Buffered and Isotonic Solutions

Physical Pharmacy Lecture 8 Buffered and Isotonic Solutions Assistant Lecturer in Pharmaceutics

Overview Buffered Solutions Definition Buffer Equation Buffer Capacity Buffer in Biological Systems Pharmaceutical Buffers Isotonic Solutions Introduction Isosmotic Solutions Isotonic Solutions Methods of Adjusting Tonicity

Learning Objectives 1. Understand the common ion effect. 2. Understand the relationship between ph, pka, and ionization for weak acids and weak bases. 3. Apply the buffer equation, also known as the Henderson Hasselbalch equation, for a weak acid or base and its salt. 4. Understand the concept of buffer capacity and its calculations. 5. Discuss the relationship between buffer capacity and ph on tissue irritation. 6. Describe the relationship between ph and solubility. 7. Describe the concept of tonicity and its importance in pharmaceutical systems. 8. Calculate solution tonicity and tonicity adjustments.

Buffered Solutions Definition Buffer Equation Buffer Capacity Buffer in Biological Systems Pharmaceutical Buffers

Definition Buffers are mixtures of compounds that resists changes in ph upon the addition of small quantities of acid or alkali. A buffer is composed of a weak acid (HA) and its salt (conjugate base A ) or a weak base (B) and its salt (conjugate acid BH + ).

Buffer Equation The ph of a buffer solution can be calculated by use of the buffer equation. E.g. When sodium acetate (NaAc) is added to acetic acid (HAc), the salt and the acid have an ion in common ( Ac ). HAc H 3 O + + Ac NaAc Na + + Ac K a for the weak acid is disturbed because the Ac supplied by the salt increases the [Ac ] term in the numerator: K a = H 3 O + [Ac ] [HAc]

Buffer Equation To reestablish the constant K a, H 3 O + is decreased, by shifting the equilibrium in the direction of the reactants (the ionization of acetic acid is repressed). HAc + H 2 O H 3 O + + Ac K a = H 3O + [Ac ] [HAc] Since the weak acid is slightly ionized, [HAc]= initial [acid] Since most Ac comes from the salt, [Ac ] = [Salt] K a = H 3O + [Salt] [acid] By rearranging the equation and using the logarithmic form: ph = pka +log [salt] [acid]

Buffer Equation The previous equation is known as the buffer equation or the Henderson Hasselbalch equation. For a weak acid (HA) and its salt (S): ph = pka +log [S] [HA] For a weak Base (B) and its salt (S): ph = pka +log [B] [S] Buffer solutions are not commonly prepared from weak bases and their salts because of: 1. Volatility and instability of the bases 2. Dependence of their ph on pk w, which is often affected by temperature changes.

Buffer Equation Example 1 What is the ph of a buffer solution containing 0.1 M acetic acid and 0.1 M sodium acetate? (Ka for acetic acid = 1.75 10 5 ) pk a = log K a pk a = log 1.75 10 5 = 4.76 ph = pka +log [S] [HA] ph = 4.76 +log [0.1] [0.1] = 4.76

Buffer Equation Example 2 How much sodium acetate (mol) should be added to 100 ml of 0.1 mol/l acetic acid solution to prepare a buffer of ph 5.2? pka for acetic acid = 4.76 ph = pka +log [S] [HA] 5. 2 = 4. 76 +log [S] [0. 1] S = 2.754 mol/l n sodium acetate = M v = 2.754 0.01 = 0.02754 mol

Buffer Equation Example 3 What is the ph of a solution containing ephedrine 0.1 M and ephedrine HCl 0.01 M? (K b for ephedrine = 2.3 10-5 ) pk b = log K a pk b = log 2.3 10 5 = 4.64 pk a = pkw pkb pk a = 14 4.64 = 9.36 ph = pka + log [B] [S] ph = 9.36 + log 0.1 0.01 = 10.36

Buffer Capacity Definition A weak acid and its conjugate base have buffering capacity because A ions remove the added H + as undissociated weak acid, while HA remove the added OH ions as water: A + H 3 O + H 2 O + HA HA + OH H 2 O + A

Buffer Capacity Definition A weak base and its conjugate base have buffering capacity because B ions remove the added H + as undissociated weak acid, while BH + remove the added OH ions as water: B + H 3 O + H 2 O + BH + BH + + OH H 2 O + B Buffer capacity is the quantity of strong acid or base that must be added to change the ph of one liter of buffer solution by one ph unit.

