Lecture Notes Pharmaceutics 2005

Pharmaceutics (Part I) – Spring 2005 Prof. C. Thompson (SB 383; 243-4643; [email protected]

Pharmaceutics Class This document contains lecture notes for Part I of the class (about 4 weeks). These lecture notes are intended as an aid for your note taking and are not comprehensive. Additional/clarifying information, select resources, and practice problems will also be provided in the classroom. Although the notes have been scanned and checked for errors, mistakes are possible so feel free to drop me a line with any corrections. Helpful Hints (really). • Relocate your general chemistry book. Briefly review sections on equilibrium, acids & bases. Answer a

few calculations of pH problems. It helps to get the feel for the numbers. • Relocate your organic and biochemistry chemistry books. Familiarize yourself with organic functional

groups, particularly carboxylic acids and amines. Amines (organic bases) are synonymous with pharmaceuticals and a brief review of their structure and properties will be helpful in this class and beyond. Carboxylic acids will be the subject of a majority of calculations in this class and worth a quick review. Your biochemistry book will have examples and problems in the area of pKa, Henderson-Hasselbalch, buffers, and ionization. Please review these sections.

• Do not ignore the first two helpful hints. Right about now you’re thinking, “Where are those chem.

books? Oh yeah, I sold them for a case of Lunch Noodles, I don’t really need them.” • Check those seldom used calculator keys! You will need to calculate log and antilog so find those fun

keys on your calculator and reacquaint yourself with them (it may have been a while). Remember that: log (100) = 2, and log (10) = 1. So, when you solve for x in the equation “log(x) = 1.7” does it makes sense that x = 50? Just for fun, push the antilog key about 50 times and let it know who is boss.

• Web Assistance: The internet can be an excellent source of information, but beware. In the area of pH,

pKa, buffers, etc., about half of them very useful, one-fourth OK and one-fourth had many incorrect statements. I am not responsible for the content in these web sites, or the vast resource of the internet but do encourage you to find instructional materials on the web that complement our topics. One example is an interactive website that allows you to visualize how the Henderson-Hasselbalch equation is graphed. You simply put in the pKa and pH boundaries, and voila, a nice S-shaped titration. Check out the following website and follow the link to Henderson-Hasselbalch weak acid or weak base.

http://www.cpb.uokhsc.edu/cc/ • Stay Current. Try to keep up with the assigned problems. This section is short and once you fall

behind, it is difficult to catch up. I encourage workgroups but I wish to emphasize the importance of being confident in your own calculations and understanding.

At the end of this document is a copy of last year’s exam. Answers will be provided in a few weeks. C Thompson

http://www.cpb.uokhsc.edu/cc/


Acids, bases, pH/pKa calculations and the Henderson Hasselbalch equation

Introductory questions:

1. What is the difference between an organic compound that is described as an acid (proton donor) and a compound described as capable of hydrogen bonding?

2. The alkaloid natural product, cocaine is a “nitrogenous base” typically isolated as the hydrochloride salt.

a. What is a nitrogenous base?

b. What is a hydrochloride salt?

c. What benefits if any would result from forming an alternate salt (e.g., HBr, HI, benzoate, maleate, and succinate) with cocaine?

3. Three common classifications of organic compounds are polar non-protic, polar protic and non polar.

a. Which class is more likely to penetrate skin? Why?

b. Which class is more likely to dissolve in saliva? Why?

c. What is DMSO and which class does it belong?

d. One compound class that is not included is non-polar protic. Can a molecule with these properties exist?

e. How do these “chemical” properties relate to lipophilic/hydrophobic or lipophobic/hydrophilic?

4. Almost all modern drugs are “organic,” that is, they contain one or more carbon atoms. Name some inorganic drugs that pharmacists MUST be knowledgeable of.

5. What percentage of drugs currently prescribed is likely to be in use in 5 yrs, 10 yrs, and 25 yrs from now? How will an understanding of basic molecular properties enable you to withstand this change?


Acids & bases Keywords: acid base buffer pH pKa ionization equilibria Henderson-Hasselbalch equation ‘salt’ Objective(s): • Restore understanding of fundamental acid-base ionization equilibria. • Relate acidity and basicity of organic compounds to that of water. • Conduct and gain confidence in ionization calculations. • Gain a basic understanding of the physiologic importance/implication of drug ionization – specifically in

relation to absorption, transport, and excretion. Definitions and descriptions: Hydrogen ions (H+): Hydrogen ions (H+), or protons, do not have electrons, but can bond to nitrogen or oxygen containing molecules because nitrogen and oxygen have “non-bonding” pairs of electrons. Hydrogen ions cannot bond to carbon since carbon does not have “non-bonding” electrons. Acids: According to the Bronsted-Lowry theory of acids and bases, an acid is a substance capable of donating a proton, or hydrogen ion (H+) to a base. Bases: A base is a substance that is capable of accepting a proton from an acid. pH: A method of expressing the hydrogen ion concentration [H+], or more correctly, hydronium ion [H3O+] concentration in solution. pH is equal to the negative log of the molar hydrogen ion concentration. pH = -log[H+] pOH: A method of expressing hydroxyl ion (OH-) concentration in solution. pOH is equal to the negative log of the molar hydroxyl ion concentration. pOH = -log[OH-]


IONIZATION OF WATER Kw and pKw: Water ionizes (dissociates) slightly to yield hydronium and hydroxyl ions:

H2O + H2O H3O+ + OH-

The dissociation constant for water is Kw: Kw = [H3O+] x [OH-] . 1.0 x 10-14

pKw is equal to the negative log of the dissociation constant of water: pKw = -logKw = pH + pOH = 14 So, how do pH and pOH correlate? Some pH examples:

[H+] pH [OH-] pOH Kw pKw

10-1 1 10-13 13 10-14 14

10-5 5 10-9 9 10-14 14

10-7 7 10-7 7 10-14 14

10-10 10 10-4 4 10-14 14

10-14 14 1 0 10-14 14

1 0 10-14 14 10-14 14

Note: At [H+] = 10-7, [OH-] = 10-7 and therefore [H+] = [OH-] and pH = pOH. This is neutral pH (pH = 7). Questions: 1. For a 0.000001 M solution of hydroiodic acid (HI) what are:

a. the H+ concentration b. the pH c. the OH- concentration d. pOH

2. As a successful pharmacist, you chlorinate your hottub on a regular basis and find the pH = 5. What is

the concentration of H+ and molarity of HCl?


ACID-BASE EQUILIBRIA AND IONIZATION OF ELECTROLYTES Acid-Base Reactions: Acid-base reactions are equilibrium reactions. That means that there is a forward reaction and a reverse reaction. At equilibrium, the forward and reverse reaction rates are the same, and the relative concentrations of reactants and products and remain constant. Strong and weak acids:

Strong acid - HCl is considered a strong acid because it has a strong tendency to ionize resulting in a large [H3O+] and the reverse reaction occurs only to a small amount. This further indicates that Cl- is very weak as a conjugate base and not likely to accept a proton. What this means is that Cl- is ‘stable’ just the way it is. This is an attribute of many strong acids – the conjugate base is stable while bearing a negative charge. This means the conjugate base happily accepts the extra electron. Strong acids are completely ionized at all pH values (pH independent ionization).

