1. Physical Pharmacy
Najmadin H Mohammad

2. ARRHENIUS THEORY OF ELECTOLYTIC DISSOCIATION
An electrolyte, when dissolved in water, breaks up into two types of charged particles, one carrying a positive charge and the other a negative charge. These charged particles are called ions. Positively charged ions are termed cations and negatively charged as anions.
AB > A+ + B-
NaCl > Na+ + CL-
K2SO > 2K+ + SO4-

3. Continue.
The process of splitting of the molecules into ions of an electrolyte is called Ionization. The fraction of the total number of molecules present in solution as ions is known as degree of ionization or degree of dissociation.
Ions present in solution constantly re-unite to form neutral molecules and, thus, there is a state of dynamic equilibrium between the ionized and non-ionized molecules, i.e. AB <----> A + B-
Then [A + ][B - ] /[AB] =K

4. Continue.
K is known as ionization constant. The electrolytes having high value of K are termed strong electrolytes and those having low value of K as weak electrolytes.
When an electric current is passed through the electrolytic solution, the positive ions (cations) move towards cathode and the negative ions (anions) move towards anode and get discharged, i.e., electrolysis occurs.
• The ions are discharged always in equivalent amounts, no matter what their relative speeds are.

5. Drugs and Ionization
Some drugs, such as anionic and cationic antibacterial and antiprotozoal agents, are more active when in the ionic state.
• Other compounds, such as the hydroxybenzoate esters (parabens) and many general anaesthetics, bring about their biologic effects as nonelectrolytes.
• Still other compounds, such as the sulfonamides, are thought to exert their drug action both as ions and as neutral molecules.

6. Ionic strength
The ionic strength of a solution is a measure of the concentration of ions in that solution.
Where ci is the molar concentration of ion, zi is the charge number of that ion, and the sum is taken over all ions in the solution.
For a 1:1 electrolyte such as sodium chloride, the ionic strength is equal to the concentration, but for MgSO4 the ionic strength is four times higher.

7. Example /Ionic Strength of a Solution
A buffer contains 0.3 mole of K2HPO4 and 0.1mole of KH2PO4 per litter of solution. Calculate the ionic strength of the solution. The concentrations of the ions of K2HPO4 are [K+] = 0.3 × 2 and [HPO4^2-] = 0.3. The values for KH2PO4 are [K+] = 0.1 and [H2PO4-] = 0.1.

8. The Debye-Hückel Theory
The Debye-Hückel Theory states that strong electrolytes are completely ionized in dilute solution and that the deviations of electrolytic solutions from ideal behaviour are due to the electrostatic effects of the oppositely charged ions.
• :the activity coefficient, z: ion valence, μ :ionic strength, A:a factor that depends only on the temperature and the dielectric constant of the medium, for water is approximately equal to 0.51

9. Example /Mean Ionic Activity Coefficient
Calculate the mean ionic activity coefficient for M atropine sulphate (1:2 electrolyte) in an aqueous solution containing 0.01 M NaCl at 25°C. For water at 25°C, A is 0.51.
• Ans./ Because the drug is a uni-bivalent electrolyte, z1z2 = 1 ×2 = 2.

10. Osmolality
Although osmotic pressure classically is given in atmospheres, in clinical practice it is expressed in terms of osmols (Osm) or milliosmols (mOsm).
A solution containing 1 mole (1 g molecular weight) of a nonionizable substance in 1 kg of water (a 1 m solution) is referred to as a 1-osmolal solution. It contains 1 osmol (Osm) or 1000 milliosmols (mOsm) of solute per kilogram of solvent.
For an electrolyte that dissociates into ions in a dilute solution, osmolality or milliosmolality can be calculated from
where i is approximately the number of ions formed per molecule and mm is the millimolal concentration.

