 2.1 at pH 7.0.

Fig. 2.1 Ionisation of a basic and an acidic drug.

Buffers

Buffers can be prepared from any weak acid or base and are used to maintain the pH of a solution in a narrow range. This is important in living systems; for example, human plasma is buffered at pH 7.4 by a carbonic acid/bicarbonate buffer system.

Buffers are used in a number of areas of analytical chemistry, such as the preparation of mobile phases for chromatography and the extraction of drugs from aqueous solution. The simplest type of buffer is composed of a weak acid or base in combination with a strong base or acid. A common buffer system is the sodium acetate/acetic acid buffer system. The most direct way of preparing this buffer is by the addition of sodium hydroxide to a solution of acetic acid until the required pH is reached. The most effective range for a buffer is 1 pH unit either side of the pKa value of the weak acid or base used in the buffer. The pKa value of acetic acid is 4.76; thus its effective buffer range is 3.76–5.76.

Calculation example 2.2

1 litre of 0.1 M sodium acetate buffer with a pH of 4.0 is required. Molecular weight of acetic acid = 60; therefore, in a litre of 0.1 M buffer there will be 6 g of acetic acid. To prepare the buffer, 6 g of acetic acid are weighed and made up to ca 500 ml with water. The pH of the acetic acid solution is adjusted to 4.0 by addition of 2 M sodium hydroxide solution, using a pH meter to monitor the pH. The solution is then made up to 1 litre with water. Calculate the concentration of acetate and acetic acid in the buffer at pH 4.0.

Using the Henderson–Hasselbalch equation:

The buffer is composed of 1 part acetic acid and 0.17 part acetate.

The buffer was prepared from 0.1 moles of acetic acid; after adjustment to pH 4.0, the amounts of acetic acid and acetate present are as follows:

Since the acetic acid and acetate are dissolved in 1 litre of water, the buffer is composed of 0.085 M CH₃COOH and 0.015 M CH₃COO⁻. Although the concentrations of acetate and acetic acid vary with pH, such a buffer would be known as a 0.1 M sodium acetate buffer.

Note: The 0.1 M buffer should not be prepared by adding acetic acid to a solution of 0.1 M NaOH, since a lot of acetic acid would be required to adjust the pH to 4.0; in fact, 0.1 moles of NaOH would require 0.57 moles of acetic acid to produce a buffer with pH 4.0 and a strength of 0.57 M.

An alternative way of producing 1 litre of 0.1 M acetate buffer would be to mix 850 ml of a 0.1 M solution of acetic acid with 150 ml of a 0.1 M solution of sodium acetate.

1 litre of a 0.1 ammonium chloride buffer with a pH of 9.0 is required. Ammonia has a pKa value of 9.25. If a precise molarity is required this buffer is best prepared from ammonium chloride. The pH of the ammonium chloride may be adjusted to pH 9.0 by addition of a solution of sodium hydroxide (assuming the presence of sodium is not a problem), to avoid adding an excessive amount of the weak base ammonia. A total of 5.35 g (0.1 moles) of ammonium chloride are weighed and dissolved in ca 500 ml of water. The pH is then adjusted to pH 9.0 by addition of 5 M NaOH. The solution is then made up to 1 litre with water.

Calculate the concentrations of and NH₃ in the buffer at pH 9.0 and indicate an alternative method for preparing the buffer.

Answer: 0.036 M NH₃ and The buffer could be prepared by mixing 360 ml of a 0.1 M ammonia solution with 640 ml of 0.1 M NH₄Cl

Some weak acids and bases have more than one buffer range; for example, phosphoric acid has three ionisable protons with three different pKa values and can be used to prepare buffers to cover three different pH ranges. The ionic species involved in the ranges covered by phosphate buffer are:

The buffering ranges of a weak electrolyte are only discrete if the pKa values of its acidic and/or basic groups are separated by more than 2 pH units. Some acids have ionisable groups with pKa values less than 2 pH units apart, so they produce buffers with wide ranges. For example, succinic acid, which has pKa values of 4.19 and 5.57, can be considered to have a continuous buffering range between pH 3.19 and 6.57.

Determine the buffer range(s) for the following compounds:

(i) Carbonic acid pKa 6.38, 10.32

(ii) Boric acid pKa 9.14, 12.74

(iii) Glycine 2.34, 9.60

(iv) Citric acid 3.06, 4.74, 5.4.