HCl + H2O H3O+ + Cl- Weak Acid – Phenol is a good example of a weak acid. In comparison to HCl, phenol has a weaker tendency to ionize and the conjugate base, phenoxide ion, is moderately strong. As a result, the equilibrium shifts further to the left and the acid (protonated) form of phenol predominates.

OH

+ H2O H3O+ +

O-

Question: When substituents are added to the aromatic ring (e.g., methyl, nitro, etc.), what effect on the ionization might they have? One example of how the phenol structure has been altered to achieve the desired pharmacologic property is the anti-bacterials:

Fun fact: Phenol causes some of the most serious and damaging chemical burns to the skin. In the laboratory phenol is used to lyse cells!

Anti-bacterial activity was increased by increasing the waxy chain to enhance membrane infiltration. Hexylresorcinol and hexachlorophene are used in hospital sterilization methods.


In general, for a weak acid:

HA + H2O H3O+ + A-

(conj base)

kf

kr Note that the product (A-) is designated as the conjugate base. The conjugate base is also known as the salt of the acid when combined with a counterion such as Na+ to give Na+A-. An equilibrium constant (K) can be defined that is equal to the forward rate constant (kf) divided by the reverse rate constant (kr). It is also equal to the product concentrations divided by the reactant concentrations:

K = kf = [A-][H3O+] kr [HA][H2O]

Ka = 55M = [A-][H3O+] [HA]

since water is constant @ 55 M

Ka is a measure of (weak) acid strength as expressed by the concentration of ionized molecules divided by the concentration of unionized molecules. Question: Alcohol are organic ‘analogs of water meaning they share the identical functional group (OH) but vary in one component, the “switch” of an H for a Me, Et, etc. Explain the increasing trend in pKa values (table at right) for the alcohols and the “unusual” drop in pKa for phenol relative to the other alcohols.

R-OH pKa H-OH (water) 15.8

MeOH (methanol) 16.4 Et-OH (ethanol) 16.8

iPr-OH (isopropanol) 17.2 t-Bu-OH (tert-butanol) 18

Ph-OH 10

Pharmaceutics (Part I) – Spring 2005 Prof. C. Thompson (SB 383; 243-4643; [email protected] Trends to consider based on the Ka equation: As acid strength increases (8): Ka increases [A-] increases [H3O+] increases [HA] decreases A helpful utility is pKa, which relates the strength of weak acids on logarithmic scale. It provides a means to compare the acidity of a class of compounds or across classes of compounds in whole numbers. Because the scale is log-based, comparisons are typically based on ten-fold increments, which is fine for most comparisons.

pKa = -log(Ka) As the acid strength increases, so does the Ka. However, the pKa decreases as the acid strength increases.

HA H+ + A-

(conj base)For weak acids, the ionization in water is: Ka = [A-][H+]

[HA]Which can be expressed as:

[HA] [HA]Ka = [H+][H+] = [H+]2 and [H+]2 = Ka[HA]

With no other ions in solution, the concentration of H+ and A- should be equal and, therefore: [H+] = Ka[HA]

[H+] = KaC

if you take the -log of each side:

-log[H+] = 1/2 (-log Ka - log C) or

pH = 1/2 (-logKa - log C)

pH = 1/2(pKa) - 1/2(logC)

Rearrange and substitute to: This equation in bold is useful for calculating the pH of a weak acid in water when you know the molar concentration and the pKa of the weak acid or drug (usually found in tables).


Question: Calculate the pH of a 0.1 M solution of a drug at 25 oC. The pKa of the acid is 4.76 at 25 oC. Weak Bases. Weak bases react with water to take up hydrogen atoms to form a conjugate acid resulting in hydroxyl ions that turn the solution alkaline (basic).

B: + H2O BH+ + OH-

(conj acid) Note that the product (BH+) is designated as the conjugate acid. The conjugate acid is also known as the salt of the base when combined with a counterion such as Cl- to give BH+Cl-. One example of a weak base is ammonia (NH3), which reacts with water to form ammonium hydroxide (NH4OH or NH4

+ / OH-). Ammonia has a weak tendency to ionize and the conjugate acid is strong: the equilibrium favors the left side of the equation. The basicity constant is defined as Kb (expressed similarly to the Ka) and is a measure of base strength. When base strength is increased the equilibrium is moved to the right. Kb is mostly unused and has been replaced by Ka. Ka and pKa are generally used for both acids and bases because pKa uses the same scale and goes in the same direction as pH. For bases, Ka and pKa indicate the strength of the conjugate acid, BH+. Ka and Kb are related: Ka x Kb = Kw = 1.0 x 10-14 pKa and pKb are also related: pKa + pKb = pKw = 14 So, if the Ka and pKa terms are used, as base strength increases: Ka decreases pKa increases


Examples of pKa Acids and bases: Acetic acid: Ka = 1.75 x 10-5; Ka = 4.76

H3C CO

OH + H2O

H3C CO

O- + H3O+

Salicylic acid: Ka = 1.06 x 10-3; Ka = 2.97

CO2H

OH

CO2

OH

salicyclic acid

+ H2O + H3O+

Phenobarbital: Ka = 3.9 x 10-8, pKa = 7.41

N H

H N

O

O

OEt

Ph

+ H2O

N-

H N

O

O

OEt

Ph

+ H3O+

Pyridine: Ka = 7.1 x 10-6, pKa = 5.15 (the pyridinium or protonated form is the acid).

N + H2O

NH+ + OH-

Question: Amines rarely give up a proton and behave like an acid. Why does the proton on phenobarbital (above) act like an acid?

Pharmaceutics (Part I) – Spring 2005 Prof. C. Thompson (SB 383; 243-4643; [email protected] Some pKa’s of interest..…please note the range in pKa, correlation with water

Compound pKa Compound pKa

sulfuric acid -9 RNH3+ (protonated primary amines) 10 to 11hydrochloric acid -7 HCO3

- 10.3 nitric acid -1.4 RSH (sulfides) 10 to 11

RCOOH (carboxylic acids) 4 to 5 water 15.7 H2CO3

6.3 ROH (alcohols) 15 to 18NH4

+ 9.2 RC(O)NH2 (amides) 17 to 18phenol 8.5 to 11 alkanes 48-50

Weak acids and bases are weak electrolytes meaning that they are only ionized to a small extent (less than 1%) in solution. Weak acids have Ka values which are << 1 and positive pKa values. Acetic acid, phenobarbital, and pyridine are examples of weak acids and bases. Most drugs are weak acids or bases.