11. Example /Calculating Milliosmolality
What is the milliosmolality of a m solution of potassium bromide? i =
0.120 m = 120 millimolal

12. Osmolarity
Osmolarity = measured osmolality × (solution density in g/ml – anhydrous solute conc. In g/ml)
• Example/ A 30-g/liter solution of sodium bicarbonate contains g/mL of anhydrous sodium bicarbonate. The density of this solution was found to be g/mL at 20°C and its measured milliosmolality was mOsm/kg. Convert milliosmolality to milliosmolarity.

13. Osmolarity = (Measured osmolality)x (Solution density in g/mL — Anhydrous solute concentration in g/mL)

14. Ionic Equilibria

15. Modern theory of Acid, Base and Salts
Acid : is a substance that liberates hydrogen ions (H+).
Base: is a substance that supplies hydroxyl ion on dissociation.
Acid and base neutralize each other
Acid and base react with each other to form salts
Acid and bases change the colours of certain indicators.

16. Bronsted- Lowry theory
According to this theory :
Acid is a substance, charged or uncharged, that capable of donating a proton
Base is a substance, charged or uncharged, that capable of accepting proton from an acid.
The relative strengths of acids and bases are measured by the tendencies of these substances to give up and take on proton

17. Cont. If we consider the following reaction: HCl + H2O  H3O+ + Cl-
CH3COOH + H2O  CH3COO- + H3O+
HCl is a strong acid in water since it gives its proton readily
Acetic acid is a weak acid because it gives its proton only to a small extent.
HCl is weak acid in glacial acetic acid and acetic acid is strong acid in liquid ammonia.

18. THE STRENGTH OF AN ACID DEPENDS NOT ONLY ON ITS ABILITY TO GIVE UP PROTON, BUT ALSO ON THE ABILITY OF SOLVENT TO ACCEPT PROTON, THIS IS CALLED BASIC STRENGTH OF THE SOLVENT.

19. Classification of solvent:
Protophilic: or basic solvent is one that is capable of accepting protons from the solute such as acetone, ether and liquid ammonia
Protogenic: is a proton- donating compound and its represented by acids such as formic acid, acetic acid and liquid HCl
Amphiprotic: act as both proton accepting and proton donating solvent, such as water, and alcohol
Aprotic: neither accepting nor donate protons, its neutral and useful for study reaction of acids and bases free of solvent effect such as hydrocarbons.

20. In the reaction of HCl with water HCl Is acid and water act as base
HCl + H2O  H3O+ + Cl-
NH3 + H2O  NH4+ + OH-
A conjugated acid-base pair consists of two species in an acid-base reaction, one acid one base, that differ by the loss or gain of proton.
Water act as AMPHIPROTIC species that act as acid or base depending on the other REACTANT.

21. Example Identify the acid base species in the following reaction
CO H2O  HCO3- + OH-

22. Relative strength of acids and bases
The strongest acids have the weakest conjugated bases and the strongest bases have the weakest conjugated acids

23. Acid- base equilibria:
Equilibrium: may be defined as a balance between two opposing forces or action.
Ionization of weak acids
Acidity constant for weak acid, HB may be expressed as the following
HB + H2O H30+ +B-
Ka= [H3O+] [B-]/ [HB]
[H3O+] = 𝐾𝑎𝑐

24. Ionization of weak bases
B- + H2O  OH- + BH+
Kb = [OH-] [BH+]/[B-]
[OH-]= 𝐾𝑏𝑐
Ex: The basicity or ionization constant Kb for morphine base is 7.4 x 10~7 at 25°C. What is the hydroxyl ion concentration of a M aqueous solution of morphine?