Buffer Capacity Approximate Equation Koppel and Spiro and Van Slyke devised an approximate equation for calculating buffer capacity (β): β = ΔB ΔpH ΔB: number of moles of strong acid or base per liter of buffer. ΔpH: change in ph. When one of the buffer components is depleted completely, the solution lose its buffering capacity and can no longer resist the change in ph

Buffer Capacity Exact Equation Koppel and Spiro and Van Slyke developed a more exact Equation for calculating buffer capacity: β = 2. 3C K a [H 3 O + ] (Ka + H 3 O + 2 C is the total buffer concentration (the sum of the molar concentrations of the acid and the salt). This equation allows the calculation of buffer capacity at any ph (even when no acid or base has been added to the buffer). The equation shows that an increase in the concentration of the buffer components ( C ) results in a greater buffer capacity (β).

Buffer Capacity Exact Equation Example At a hydrogen ion concentration of 1.75 10 5, what is the capacity of a buffer containing 0.10 mole each of acetic acid and sodium acetate per liter of solution? (Ka = 1.75 10 5 ) C = [Acid] + [Salt] = 0.1+ 0.1 = 0.20 mole/liter K a [H 3 O + ] β = 2. 3C (Ka + H 3 O + 2 β = 2. 3 0.2 (1.75 10 5 )(1.75 10 5 ) = 0.115 mol/l (1.75 10 5 ) + (1.75 10 5 ) 2

Buffer Capacity Maximum Buffer Capacity The buffer capacity depends on: (a) the value of the ratio [Salt]/[Acid], (buffer capacity increases as the ratio approaches 1) (b) The total buffer concentration (buffer capacity increases as the salt and acid concentrations are increased).

Buffer Capacity Maximum Buffer Capacity The maximum buffer capacity occurs where ph = pk a, or, in equivalent terms, where ph = p K a (or [H3O+] = K a ). β max = 0. 576 C Where C is the total buffer concentration Buffer capacity for buffer solution of weak acid with pka of 4.76

Buffer Capacity Maximum Buffer Capacity Example What is the maximum buffer capacity of an acetate buffer with a total concentration of 0.020 mole/liter? β max = 0. 576 C β max = 0.576 0.02 = 0.012

Buffer in Biological Systems Some body fluids have natural buffer capacity: 1. ph of tears is 7-8 with higher buffer capacity so that a reasonably wide ph range of medicines can be tolerated. 2. ph of blood is maintained at approximately 7.4 by buffer component in the plasma (bicarbonate and phosphate buffers) and erythrocytes (hemoglobin and phosphate buffers). H 2 CO 3 HCO 3 + H + H 2 PO 4 HPO 4 2 + H + HbH + + O 2 + O 2 Hb + H + Hb: hemoglobin, O 2 Hb: oxyhemoglobin

Pharmaceutical Buffers Buffer solutions are widely used to adjust ph of aqueous pharmaceutical solutions to ensure: 1. Tissue irritation prevention 2. Optimum therapeutic effect 3. Maximum drug stability 4. Maximum drug solubility

Pharmaceutical Buffers Tissue Irritation Prevention Solutions to be applied to delicate tissues or administered parenterally are liable to cause irritation if their ph is greatly different from the normal ph of the relevant body fluid. In case there is a large difference in ph, Tissue irritation can be minimized if the volume and buffer capacity of the solution is lower than the volume and buffer capacity of the physiologic fluid

Pharmaceutical Buffers Maximum Therapeutic Effect vs Stability The undissociated form of a weakly acidic or basic drug often has a higher therapeutic activity than that of the dissociated salt form because they can penetrate body membranes readily due to their lipid solubility. The ph for maximum stability of a drug for ophthalmic use may be far below that of the optimum physiologic effect. Under such conditions, the solution of the drug can be buffered at a low buffer capacity and at a ph that is between that of optimum stability and that for maximum therapeutic action. When the solution is instilled in the eye, the tears bring the ph to about 7.4, converting the drug to the physiologically active form.