Henderson-Hasselbalch Equation - Utility The Henderson-Hasselbalch equation can be used to determine pH, pKa and/or the relative amounts of acid and base in a solution. It can be used for both weak acids and weak bases and primarily for the following: calculating the ratio of base to acid when the pKa and pH of the solution are known. [This information can be used to prepare buffer solutions]. calculating the pH of a solution if the ratio of base to acid and the pKa are known. calculating the pKa if the ratio of base to acid and pH of the solution are known.

One of the more important features of the HH equation in Pharmacy is the ability to predict the proportions of ionized and unionized forms of a drug at a given pH. Since pH varies in vivo depending on the system/location and a one (1) pKa unit difference results in a ten-fold change in the ionization (log scale), small changes in the local environment can have a dramatic effect on uptake, distribution, net effect, metabolism, and excretion. Also important to consider is that the ionized form of a drug, for example, may not bind the receptor or target whereas the neutral form of the drug does. Careful use of the HH equation, therefore, may permit better prediction of the physiologic effect.

Pharmaceutics (Part I) – Spring 2005 Prof. C. Thompson (SB 383; 243-4643; [email protected] From the equilibrium and Ka equation:

Ka = [A-][H3O+]

[HA]

HA + H2O H3O+ + A-

(conj base) the following equation can be derived:

Ka = [A-]

[H3O+] [HA]which in log form is converted to:

log Ka = log [A-]

[H3O+] [HA]or

logKa - log[H3O+] = log [A-]

[HA]

log[H3O+] = - logKa + log [A-]

[HA]orpH = pKa + log [A-]

[HA] finally, in descriptive terms,

pH = pKa + log [conj base] [acid]

Pharmaceutics (Part I) – Spring 2005 Prof. C. Thompson (SB 383; 243-4643; [email protected] Key points:

One important thing to keep in mind is that the weak acid is the proton donor. That means the “form” or “structure” of the acid may vary as a carboxylic acid, alcohol, sulfide or even amine hydrochloride, for example.

You can rearrange the equation to solve a number of calculations, but the real issue is to keep “trends” and ratios at the forefront. For example, rearrange the equation to get an understanding of the acid to base ratio as follows:

pH - pKa = log [conj base] [acid]

In this instance, the difference between the pH and the pKa (typically defined for the drug in question) can tell you how much of one form the drug is in (ionized or unionized). Important: this difference is logarithmic! [base] % conj. pH - pKa [acid] % acid base % unionized % ionized 4 10,000/1 0.01 99.99 0.01 99.99 3 1000/1 0.1 99.9 0.1 99.9 2 100/1 1 99 1 99 1 10/1 9 91 9 91 0 1/1 50 50 50 50 -1 1/10 91 9 91 9 -2 1/100 99 1 99 1 -3 1/1000 99.9 0.1 99.9 0.1 -4 1/10,000 99.99 0.01 99.99 0.01 Note that when the pKa = pH, there will be an equal amount of ionized and ionized (because the log of zero = 1). Using the table above, one could get a “sense” for the amount of acid and base forms. So, if a drug contained a carboxylic acid group that had a pKa of 4.2, one could determine that the drug was deprotonated at physiologic pH (approx. 6.8-7.5). That is, the carboxylic acid group lost its proton to form the ionized carboxylate. However, in the stomach, at pH 1, the drug would remain in the protonated, acid form. Question: Which form of a carboxylic acid (carboxylic acid or carboxylate anion) is more likely to traverse cell membranes?


Salts of Weak Acids In chemistry classes we learn that metals like sodium, potassium and calcium are common counterions for salts of drugs that are weak acids. Salts of weak acids have the general form R-M+ where R is a carboxylic acid, phenol, imide or other ionizable group.

O

OHR OH

R N H

R

O O

Question: Why is lithium not used as a countercation for drugs that are weak acids in salt formation? Example of drug salt formation:

NH

NHO

O

OEt

Ph

NaOH

N

NHO

O

OEt

PhNa

Phenobarbital Phenobarbitalsodium

CO2HI

I

I

H3C(OC)HN C(O)NHCH3

NaOH

CO2I

I

I

H3C(OC)HN C(O)NHCH3

Na

Iothalamate Iothalamatesodium

radiologiccontrastagent

In the case of the radiologic (x-ray) contrast agent, iothalamate, the formation of the sodium salt adds significant water solubility, which allows the contrast agent to distribute evenly throughout extracellular space. There is no significant penetration into intracellular matrices and moreover, the contrast agent is cleared rapidly from the body (t ½ = 2 h).

Fun fact: There are five radiologic x-ray densities: air, fat, fluid, bone and metallic, all of which show up differently on “film.” And, as many know, many organs and tissues do not show up well on traditional x-ray, e.g., liver, spleen, kidneys, intestines, bladder, and abdominal muscles (all have similar density and ‘shadow’ each other). Certain compounds increase the radiographic contrast b/c they have x-ray absorption characteristics that afford a ‘bone-like’ or metal-like’ density. Compounds that fall into this class are iodinated aromatic compounds (bone-like density) and barium salts (metal like density). Talk about high doses! Contrast agents are administered as 100 mL of a 60% solution. Osmotoxicity!

Examples of other drugs that are salts of weak acids and their counterions: 1. Warfarin sodium Na+ 3. Fenoprofen calcium Ca+2 2. Penicillin G potassium K+


Salts of Weak Bases Most of the weak bases of interest to pharmacists are the amines and anilines. Both classes are capable of being protonated by an acid to form an amine or aniline salt. The acid used to do the protonation contains a conjugate anion that forms the anion after the amine has been protonated.

NH2

R1 NR2

R3

R1 NH2

R1 NH

R2

HA HA HA HA

R1 NH3 AR1 N

R2

H H

R1 NR2

R3H AA

NH3 A

primaryamine

secondaryamine

tertiaryamine

aniline

where R1, R2 and R3 are various alkyl/aryl groups and A is the conjugate base, e.g., Cl, Br, etc.

Amines have a pKa ranging from 9-11 and anilines have a pKa 4 to 5. Question: Why are the pKa values so different for amines and anilines? Another example of a base is the aromatic heterocycle pyridine. Its structure appears in many drugs and it readily forms salts with mineral and weak acids.

NHCl N H Cl

pyridine pyridinehydrochloride

Examples of drugs that are salts of weak bases and their counterions: Meperidine hydrochloride Cl- Dextromethorphan hydrobromide Br- Atropine sulfate SO4-2

Codeine phosphate PO4-3

Biperiden lactate CH3CHOHCOO-

Ergotamine tartrate -OOCCHOHCHOHCOO- Chlorpheniramine maleate -OOCCH=CHCOO- Benztropine mesylate CH3SO3-

Pharmaceutics (Part I) – Spring 2005 Prof. C. Thompson (SB 383; 243-4643; [email protected] Lets examine the “organic” salts a little closer.