25. Ionization of water Water undergo self-ionisation Autoprotolysis
Since water act as acid and base
H2O + H2O  H3O+ + OH-
The extent of this reaction is very small
The equilibrium constant expression for this reaction is:
Kc= [H3O+][OH-]/[H2O]2
Since The concentration of water is essentially constant
[H2O]2 Kc = Kw (constant)
so: Kw = [H3O+] [OH-]

26. The equilibrium value of the ion product [H3O+] [OH-] are called ion-product constant of water
Kw= [H3O+] [OH-]= 1.0*10^-14 at 25C0
By using Kw we can calculate the concentration of [H3O+] [OH-] in pure water
[H3O+] [OH-]= 1.0*10^-14
But in pure water [H3O+]=[OH-]
Therefore [H3O+]=[OH-] = 1.0 *10^-7 M
Note that if u add acid or base to water the concentration of [H3O+] [OH-] will no longer be equal.

27. Relationship b/w Ka and Kb
We said : HB + H2O H30+ +B-
Ka= [H3O+] [B-]/ [HB]
And for : B + H2O  OH- + BH+
Kb = [OH-] [BH+]/[B-]
By multiplying
Ka * Kb = [H3O+] [B−] [HB] * [OH−] [BH+] [B−]
=[OH-][H3O+]= Kw
Therefor Ka * Kb = Kw

28. Example
Ammonia has a Kb of 1.74 x 10-5 at 25°C Calculate Ka for its conjugate acid, NH4+.
Ka * Kb = Kw
Ka = kw/kb
= 1.0* 10^-14 / 1.74 * 10 ^-5 = 5.57* 10^-10

29. Ionization of polyprotic acids and bases
Acids that donate a single proton and bases that accept a single proton are called MONOPROTIC ELECTROLYTES.
A POLYPROTIC (POLYBASIC) ACID is one that is capable of donating two or more protons, and a polyprotic base is capable of accepting two or more protons.
Ex: H3PO4+ H2O  H3O+ +H2PO4-

31. IN GENERAL, THE SECOND IONIZATION CONSTANT, KA2, OF POLYPROTIC ACID IS MUCH SMALLER THAN THE FIRST KA1
IN THE TRIPROTIC ACIDS, THE THIRD KA3 IS MUCH SMALLER THAN KA2

32. Calculating pH
pH is defined as the negative logarithm of the molar hydronium-ion concentration
pH= -log [H+]
In the same manner we can also define POH:
pOH= -log[OH-]

33. Example
Ex: the hydronium ion concentration of a 0.05 M solution of HCl is M what is the PH of this solution?

34. pK and pOH We know that Kw= [H3O+] + [OH-]= 1.0* 10^-14 Therefore
-log {[H+][OH-]}= -log 1* 10^-14
-log [H+]+ -log [OH-]= -log 1* 10^-14
PH+POH= 14.00

35. Acidity Constant
Is One of the most important property of the a drug molecules
This can be related to the physiologic and pharmacologic activity, solubility, rate of solution, extent of binding, and rate of absorption of the drug

36. Buffered and Isotonic solution

37. Buffer is a compounds or mixture of compounds, that by their presence in solution resist changes in PH upon addition of small quantity of acid or alkali.
The resistance to a change in PH is known as buffer action
If to water or a solution of NaCl, a small amount of strong acid or base is added, the PH is altered considerably, such system have no buffer action.
A combination of a weak acid and its conjugate base (its salt), or a weak base and its conjugate acid act as buffer.

38. Example
If we add 1 ml of 0.1 N HCl into 100ml of water the pH is reduced from 7 → 3.
If strong acid is added to 0.01M solution containing equal amount of acetic acid and sodium acetate, the PH is changed only 0.09pH unit, because the base Ac- ties up the hydrogen ions according to this reaction:
Ac- + H3O+  HAc + H2O.