Pharmaceutical Buffers Maximum Therapeutic Effect vs Solubility The ph of the solution can affect the solubility of the drug. At a low ph, a base is predominantly in the ionic form, which is usually very soluble in aqueous media. As the ph is raised, more undissociated base is formed, which has poor water solubility, leading to precipitation of this form from solution. Therefore, the solution should be buffered at a sufficiently low ph so that the concentration of the free base is less than its solubility. When the solution is instilled in the eye, the tears bring the ph to about 7.4, converting the drug to the physiologically active form.

Isotonic Solutions Introduction Isosmotic Solutions Isotonic Solutions Methods of Adjusting Tonicity

Introduction In addition to carrying out ph adjustment, pharmaceutical solutions that are meant for application to delicate body membranes 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 and produce no discomfort when instilled in the eye, nasal tract, blood, or other body tissues. Isotonic sodium chloride NaCl (0.9%) is a familiar pharmaceutical example of such a preparation.

Osmolality and osmolarity Osmolality and osmolarity are colligative properties that measure the concentration of the solutes independently of their ability to cross a cell membrane. The unit to express the amount of osmotically active particles in a solution is the osmole or milliosmole: 1 Osmol = 1 mol n Where n is the number of species into which the solute is dissolved 1 Osmol = 10 3 mosmol Osmolarity is the number of osmoles of solute per L of solution Osmolality is the number of osmoles of solute per kg of solvent

Isosmotic Solutions When two solutions are separated by a perfect semipermeable membrane and there is no net movement of solvent molecules across the membrane, the solutions are isosmotic (i.e. have equal osmotic pressure or osmolarity). Perfect semipermeable membrane is permeable only to solvent molecules.

Isosmotic Solutions Biological membranes do not always function as perfect semipermeable membranes; some solutes pass through and some interact with the membrane. These solutes are regarded as solvent and they do not exert an osmotic pressure on the membrane (the solutions are isosmotic but not isotonic)

Isotonic Solutions When two isosmotic solutions contain solutes that can not cross the biological membrane, they are described as isotonic with respect to that membrane. Tonicity is the concentration of only the solutes that cannot cross the membrane since these solutes exert an osmotic pressure on that membrane.

Methods of Adjusting Tonicity Class 1 method: Sodium chloride concentration added to the drug to make an isotonic solution is calculated (1) Cryoscopic method (2) Sodium chloride equivalent method Class 2 method: Amount of water added to the drug to make an isotonic solution is calculated (1) White-Vincent Method (2) Sprowls Method

Methods of Adjusting Tonicity Class I Method: Cryoscopic Method The freezing point depressions T f of drug solutions can be determined theoretically from the equation: ΔT f = ikfc For solutions of electrolytes, a new factor, (L = ikf) is used: ΔT f = LC The L value for solutions that is isotonic with body fluids is written as L iso If the concentration of the drug is chosen to be 1%, then T f of can be theoretically by the equation: Tf = L iso 10 M. Wt

Methods of Adjusting Tonicity Class I Method: Cryoscopic Method Each class of electrolytes has certain L iso value: 1.9 for nonelectrolytes (e.g. sucrose) 2.0 for weak electrolytes (e.g. Boric acid) 2.0 for bi-bivalent (e.g. ZnSO 4 ) 3.4 for uni-univalent (e.g. NaCl) 4.3 for uni-bivalent (e.g. Na 2 SO 4 ) 4.8 for bi-univalent (e.g. CaCl 2 ) 5.2 for uni-trivalent (e.g. NaPO 3 ) 6 for tri-univalent (e.g. AlCl 3 ) 7.6 for tetraborate (e.g. Na borate)

Methods of Adjusting Tonicity Class I Method: Cryoscopic Method Example Calculate the Tf for 1% apomorphine HCl solution (M.Wt = 312.79 g/mol). (L iso value of 2.6) 10 Tf = L iso M. Wt Tf = 2. 6 10 = 0. 08 C 312. 79

Methods of Adjusting Tonicity Class I Method: Cryoscopic Method In the cryoscopic method, 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.