H3C CO2H

OH

CO2H

OHHO2C

OHCO2HHO2C

lactic acid tartaric acid maleic acid

from $10-100/kg These carboxylic acids serve as proton donors to weakly basic amines. Amines comprise one of the largest classes of organic molecules used as pharmaceutically important drugs. The amine and one of these acids (for example) combine to form an amine salt, which improves its water solubility. Some other important considerations in the use of these acids include: (a) low cost, (b) high chemical purity, (c) high stereochemical purity, and (d) potential to serve as ‘dual’ or as a twofold proton donor (e.g., tartaric acid and maleic acid). Maleic acid is the cis isomer – the trans isomer is called succinic acid (succinate). You are likely to see all of these acids in advertisements for pharmaceuticals.


BUFFERS

Keywords/Concepts:

buffer buffer equation buffer capacity Objective(s): • Restore understanding of buffer systems and establish importance for use. • Develop the physicochemical relationship between weak acids and bases and buffer systems. • Gain confidence in selecting pharmaceutical buffers. • Define preliminary physiologic buffer systems. Definitions and descriptions: Buffer – A buffer is a solution that resists pH changes when acids or bases are added to the solution. Most buffer solutions consist of a weak acid and its conjugate base (salt of the weak acid). Common Examples of Weak Acid Buffer Systems.

Acetic acid Sodium acetate

Boric acid Sodium borate

Citric acid Sodium citrate

Phosphoric acid Potassium phosphate Examples of Weak Acid Buffer System.

tris(hydroxymethyl)aminomethane tris(hydroxymethyl)aminomethane hydrochloride "Tris" "Tris HCl"

OH

OH

HO ClNH2

OH

OH

HO NH3

Pharmaceutics (Part I) – Spring 2005 Prof. C. Thompson (SB 383; 243-4643; [email protected] The pH of some ‘well known’ liquids containing buffer systems: Product pH (room temp) Apple juice 3.7 Club Soda (Schwepps) 5.1 Coffee (instant) 4.7 Diphenhydramine 5.0 Distilled Vinegar 2.6 Gatorade 3.0 Ipecac 1.7 Listerine 3.9 Pepsi 2.6 Robitussin DM 2.5 Saline USP 6.8 Visine 6.3 Buffer Properties and Function: When acids or bases are added to pure water, they immediately produce H3O+ or OH- ions that decrease or increase the pH, respectively. Buffer systems resist large pH changes because added acids or bases are neutralized by the existing HA/A- system (equilibrium). Added acids are neutralized by the conjugate base (A-) which is converted to the acid (HA). Added bases are neutralized by the acid (HA), which is converted to the conjugate base (A-). Addition of acids or bases therefore change the HA/A- ratio. However……

- we know that from the Henderson-Hasselbalch (buffer) equation, changes in the HA/A- ratio will

also change the pH. However, the pH change is related to the log of the change in the HA/A- ratio. Therefore, the pH change is relatively small.

Also, the assumption here is that the concentration of buffer salts exceeds that of the acid being introduced to the solution and therefore, there is a large number of “solvation ions” (halleluiah) to handle.

Pharmaceutics (Part I) – Spring 2005 Prof. C. Thompson (SB 383; 243-4643; [email protected] The Buffer Equation(s): The buffer equation is the Henderson-Hasselbalch equation adapted to consider acids and their conjugate bases leading to solutions that are resistant to pH change. The buffer equation can be used to calculate: • the pH of a buffer solution when the HA/A- ratio is known. • the HA/A- ratio required to give a buffer of a given pH. • the pH change which results from the addition of an acid or base to a buffer solution In most cases, you will need to know the pKa of the weak acid to conduct these calculations. The following forms of the buffer equation are useful for buffer calculations: pH = pKa + log B/A (useful to calculate the pH of the buffer solution)

pH - pKa = log B/A (useful to calculate the ratio of base to acid ) pKa = pH + log A/B (useful to calculate the pKa of a buffer at a known pH) pKa - pH = log A/B (useful to calculate the ratio of acid to base) Question: Calculate the pH of a buffer solution prepared by dissolving 242 mg of Tris in 10 mL of 0.170 M HCl and diluting to 100 mL with water. [Tris: mw 121 g/mol and pKa = 8.08 for the conjugate acid] Buffer Capacity: The ability of a buffer system to resist pH changes is its buffer capacity and indicated by the buffer index (β):

β = ∆B/∆pH where: B = strong base (in molarity) ∆ = change (delta)

Buffer capacity is defined as the number of equivalents of strong base (∆B) required to cause a one-unit change in pH (∆pH ) in 1 L of solution. The greater the buffer capacity, the smaller the change in pH from the addition of a given amount of strong acid or base. Buffer capacity is dependent on the total concentration of the buffer system and on the HA/A- ratio. The buffer index number is generally experimentally derived in a manner like a titration (see Table below). For example, when 0.03 moles of NaOH was added to an acetate buffer system prepared at 0.1 M, the pH expectedly increased from 4.76 to 5.03; a change of 0.27 pH units. Therefore, the equation β = δB/∆pH = 0.03/0.27 = 1/9 = 0.11

Pharmaceutics (Part I) – Spring 2005 Prof. C. Thompson (SB 383; 243-4643; [email protected] Note that the buffer capacity is highest when the smallest number of moles of NaOH are added. Also, the buffer equation can be adapted for the addition of acids, except the pH would decrease in these experiments.

Buffer capacity is increased by the following factors:

- increasing the concentration of the buffer system components (e.g. doubling the total molar concentration of the buffer system will double the buffer capacity at a given pH).

- using equimolar concentrations of the acid (HA) and its conjugate base (A-). Buffer capacity is maximal when pH = pKa ([HA]=[A-]). Why use buffers in pharmacy? Solubility - The ionized form of a drug is more water soluble than the unionized form. Buffers can be used to maintain a drug in its ionized (salt) form for aqueous solutions. Absorption - The unionized form of a drug is more lipid soluble than the ionized form. The unionized form therefore penetrates biological membranes much more efficiently than the ionized form. buffers can also be used to maintain the drug in its unionized form. Stability - pH can affect the stability of a drug in an aqueous solution. For example, ester drugs are very susceptible to hydrolytic reactions. Buffering formulations at low pH (pH 3-5) can reduce the rate of hydrolysis. Tissue irritation - High or low pH can cause tissue irritation. Buffering a formulation to near neutral pH can reduce tissue irritation. Ophthalmic products are least irritating at pH 7-9.

Pharmaceutics (Part I) – Spring 2005 Prof. C. Thompson (SB 383; 243-4643; [email protected] Buffers and the Body:

Body fluids contain buffering agents and buffer systems that maintain pH at or near pH=7.4. Important endogenous (natural) buffer systems include carbonic acid/sodium bicarbonate and sodium phosphate in the plasma and hemoglobin, and potassium phosphate in the cells. An in vivo value of pH < 6.9 or pH > 7.8 can be life threatening.

Pharmaceutical solutions generally have a low buffer capacity in order to prevent overwhelming the bodies own buffer systems and significantly changing the pH of the body fluids. Buffer concentrations of between 0.05 and 0.5 M and buffer capacities between 0.01 to 0.1 are usually sufficient for pharmaceutical solutions.