39. The buffer equation For weak acid and its salt:
the PH of buffer solution and change in the PH upon addition of acid or base can be calculated by buffer equation
Ka = [
[H3O+]= Ka by taking log of both sides
-log [H3O+]= -log Ka- log [acid] + log [salt]
PH=Pka+log {Handerson-Hasselbalch equation} [for buffer]

40. pH= ½ Pka – ½ log C for pH of acid only
Example: what is the pH of 0.1M acetic acid solution, Pka =4.76?
What is pH after enough sodium acetate has been added to make the solution 0.1M with respect to this salt?
pH= ½ Pka – ½ log C = 2.88
pH = Pka + log [salt]/[acid]
pH= log 0.1/0.1 = 4.76

41. Buffer equation for a weak base and its salt
Buffer solution 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 often affected by temp changes.
PH = PKw-PKb + log [base]/[acid]

42. Example:
example/ what is the PH of a solution containing 0.1 mole of ephedrine and 0.01 mole of ephedrine hydrochloride per litter of solution, the PKb of ephedrine = 4.64?

43. Activity coefficients and the buffer equation
A more exact treatment of buffers begins with the replacement of concentrations by activities in the equilibrium of a weak acid
PH=Pka + log [salt]/[acid] + log ƳAc-
For aqueous solution of a univalent ion at 25C0 having an ionic strength not greater than about 0.1 or 0.2, A= 0.51 so
PH=Pka + log [salt]/[acid] -

44. Example
A buffer contain 0.05 mole per litter of formic acid and 0.1 mole per litter of sodium formate, the Pka of formic acid is 3.75, the ionic strength of the solution is 0.1, compute the PH (a) with and (b) without consideration of the activity coefficient correction?
A- PH=Pka + log [salt]/[acid] -
PH=3.75+ log 0.1/0.05 – 0.51/1+
PH= 3.93
B- PH=Pka + log [salt]/[acid]
PH= 4.05

45. Some factors affect the PH of buffer solution
addition of neutral salts to buffers changes the PH of the solution by altering the ionic strength
dilution change both ionic strength and PH of buffer
Addition of water in moderate amount while not changing in the PH cause a small +ve or –ve deviation bcoz itself can act as a weak acid or base
Temperature also influences the PH of buffer

46. Buffer capacity
The magnitude of the resistance of a buffer to PH changes is referred to as the buffer capacity β
Buffer capacity is a ratio of 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 per liter of strong base added to the buffer solution to produce a PH change of ∆PH.

47. Approximate calculation of buffer capacity:
the changes in concentration of the salt and the acid by the addition of a base are represented in the buffer equation as the following: PH= Pka + log 𝑠𝑎𝑙𝑡 +[𝑏𝑎𝑠𝑒] 𝑎𝑐𝑖𝑑 −[𝑏𝑎𝑠𝑒]

48. A more exact equation for buffer capacity
β= 2.3 C Ka [H3O+]/{Ka+ [H3O+]}^2
{C= is the sum of molar conc. Of weak acid and its salts}
Ex:At a hydrogen ion concentration of 1.75 x 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 mole/liter, and the dissociation constant is 1.75 x 10-5

49. Buffer in the pharmaceutical and biological system
In vivo biologic buffer system
Blood maintained at a PH of about 7.4 by the so-called primary buffer in the plasma and the secondary buffers in the erythrocytes.
In plasma: carbonic acid/ bicarbonate and sodium salt of phosphoric acid as buffers
In erythrocytes: the two buffer system consist of haemoglobin/ oxyhemoglobin and acid/alkali potassium salts of phosphoric acid.

50. Usually when PH of blood goes below 6. 9 or above 7
Usually when PH of blood goes below 6.9 or above 7.8, life is in serious danger. As in case of diabetic coma.
Lacrimal fluid or tear, have great degree of buffer capacity , which allow dilution 1:15 with neutral D.W before an alteration of PH noticed
PH of tear is about 7.4, with range of 7 to 8 or slightly higher.

51. Urine : in 24 hr collection of urine in adult have PH of 6
Urine : in 24 hr collection of urine in adult have PH of 6.0 units, it may be as low as 4.5 or as high as 7.8.
When PH is low means hydrogen ion is excreted by the kidney and conversely when PH of urine above 7.4 hydrogen ion retained by action of kidney in order to retain the PH in normal range.