Methods of Adjusting Tonicity Class I Method: Cryoscopic Method Example How much w/v% of NaCl is required to render 100 ml of a 1% solution of apomorphine HCl isotonic with blood serum? 1% solution of the drug has a T f of 0.08 C. 1% solution of NaCl has a T f of 0.58 C. To make this solution ( T f = 0.08 C) isotonic with blood ( T f = 0.52 C), sufficient NaCl must be added to reduce the freezing point by an additional 0.44 C (0.52 C - 0.08 C). 1% 0. 58 C = X = 0. 76% X 0. 44 C Thus, 0.76% NaCl will lower the freezing point by 0.44 C and will render the solution isotonic.

Methods of Adjusting Tonicity Class I Method: NaCl Equivalent Method The sodium chloride equivalent (E) of a drug is the amount of sodium chloride that has the same osmotic effect of 1 g of the drug. E value can be obtained theoretically from L iso Molecular weight of the drug E = 17 L iso M. Wt value and

Methods of Adjusting Tonicity Class I Method: NaCl Equivalent Method Example 1 Calculate the approximate E value for ephedrine sulfate (M.Wt =428.54) (L iso = 5.8) E = 17 L iso M. Wt E = 17 5.8 428.54 = 0.23

Methods of Adjusting Tonicity Class I Method: NaCl Equivalent Method In the NaCl equivalent method, sodium chloride or some other substance is added to the solution of the drug to make the concentration of the solution equivalent to 0.9% of NaCl and thus make it isotonic with body fluids.

Methods of Adjusting Tonicity Class I Method: NaCl Equivalent Method Example 2 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? E value for the drug is 0.23 The quantity of the drug is multiplied by its NaCl equivalent, E : Ephedrine sulfate: 1 g 0.23 = 0.23 g The ephedrine sulfate has contributed a weight of material osmotically equivalent to 0.23 g of NaCl. Because a total of 0.9 g of NaCl is required for isotonicity, 0.67 g (0.90-0.23 g) of NaCl must be added.

Methods of Adjusting Tonicity Class II Method: White-Vincent Method The class II methods of computing tonicity involve the addition of water to the drugs to make an isotonic solution, followed by the addition of an isotonic-buffered diluting vehicle to bring the solution to the final volume. White and Vincent developed a simplified equation for calculating the volume V (mls) of isotonic solution prepared by mixing the drug with water. V = w E 111. 1 V : w : weight (g) of the drug. E : NaCl equivalent

Methods of Adjusting Tonicity Class II Method: White-Vincent Method Example 1 How to make 30 ml of a 1% solution of procaine HCl isotonic with body fluid? NaCl equivalent for procaine HCl is 0.21 Weight of the drug = 30 1% = 0.3 g V = w E 111. 1 V = 0.3 0.21 111.1 = 7 ml 7 ml of water is added to the drug to make it isotonic, then enough isotonic diluting solution is added to make 30 ml of the finished product.

Methods of Adjusting Tonicity Class II Method: White-Vincent Method Example 2 Make the following solution isotonic with body fluid: Phenacaine HCl.0.06 g Boric acid 0.30 g DW, enough to make 100 ml E for Phenacaine HCl and Boric acid are 0.2 and 0.5 respectively V Phenacaine HCL = 0.06 0.20 111.1 = 1.33 ml V Boric acid = 0.3 0.5 111.1 = 16.66 ml V Total = 1.33 + 16.66 = 18 ml The drugs are mixed with water to make 18 ml of an isotonic solution, and the volume of the preparation is completed to 100 ml by adding an isotonic diluting solution.

Methods of Adjusting Tonicity Class II Method: Sprowls Method Sprowls method is a simplification of White-Vincent Method in which V values for drugs of fixed weights (0.3 g) are computed and constructed as a table.

References Sinko, P. J. M. A. N. 2006. Martin's physical pharmacy and pharmaceutical sciences: physical chemical and biopharmaceutical principles in the pharmaceutical sciences, Philadelphia, Lippincott Williams & Wilkins.