Choosing the Right Pharmaceutical Buffer : • Choose a weak acid with pH » pKa. • Use buffer equation to calculate ratio of acid/base needed to give required pH. • Choose concentration needed to give suitable buffer capacity. • Choose available ingredients considering sterility, stability, cost, toxicity. • Use pH meter or, at least, pH indicator paper.


PHARMACEUTICAL BUFFER EXAMPLES The following are some buffer systems that are used in the formulation of pharmaceuticals. Most pharmaceutical buffers are composed of ingredients that are found in the body (e.g. acetate, phosphate, citrate):

Buffer System pKa buffer pH range

Acetic acid/sodium acetate 4.76 3.8 to 5.6 Phosphoric acid/sodium phosphate H3PO4/NaH2PO4 2.1(pK1) NaH2PO4/Na2HPO4 7.2(pK2) 5 to 8 Na2HPO4/Na3PO4 12.3(pK3) Citric acid/sodium citrate 3.1(pK1)

4.8(pK2) 1.2 to 6.6 9.2(pK3) Boric acid/sodium borate 9.2 7.8-10.6


BONDING Bonding and molecular interactions: Understanding the basic forces that hold molecules together is important for understanding how molecules interact with each other. These interactions affect solubility, stability and other properties of drugs.

BOND TYPES

Covalent Bonds - the strongest bond (arrows), worth anywhere from -40 to -110 kcal/mol in stability. Bonds in the molecule at right are all covalent even though some are single, double or triple bonds. Not shown are the carbon-hydrogen bonds that are also covalent. Although most drug interactions are non-covalent, alkylation of DNA for example in cancer chemotherapy is an example of covalent bond formation by drugs.

Ionic Bonds – a.k.a. Electrostatic interactions can be attractive or repulsive, depending on the relative charges. Like charges repel each other and opposite charges attract. For ionic bonds, opposite charges hold ions together. Electrons are not shared, but are transferred. As an example, at physiologic pH basic side chains of certain amino acids like arginine, lysine, and, in part, histidine are protonated and therefore provided a cationic environment. Acidic groups like aspartic and glutamic acid are deprotonated to give negatively charged groups. Therefore, drugs and their target receptors can be mutually attracted by opposite charges on their surfaces, e.g., cationic drug with anionic receptor site. A simple ionic attraction can provide as much as -5.0 kcal/mol and declines as the square of the distance between the charged sites. If an ionic attraction is reinforced by other interactions (H-bonding, etc) is can provide up to -10.0 kcal/mol. Ion-dipole and dipole-dipole Interactions - by virtue of varied electronegativity values when compared to carbon certain groups have an asymmetric distribution of electrons that produce dipoles. Dipolar interactions refer to unequal distribution of charge within a bond or a molecule. Dipoles exist where one atom (X) is more electronegative than another (H). The dipoles in a drug molecule, therefore, can find complementary dipole interactions in the receptor. However, since the net charge on a dipole is less than on a full ion, the interaction is generally weaker (G ranges from -1 to -7 but is usually -1 to -2).

OF O

δ+

δ+ Hδ+

δ-

δ-

δ-δδ+

δδδ+


he

pproach

.1

Hydrogen Bonds - simply a variation of a dipole-dipole interaction formed between the proton of one group and an electronegative atom of another group. In molecules of biological interest the hydrogen bonds most widely experienced are when X = O, N, S. Hydrogen bonds are usually denoted by a dashed or dotted line as in the case of the DNA bases at right sharing two hydrogen bonds:

N

NNH

N

O

NH2

H

N N

O

O

HH

What makes hydrogen bonds so unique is that hydrogen is the only atom that can carry a positive charge at physiologic pH while remaining covalently attached to a molecule. Further, hydrogen is small enough to permit close approach by another electronegative atom or molecule to accept the hydrogen bond.

There are two forms of hydrogen bonds - intramolecular (within a single molecule) and intermolecular (between two molecules). Hydrogen bonds for example, are important in maintaining the structural integrity of peptides and proteins - -helical structure and -sheets derive their structure from hydrogen bonds. The G for hydrogen bonding ranges from -1 to -7 kcal/mol but in biosystems is usually between -3 and -5 kcal/mol. Charge Transfer Complexes - this attraction comes about when molecules containing a good electron donor group (on an aromatic ring) interact with a molecule that contains an electron-withdrawing (acceptor) group. The interaction involves transfer of charge from donor to acceptor. Electron donor groups or molecules usually contain a -bond (alkene, alkyne, aromatic moiety) that bears a good electron donating group (alkyl, methoxy, OR, SR, etc.). Acceptor groups or molecules usually contain a π-bond with electron withdrawing groups attached (e.g., cyano, carbonyl, halogen, etc.). The interaction is rare and most likely found with the tyrosine residue on certain receptors (as the donor). TG for charge-transfer complexes range from -1 to -7 kcal/mol.

Hydrophobic Interactions - involve a non-polar/non-polar interaction between receptor and drug. Usually alkyl chains are found to stack as in the lipid bilayer except this is for a single interaction. It is important to consider in this interaction that each receptor and drug molecule aliphatic chain is surrounded by water molecules, and as the chains approach each the water molecules are squeezed out to permit the non-polar chains to align with each other in a "side by side" position. This orienting results in a drop in the free energy of about 0.1 to -2.0 kcal/mol.

Van der Waals or London Forces - Some atoms on mostly non polar molecules experience or exert a temporary non-symmetric distribution of electrons to cause a dipole. As these atoms aeach other in different molecules, the opposite dipoles attract forming an interaction. The G varies greatly and values from -0to -2.0 are reasonable.

R1 R2

OO

NHRR4HN

R3

δ +

δ +

δ −δ −

drug molecule

receptor target


SOLUBILITY

Solubility is the maximum concentration that can be attained by a solute in a specific solvent. A saturated solution is one in which the solvent has dissolved all of the solute that it can, and addition of further solute will not increase the concentration of the solute in the solvent. When excess solute is added to a saturated solution, an equilibrium will exist between the solid solute and the solute in solution. Solutions are the result of intermolecular attractive forces between solutes and solvents. Electrostatic, dipole-dipole (van der Waals) and hydrogen-bonding forces may be involved in solute-solvent interactions. In order for a solute to dissolve in a solvent, the intermolecular attractive forces between solute molecules and between solvent molecules must be overcome and replaced with solute-solvent attractive forces.