52. Pharmaceutical buffer
Buffer is more commonly used in the preparation of ophthalmic solution
Solutions to be applied to tissues or administered parentally are liable to cause irritation if their pH is greatly different from the normal pH of the relevant body fluid.
Tissue irritation, due to large pH differences between the solution being administered and the physiologic environment in which it is used

53. Buffered Isotonic Solutions
pharmaceutical solutions that are meant for application to delicate membranes of the body should also be adjusted to approximately the same osmotic pressure as that of the body fluids.
Isotonic solutions cause no swelling or contraction of the tissues with which they come in contact and produce no discomfort when instilled in the eye, nasal tract, blood, or other body tissues.

54. The need to achieve isotonic conditions with solutions to be applied to delicate membranes is dramatically illustrated by mixing a small quantity of blood with aqueous sodium chloride solutions of varying tonicity.
For example, if a small quantity of blood, defibrinated to prevent clotting, is mixed with a solution containing 0.9 g of NaCl per 100 mL, the cells retain their normal size.
The solution has essentially the same salt concentration and hence the same osmotic pressure as the red blood cell contents and is said to be isotonic with blood.

55. If the red blood cells are suspended in a 2
If the red blood cells are suspended in a 2.0% NaCl solution, the water within the cells passes through the cell membrane in an attempt to dilute the surrounding salt solution until the salt concentrations on both sides of the erythrocyte membrane are identical.
This outward passage of water causes the cells to shrink and become wrinkled.
The salt solution in this instance is said to be hypertonic with respect to the blood cell contents

56. Finally, if the blood is mixed with 0
Finally, if the blood is mixed with 0.2% NaCl solution or with distilled water, water enters the blood cells, causing them to swell and finally burst, with the liberation of hemoglobin
This phenomenon is known as hemolysis, and the weak salt solution or water is said to be hypotonic with respect to the blood.
the red blood cell membrane is not impermeable to all drugs; that is, it is not a perfect semipermeable membrane.

57. RBC: will permit the passage of not only water molecules but also solutes such as urea, ammonium chloride, alcohol, and boric acid.
A 2.0% solution of boric acid has the same osmotic pressure as the blood cell contents when determined by the freezing point method and is therefore said to be isosmotic with blood. The molecules of boric acid pass freely through the erythrocyte membrane, however, regardless of concentration. As a result, this solution acts essentially as water when in contact with blood cells.

59. Osmolality and osmolarity are colligative properties that measure the concentration of the solutes independently of their ability to cross a cell membrane.
Tonicity is the concentration of only the solutes that cannot cross the membrane since these solutes exert an osmotic pressure on that membrane.
Tonicity is not the difference between the two osmolarities on opposing sides of the membrane.

60. A solution might be hypertonic, isotonic, or hypotonic relative to another solution.
For example, the relative tonicity of blood is defined in reference to that of the red blood cell (RBC) cytosol tonicity.
As such, a hypertonic solution contains a higher concentration of impermeable solutes than the cytosol of the RBC; there is a net flow of fluid out of the RBC and it shrinks
The concentration of impermeable solutes in the solution and cytosol are equal and the RBCs remain unchanged, so there is no net fluid flow
A hypotonic solution contains a lesser concentration of such solutes than the RBC cytosol and fluid flows into the cells where they swell and potentially burst.

61. Measurement of Tonicity
The tonicity of solutions can be determined by one of two methods.
First, in the hemolytic method, the effect of various solutions of the drug is observed on the appearance of red blood cells suspended in the solutions.
The second approach used to measure tonicity is based on any of the methods that determine colligative properties by Calculating Tonicity Using Liso Values

62. ∆Tƒ= L C { L = Liso for isotonic solution}
The freezing point of the blood is -0.52
For 0.9% NaCl which is isotonic with body fluid
∆Tƒ= Liso C
0.9g ml
x ml  x = 9g
N= 9/58.5 g/mol = 0.154
Nacl is univalent ion Liso =3.4
∆Tƒ= 3.4* 0.154= 0.52 so FP= so 0.9% NaCl is isotonic with blood.