Answers: (i) 5.38–7.38, 9.32–11.32. The lower range is not useful because of the ease with which CO₂ is lost from solution; (ii) 8.14–10.14, 11.74–13.74; (iii) 1.34–3.34, 8.6–10.6; (iv) Continuous buffering range 2.06–6.4

Sometimes a salt of a weak acid with weak base is used in a chromatographic mobile phase to, apparently, set the pH at a defined level, e.g. ammonium acetate or ammonium carbonate. These salts are marginally more effective than a salt of strong acid with a strong base at preventing a change in pH but they are not truly buffers. Such salts have buffering ranges ca 1 pH unit either side of the pKa values of the weak acid and weak base composing them. For example, the pH of a solution of ammonium acetate is ca 7.0 but it does not function effectively as a buffer unless the pH either rises to ca 8.25 or falls to ca 5.76.

A buffer is most effective where its molarity is greater than the molarity of the acid or base it is buffering against.

Calculation example 2.3

If 10 ml of 0.05 M HCl is added to 100 ml of 0.2 M sodium acetate buffer pH 4.5, the resultant pH can be calculated as follows:

Molarity × volume = mmoles

Number of mmoles of acetate + acetic acid in 100 ml of buffer = 0.2 × 100 = 20 mmoles.

Using the Henderson–Hasselbalch equation:

The buffer contains 1 part CH₃COOH and 0.55 parts CH₃COO⁻.

When HCl is added the following reaction occurs:

The amount of HCl added is 10 × 0.05 = 0.5 mmoles

Therefore, after addition of HCl, the amount of acetate remaining is:

7.1 − 0.5 = 6.6 mmoles

The amount of acetic acid now present is:

12.9 + 0.5 = 13.4 mmoles

Therefore, the pH of the buffer after addition of HCl is determined as follows (amounts may be substituted in the equation instead of concentrations since CH₃COO⁻ and CH₃COOH are present in the same volume):

The new pH of the buffer is 4.45.

The molarity of the buffer has also changed, since the total amount of CH₃COO⁻ and CH₃COOH is now contained in 110 ml instead of 100 ml, giving a new molarity of

If 10 ml of 0.05 M HCl were added to 100 ml of water, the pH would be determined as follows:

10 ml of 0.1 M HCl is added to 20 ml of a 0.5 M sodium acetate buffer with a pH of 4.3. Calculate: the pH of the buffer after addition of the HCl, the molarity of the buffer after addition of the HCl, the resultant pH if the HCl had been added to 20 ml of water.

Answer: pH = 4.04, new molarity = 0.33. Addition of 10 ml of 0.1 M HCl to 20 ml of water would give a pH of 1.48

Salt hydrolysis

When the salt of a strong acid and a strong base is dissolved in water it produces a pH of ca 7.0. When salts of a weak acid and a strong base or of a strong acid and a weak base are dissolved in water, they will produce, respectively, alkaline and acidic solutions.

When sodium acetate is dissolved in water, the acetate ion behaves as a base, removing protons from solution. For a weak electrolyte in water, Kb × Ka = Kw. If a 0.1 M solution of sodium acetate in water is considered:

Regarding the change in the concentration of water as not affecting the equilibrium and regarding the [CH₃COO⁻] as being relatively unchanged by hydrolysis.

Calculate the pH of a 0.1 M solution of NH₄Cl. Here salt hydrolysis increases [H⁺] and the equilibrium in this case is:

The Ka for this reaction is 5.6 × 10⁻¹⁰.

Answer: pH = 5.13

Activity, ionic strength and dielectric constant

The activity of ions in a solution is governed by the dielectric constant of the medium they are dissolved in and by the total concentration of ions in solution. For solutions of electrolytes in water with concentrations < 0.5 M, the activity of the ions present in solution is usually approximated to their individual concentrations. The mean activity coefficient for an ion in solution is defined as:

Although activity is regarded as 1 in dilute solutions this is still an approximation. The activity of an electrolyte solution in water can be estimated from the following equation:

where −0.509 is a constant related to the dielectric constant of the solvent used to prepare the electrolyte solution and to temperature; z is the charge on a particular ion; I is the ionic strength of the solution; and m is the molality (moles per kg of solvent) of a particular ion in solution.

Using this equation, the activity of H⁺ in 0.1 M HCl can be calculated to be 0.69. Thus, the true pH of 0.1 M HCl is calculated as follows:

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Tags: Pharmaceutical Analysis A Textbook for Pharmacy Students

Jun 24, 2016 | Posted by admin in PHARMACY | Comments Off

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