Pharmaceutics (Part I) – Spring 2005 Prof. C. Thompson (SB 383; 243-4643; [email protected] Solubility can be affected by temperature, pH, and the physical form of the solute. • temperature: Increasing temperature generally increases solubility. Most solutes absorb heat upon dissolution and have a positive heat of solutions. In comparison, a few solutes have a negative heat of solution which means that they give off heat during dissolution and solubility decreases with increasing temperature. • pH: Weak acids and bases are more soluble in their ionized (conjugate base or conjugate acid) forms than their unionized forms, so pH is very important for the solubility of weak electrolytes. • physical form: Non-crystalline (amorphous) solids are generally more soluble than crystalline solids. Also, different polymorphic forms, solvates and hydrates will have different solubilities. These terms will be discussed later. Water as solvent. Why is water important toward an understanding of drug action? It is the universal solvent, delivery agent, dispersing agent, nucleophile, and primary source of H+ and OH- in vivo. Therefore, water is not inert. It also is responsible for certain aspects of protein structure (conformation), the formation astructure of the lipid bilayer and other important intra- and extracellular domains, and other phenomena where H-bonding plays a role. In the figure below, some basic properties of water are illustrated: bond angles, charge distribution and H-bonding. The water molecule is not linear. Aggregates of water form easily based upon hydrogen bonding. Ice takes on a tetrahedral geometry and although the structure of ice does not play a significant role in vivo, the structure of water does vary depending on its environment and local temperature. Packets of water exist in flux, a vibrating, rotating, rapidly reorganizing environment.

nd

Solubility (water as a solvent): In part, the flux adds to the "solvent" properties of water. Due to a high dipole, water has a very high dielectric constant (Debeye = D = 80) as compared to the organic solvents acetone (D = 21) or hexane (D = 1 to 2). [Note: highly polarized solvents favor "polarized" transition states.] With a high dielectric constant, water is capable of breaking the electrostatic attraction of ions (crystal lattices) to favor hydration. When the heat of solvation is favored over the energy stored in the stability of the crystal lattice, the compound dissolves (e.g., NaCl). Recall that certain organic compounds forms "ions," for example, carboxylic acids (oxyanions) and amines (nitrogen cations). Charged species are generally more soluble in water and neutral compounds. For some neutral organic molecules a different physicochemical process is involved. As a result of solvation, the structure of the water cluster also is altered to form an envelope or "breaker" in the structure. Think of a microscopic droplet of oil surrounded by water (as represented by an organic compound). The water exerts a solvation effect on the compound and the compound exerts a "disturbance" on the H-bonded water environment. Such complexes between a solvated species and the water are called "clathrates." Compare this phenomena to a truly ionic compound in solution that is fully ionized, H-bonded and intermingled. Within this description, some organic molecules have an ionic character and fall somewhere in between an "inclusion" complex and solvation. To effect a spectrum of solubility characteristics, we ask the following question:

Pharmaceutics (Part I) – Spring 2005 Prof. C. Thompson (SB 383; 243-4643; [email protected] To what extent are organic molecules soluble in water? Consider the body as a huge separatory funnel, filled with water (aqueous layer; e.g., serum) and fat (organic layer). Where will the drug go? Why is this important? This question is important because all biochemistry processes are based upon small amounts of organic molecules dispersed in the aqueous or lipid (organic) phase or both. Circulating fluids (blood) have aqueous characteristics, but tissues (e.g., tissue membranes) and the cavities of certain proteins have an organic environment. Some trends for organic molecule solubility in water:

a. up to five carbons with one functional group are usually soluble, b. molecule with branched chains are more soluble than linear chain c. water solubility deceases with increase in molecular weight

Some nomenclature used to describe solubility that you should be familiar with:

o Hydrophobic (a.k.a lipophilic); fat phase loving, aqueous phase hating o Hydrophilic; aqueous phase loving, fat phase hating o Amphiphilic; has atomic characteristics to permit solubility in both phases

Solvent-solute interactions: As a general rule: like dissolves like so polar solvents dissolve polar solutes and non-polar solvents dissolve non-polar solutes.

Polar solvents • Polar solvents (e.g. water, methanol) readily dissolve polar solutes such as salts (ionic compounds), sugars and other polyhydroxy compounds. Water can also dissolve (to some extent) other O- and N-containing compounds including alcohols, amines, ketones, aldehydes, phenols. These compounds contain polar groups which can hydrogen bond with water. Water can also dissolve the salt (ionic) forms of weak acids and weak bases (the conjugate base and conjugate acid forms of weak acids and weak bases). Water is miscible with other liquids such as ethanol and methanol meaning that they can be mixed in any (unlimited) proportion to form a homogeneous liquid.

Non-polar solvents: • Ionic and polar solutes are poorly or not soluble in non-polar solvents because the solvent cannot reduce the attractive forces between the solute molecules (ionic or dipolar or hydrogen bonding). • Non-polar solvents can dissolve non-polar solutes through induced dipolar interactions (non-polar solutes are held together weakly unlike polar solutes)


olubility Terms:

acetone = ketone solvent.

S

Pharmaceutics (Part I) – Spring 2005 Prof. C. Thompson (SB 383; 243-4643; [email protected] THINGS TO NOTE IN TABLE 6.2.

• “Alcohol” is synonymous with ethanol.

• The salt of a drug is more

soluble in water than in ethanol

• The “neutral” form of a

drug is usually more soluble in ethanol than in water.

Lipid vs. aqueous solubility: Drug substances must exhibit some degree of both lipid and aqueous solubility. Lipid solubility for crossing biological membranes and aqueous solubility for distribution into the systemic circulation and other biological fluids. Substances that are relatively insoluble in water may exhibit poor absorption characteristics such as erratic or incomplete absorption. The graph below shows the correlation between lsolubility (log get to this later) andrug effect.

ipid/aqueous P, we’ll

d

Pharmaceutics (Part I) – Spring 2005 Prof. C. Thompson (SB 383; 243-4643; [email protected] Modification of aqueous solubility: • salt formation or structural modification • adjust pH if a liquid formulation

• use co-solvent • complexation Rate of solution: While solubility is constant at a given temperature and pressure, rate of solution can vary depending on the particle size of the solute and the agitation rate of the solution. pH, pKa and solubility: Solubility is influenced by the degree of ionization of the substance. Total aqueous solubility of an ionizable substance can be expressed as: ST = [HA] + [A-] for a weak acid ST = [B] + [BH+] for a weak base or ST = SHA + [A-] for a weak acid ST = ST + [BH+] for a weak base

where ST is the total solubility and SHA and SB are the solubilities of the unionized weak acid or weak base. For a non-ionizable substance, a non-electrolyte:

ST = SNE For a weak acid, the following equation can be derived: ST = SHA + KaSHA

[H3O+] which indicates that solubility of a weak acid increases with increasing pH (decreasing H3O+ concentration): so as ↑ pH then ↓HA, ↑A-, and ↑ST Correlation with Henderson Hasselbalch: Maximum aqueous solubility for weak acids is attained at pH-pKa » 2, where 99% is in the ionized (A-) form. Minimum solubility is at pH-pKa » -2 where 99% is in the unionized (HA) form. The logarithmic form can be used to predict the pH (pHp) below which the unionized weak acid would precipitate from solution: pHp = pKa + log ST-SHA

SHA


For a weak base, the following equation can be derived: ST = SB + [H3O+]SB

Ka which indicates that solubility of a weak base increases with decreasing pH (increasing H3O+

concentration): as the ↓ pH then the ↓B, ↑BH+, and ↑ST Maximum aqueous solubility for weak bases is attained at pH-pKa » -2, where 99% is in the ionized (BH+) form. Minimum solubility is at pH- pKa » 2 where 99% is in the unionized (B) form. The logarithmic form can be used to predict the pH (pHp)above which the weak base would precipitate from solution: pHp = pKa + log SB

ST-SB Solubility of weak acids and weak bases in buffers vs. water: • The solubilities of a weak acid and its conjugate base are identical in a buffer solution. The pH of the

buffer solution determines the B/A ratio and the solubility will be the same for either form (weak acid or conjugate base). The same is true for a weak base and its conjugate acid.