63. example
What is the freezing point lowering of a 1% solution of sodium propionate (molecular weight 96)? Because sodium propionate is a uni-univalent electrolyte, its Liso value is 3.4. The molar concentration of a 1% solution of this compound is
= 3.4 * = 0.35 C

64. Methods of Adjusting Tonicity and pH
the methods are divided into two classes:
In the class I methods:
Sodium chloride or some other substance is added to the solution of the drug to lower the freezing point of the solution to -0.52°C and thus make it isotonic with body fluids.
Under this class are included:
the cryoscopic method and thesodium chloride equivalent method.

65. In the class II methods
water is added to the drug in a sufficient amount to form an isotonic solution. The preparation is then brought to its final volume with an isotonic or a buffered isotonic dilution solution.
Included in this class are:
the White–Vincent method and the Sprowls method.

66. Class I Methods Cryoscopic Method
How much sodium chloride is required to render 100 mL of a 1% solution of apomorphine hydrochloride isotonic with blood serum? From Table it is found that a 1% solution of the drug has a freezing point lowering of 0.08°C. To make this solution isotonic with blood, sufficient sodium chloride must be added to reduce the freezing point by an additional 0.44°C (0.52°C °C). In the freezing point table, it is also observed that a 1% solution of sodium chloride has a freezing point lowering of 0.58°C. By the method of proportion,
1%/x = 0.58C/0.44C x = 0.76%.
Thus, 0.76% sodium chloride will lower the freezing point the required 0.44°C and will render the solution isotonic. The solution is prepared by dissolving 1.0 g of apomorphine hydrochloride and 0.76 g of sodium chloride in sufficient water to make 100 mL of solution.

67. Sodium Chloride Equivalent Method
The sodium chloride equivalent E or, as referred to by these workers, the ―tonicic equivalent‖ of a drug is the amount of sodium chloride that is equivalent to (i.e., has the same osmotic effect as) 1 g, or other weight unit, of the drug.
E = 17 Liso /MW

68. A solution contains 1. 0 g of ephedrine sulfate in a volume of 100 mL
A solution contains 1.0 g of ephedrine sulfate in a volume of 100 mL. What quantity of sodium chloride must be added to make the solution isotonic? How much dextrose would be required for this purpose? The quantity of the drug is multiplied by its sodium chloride equivalent, E, giving the weight of sodium chloride to which the quantity of drug is equivalent in osmotic pressure:
Ephedrine sulfate= 1.0g* 0.23= 0.23g
The ephedrine sulfate has contributed a weight of material osmotically equivalent to 0.23 g of sodium chloride. Because a total of 0.9 g of sodium chloride is required for isotonicity, 0.67 g ( g) of NaCl must be added.

69. If one desired to use dextrose instead of sodium chloride to adjust the tonicity, the quantity would be estimated by setting up the following proportion. Because the sodium chloride equivalent of dextrose is 0.16,

70. Class II Methods White–Vincent Method
computing tonicity involve the addition of water to the drugs to make an isotonic solution, followed by the addition of an isotonic or isotonic- buffered diluting vehicle to bring the solution to the final volume.
V = w * E * 111.1
where V is the volume in milliliters of isotonic solution that may be prepared by mixing the drug with water, w is the weight in grams of the drug given in the problem, and E is the sodium chloride equivalent obtained from Table. The constant, 111.1, represents the volume in milliliters of isotonic solution obtained by dissolving 1 g of sodium chloride in water

71. Make the following solution isotonic with respect to an ideal membrane:
The drugs are mixed with water to make 18 mL of an isotonic solution, and the preparation is brought to a volume of 100 mL by adding an isotonic diluting solution.

72. The Sprowls Method
This method depend on V which obtained from White–Vincent Method. This method usually used for ophthalmic and parenteral preparation