• The solubilities of a weak acid and its conjugate base are different in water. Addition of the weak acid

makes the water acidic so there is more of the unionized weak acid form present and solubility is low. Addition of the conjugate base makes the water basic so there is more of the ionized form present and solubility is high.

• The solubilities of a weak base and its conjugate acid are also different in water for a similar reason.

Addition of the weak base makes the water basic so there is more of the unionized weak base form present and solubility is low. Addition of the conjugate acid makes the water acidic so there is more of the ionized form present and solubility is high.


PARTITION COEFFICIENTS Distribution/partition coefficients: In order to produce a biological response, a drug molecule must cross a mostly lipid biological membrane. Its ability to do this depends on its lipophilicity/hydrophilicity balance and, if ionizable, its degree of ionization. Distribution coefficients: The partition coefficient (P, K) is also known as the distribution coefficient or distribution ratio. Higher P or K values indicate higher lipid solubility relative to aqueous solubility. The true distribution coefficient (K) is the concentration of a substance in an oil phase ([HA]O), such as octanol, divided by the concentration of unionized substance (e.g. weak acid or weak base) in the aqueous phase ([HA]W): K = [HA]O

[HA]W Since substances do not ionize significantly in the oil phase: [HA]O = CO (where CO is the total concentration in the oil phase). The expression can then be written: K = CO

[HA]W The apparent distribution coefficient (K’) is the experimentally observed ratio, and it includes the ionized fraction of the substance in the water phase. K’ = [CO]

[HA]W + [ A-]W

The total concentration in the aqueous phase (CW) is: CW = [HA]W + [A-]W so the apparent distribution coefficient can be expressed: K’ = CO/CW


Degree of ionization: The degree of ionization is the fraction ionized (α) in solution and is equal to: α = [A-]W [HA]W + [A-]W The degree of ionization depends on the pH of the aqueous phase and the pKa of the substance. The fraction unionized (1-α) is equal to: 1- α = [HA]W [HA]W+[A-]W The true distribution coefficient for non-ionizable substances may be expressed: K = Co/CW (since K and K’ are the same for non-ionizable substances)


DISSOLUTION The dissolution rate is the time required for a drug substance to dissolve in the fluids at the absorption site. Dissolution rate is often the rate-limiting step in the absorption process and can control the overall bioavailability of the drug from the dosage form. Dissolution is important for the bioavailability of solid dosage forms including oral capsules, tablets and suspensions and intramuscular suspensions. Methods for increasing dissolution rates: • Decrease particle size. This increases the available surface area to the dissolving fluid. [Note: In rare cases, agglomeration of the particles may occur leading to decreased dissolution rates.] • Increase solubility in the diffusion layer. The ionized form of the drug (salt of the weak acid or salt of the weak base) will have greater solubility in the diffusion layer than the unionized weak acid or weak base. (e.g. penicillin V potassium will dissolve faster than penicillin V itself). • Alter pH of dissolution medium (e.g. buffered aspirin). • Increase agitation of dissolution medium (e.g. effervescent, buffered aspirin). Noyes-Whitney equation: The Noyes-Whitney equation shows how factors such as solubility and surface area can affect dissolution rate: dM = DS(CS-C) dt h where: dM/dt = the dissolution rate D = the diffusion coefficient of the solute in the solution S = the surface area of the exposed solid h = the thickness of the diffusion layer CS = the concentration of the drug in the diffusion layer (solubility of the drug since diffusion layer is assumed to be saturated) C = the drug concentration in the bulk solution at time t. Dissolution is therefore driven by surface area (S) and the concentration gradient (CS-C).

Pharmaceutics (Part I) – Spring 2005 Prof. C. Thompson (SB 383; 243-4643; [email protected] Dissolution Equations, Calculations and Relationships - A Brief Survey. In 1897, Noyes and Whitney described the quantitative analysis that correlated the amount of time it took to dissolve a drug from solid particles. The current version of the equation is slightly modified from the original but remains based on a diffusion layer model of dissolution (Scheme 1) of drug from a particle into a large excess bulk medium. Noyes-Whitney Equation:

The rate of dissolution (dM/dt)…. dM = DS(CS-Cb) dt h where:

M = amount of drug (material) dissolved (usually mg or mmol) t = time (seconds) D = diffusion coefficient of the drug (cm2/s) S = surface area (cm2) H = thickness of the liquid film Cs & Cb = concentrations of the drug at the surface of the particle (surface = Cs) and the bulk

medium (bulk medium = Cb)

Bulk SolutionDiffusion LayerDrugParticle

h

Cs

Cb

Scheme 1. Dissolution of drug particles according to diffusion layer model.

Pharmaceutics (Part I) – Spring 2005 Prof. C. Thompson (SB 383; 243-4643; [email protected] Problem 1: Calculate the dissolution rate of a hydrophobic drug having the following physicochemical characteristics: surface area = 2.5 x 103 cm2 saturated solubility = 0.35 mg/mL (at room temperature)

diffusion coefficient = 1.75 x 10-7 cm2/s thickness of diffusion layer = 1.25 µm

[Note: need to convert to cm, so 1 µm = 1 x 10-4 cm and 1.25 x 10-4 cm] conc of drug in bulk = 2.1 x 10-4 mg/mL

dM = DS(CS-Cb) dt h dM = (1.75 x 10-7)(2.5 x 103)( 0.35 mg/mL - 2.1 x 10-4 mg/mL) dt 1.25 x 10-4 dM = (1.75 x 10-7)(2.5 x 103)( 0.349 mg/mL) dt 1.25 x 10-4 dM = (1.75 x 10-7)(2.5 x 103)( 0.35 mg/mL - 2.1 x 10-4 mg/mL) dt 1.25 x 10-4 dM = 1.53 x 10-4 = 1.22 mg/sec dt 1.25 x 10-4 NOTE!!!! The concentration in bulk solution is generally much lower than the saturated solubility. See for example above, (0.35mg/mL – 0.00021 mg/mL), and the Cb term can sometimes be ignored. Problem 2: What would the rate of dissolution be in Problem 1 if the surface area was increased to 4.3 x 104 cm2?

Answer: 4.3 x 104 cm2 is a 17.2-fold increase over the prior surface area (2.5 x 103 cm2). Therefore, multiply the rate 1.22 mg/sec x 17.2 = 21 mg/sec (approx). Or you can recalculate the rate by substituting all the values into the equation.


Diffusion - Through a Membrane: A membrane is defined as a physical barrier that separates two or more regions. For our purposes, we consider a cell membrane that is composed of phospholipids and proteins. Absorption of drugs across the stomach lining/mucosa and the blood/brain barrier are two representative examples and important for understanding drug transport into the circulation and CNS, respectively. Skin is another great example of a membrane for the entry of drugs. The transport of drug molecules through a non-porous membrane occurs by diffusion. Transport through porous membranes occurs by diffusion and convection (not covered here). Similar to dissolution, the diffusion equation is… dM = DAK(C1-C2) dt h

M = amount of drug (material) dissolved (usually mg or mmol) t = time (seconds) D = diffusion coefficient of the drug (cm2/s) A = surface area of membrane (cm2) K = oil/water partition coefficient h = thickness of the liquid film C1 - C2 = concentration gradient where C1 is the concentration of drug at donor side of membrane

and C2 is the concentration of drug in the membrane at receptor side.

C1

CrC2

Cd

h

donor compartment(high drug conc)

receptor compartment(low drug conc)

Scheme 2. Diffusion model for Fick’s Law.

However, C1 and C2 are not measured since these are values within the membrane. Typically, the gradient is measured as Cd - Cr, representing the partition at each phase, namely Ko/w = C1/Cd and Ko/w = C2/Cr. The Integrated Studies uses C1 and C2 where Cd and Cr should be used, but it doesn’t really matter for the calculation so long as you know what the gradient is. Notes: The rate of drug transport into diffusional system is predominantly dependent upon the magnitude of the concentration gradient – the other parameters are constant.

Pharmaceutics (Part I) – Spring 2005 Prof. C. Thompson (SB 383; 243-4643; [email protected] Pharmaceutics 331 Exam 1 (February 2004) Read each question carefully and for choice questions, circle only one answer. Relevant equations are provided on page 4. Double check your calculations and write out all formulae used to answer a question. Point value in parentheses. Good luck.

R-OH pKa H-OH (water) 15.8

MeOH (methanol) 16.4 Et-OH (ethanol) 16.8

iPr-OH (isopropanol) 17.2 t-Bu-OH (tert-butanol) 18

Ph-OH 10

1. Alcohols are organic ‘analogs’ of water meaning they share the identical functional group (OH) but vary in one component, the “switch” of an H for a Me, Et, etc. Explain in one or two statements, the increasing trend in pKa values (table at right) for the alcohols (5). 2. Phenols are aromatic analogs of alcohols, and their pKa values are dramatically lower than alcohols (table above). If strong electron withdrawing substituents (e.g., nitro) were added to the phenyl ring of phenol, the pKa value of the substituted phenol would: INCREASE (LESS ACIDIC) / DECREASE (MORE ACIDIC). Choose one of the underlined words (4). 3. As the strength of a weak acid increases the (4): (a) Ka increases (b) [A-] increases (c) [HA] increases (d) answers (a) and (b) (e) answers (b) and (c) 4. For a thiol-containing drug (weak acid donor), when the pH of the solution is adjusted to equal the pKa, there will be (4):

(a) precipitation of the drug (b) a greater amount of the ionized form of the drug (c) an equal amount of ionized and ionized (d) a greater amount of the unionized form of the drug

5. Tartaric acid is frequently used to react with an amine to form a “tartrate salt.” List three possible advantages to using tartaric acid rather than a mineral acid (e.g., HCl, H2SO4, etc.) form amine salts. The structure of tartaric acid is (6):

CO2H

OHHO2C

OHtartaric acid

Pharmaceutics (Part I) – Spring 2005 Prof. C. Thompson (SB 383; 243-4643; [email protected] 6. Plausible reasons that buffers resist change in pH upon addition of acids or bases are (4): (a) added acids and bases are too dilute to effect change

(a) the buffer anions (conjugate bases) are typically bigger and neutralize acids quickly (b) added acid/base change the pH but the change is logarithmically related (c) answers (a) and (b) (d) answers (b) and (c)

7. Calculate the buffer capacity for a 0.3 M phosphoric acid/sodium phosphate buffer that has an initial pH 2.1 and changes to pH 2.2 upon the addition of 50 mmoles of NaOH (5). 8. Why is the molarity of the buffer system not found in the calculation of buffer capacity? Is it unimportant (5)?

9. Three classifications of organic compounds are polar non-protic, polar protic and non polar (6).

(a) Which class is most likely to cross membranes? _____________________

(b) Which class is most likely to dissolve in saliva and serum? ___________________

(c) Do these “classifications” directly correlate with lipophilic/hydrophobic properties? Briefly explain.

10. Deleted 11. An example of a chemical bond that uses the principle of dipole-dipole interaction is (4):

(a) covalent bonding (b) hydrogen-bonding (c) van der Waals bonding (d) answers a and b (e) answers b and c

12. The solubility of amine salts in aqueous systems is effected by the (4):

(a) temperature (b) buffer/pH (c) structure of the anion (e.g., Cl-, sulfate, maleate, etc.) (d) all of the above


18.

13. The unionized form of a drug (4): (a) is less water soluble (b) is less able to traverse membranes (c) has a Ko/w > 1. (d) answers (a) and (b) (e) answers (a) and (c)

14. If sulfadiazine is sparingly soluble in water, sulfadiazine sodium is likely to be ____________ in water (4):

(a) insoluble (b) slightly soluble (c) freely soluble (d) sparingly soluble

15. The maximum aqueous solubility for weak acids is attained when (4):

(a) the pH value exceeds the pKa value by more than 2 (b) the pH = pKa (c) the pH value is less than the pKa by more than 2

16. When 1.0 g of loratidine was added to a separatory funnel containing 50 mL of water and 50 mL of octanol, it completely dissolved in the mixture. Analysis of the octanol solution showed a concentration of loratidine = 16 mg/mL. Calculate the partition coefficient (K) for loratidine (5):

17. Calculate the dissolution rate of cimetidine having the physicochemical characteristics (6): surface area = 2.5 x 104 cm2 saturated solubility = 0.60 mg/mL (at room temperature) diffusion coefficient = 2.5 x 10-7 cm2/s thickness of diffusion layer = 1.25 µm (1.25 x 10-4 cm) conc of drug in bulk = 2.1 x 10-4 mg/mL

If the drug is a weak acid and the pKa is more than the pH, the ionized species is more / less than 50% (4). 19. Calculate the theoretical solubility S of morphine (S0 = 0.0005 M) at pH 3. (morphine pKa = 7.87; Ka 1.35 x 10-8) (6):

Lecture Notes Pharmaceutics 2005

Documents

Transcript of Lecture Notes Pharmaceutics 2005