Buffer Solutions PDF

# Buffer Solutions ## Prof. Ehab Ahmed ## Definition Buffers are compounds or mixtures of compounds that, by their presence in solution, resist changes in pH upon the addition of small quantities of acid or alkali. The resistance to change is known as buffer action. ## Buffer Equation A combina...

# Buffer Solutions ## Prof. Ehab Ahmed ## Definition Buffers are compounds or mixtures of compounds that, by their presence in solution, resist changes in pH upon the addition of small quantities of acid or alkali. The resistance to change is known as buffer action. ## Buffer Equation A combination of a weak acid and its conjugate base, e.g., its salt, or a weak base and its conjugate acid act as buffers. - **Buffer Equation for a Weak Acid and Its Salt**: $pH = pK_a + log [salt]/[acid]$ - **Buffer Equation for a Weak Base and Its Salt**: $pH = pK_w - pK_b + log [base]/[salt]$. The brackets [] means moles or molar concentrations. Buffer solutions are NOT ordinarily prepared from weak bases and their salts because of the volatility and instability of the bases and because of the dependence of their pH on $pK_w$, which is often affected by temperature changes. ## Numerical Example (1) Calculate the pH of an acidic buffer of a weak acid having pKa = 4 using the following data: 1. The molar concentration of acid is 0.01 2. The molar concentration of sodium salt is 0.05 The pH of the buffer can be calculated using the Henderson-Hasselbalch equation: $pH = pK_a + log [salt]/[acid]$ Substituting the values, we get: $pH = 4 + log (0.05)/(0.01)$ $pH = 4.7$ ## Buffering Action If a small amount of a strong acid or base is added to water, the pH is altered considerably; such systems have no buffer action. For example, if 1 ml of a 0.1 N HCl solution is added to 100 ml of pure water, the pH is reduced from 7 to 3. Considering acetic acid (HAc) as a weak acid, it will partially dissociate with ionization constant Ka as follows: $HAc \rightleftharpoons H^+ + Ac^-$ $Ka = [H+][Ac^-] / [HAc]$ Or it can be written in the presence of water as: $HAc + H_2O \rightleftharpoons H_3O^+ + Ac^-$ If sodium acetate is added to acetic acid, what will happen? Sodium acetate will completely dissociate to give a high concentration of acetate anion ($Ac^-$), which is a common ion with that produced from (HAc). This will affect the dissociation equation toward the left side, and the system is now ready to work as acetate buffer. If a small amount of a strong acid ($H^+$ or $H_3O^+$) or a strong base ($OH^-$) is added to acetate buffer: $Ac^- + H_3O^+ \rightleftharpoons HAc + H_2O$ (with very little decrease in pH) $HAc + OH^- \rightleftharpoons H_2O + Ac^-$ (with very little increase in pH) ## Numerical Example (2) Calculate the change in pH of an acetate buffer prepared from 0.1 mole acetic acid and 0.1 mole sodium acetate if: 1. 0.01 mole HCl was added or 2. 0.01 mole NaOH was added The pKa of acetic acid is 4.76. We can use the Henderson-Hasselbalch equation to calculate the pH for each scenario: $pH = pK_a + log [salt]/[acid]$ * **When no acid or base added**: $pH = 4.76 + log (0.1/0.1)$ $pH = 4.76$ * **When 0.01 M HCl was added**: $pH = 4.76 + log (0.09/0.11)$ $pH = 4.673$ * **When 0.01 M NaOH was added**: $pH = 4.76 + log (0.11/0.09)$ $pH = 4.847$ ## Factors Influencing the pH of Buffer Solutions - **The addition of neutral salts to buffers changes the pH of the solution by altering the ionic strength**: Increasing µ, decreases pH. - **The addition of water in moderate amounts may cause a small positive or negative deviation**. - **Temperature influences the change in pH**: The pH of acetate buffers was found to increase with temperature, whereas the pH of boric acid, sodium borate buffers decreased with temperature. ## Drugs as Buffers * Salicylic acid (sodium salicylate) * Ephedrine base (ephedrine hydrochloride) ## Universal Buffer A mixture of weak acids, whose pKa values are sufficiently alike (differing by no more than about 2 pH units) so that their buffer regions overlap, can be used as a universal buffer over a wide range of pH. A buffer of this type was prepared using citric acid, the three stages of citric acid. * $pK_1 = 3.15$ * $pK_2 = 4.78$ * $pK_3 = 6.40$ ## Buffer Capacity The magnitude of the resistance of a buffer to pH changes is referred to as the buffer capacity. It is also known as buffer efficiency, buffer index, and buffer value. The buffer capacity (β) is defined as: $β = ΔB/ΔpH$ where ΔB is the small increment in gram equivalents (g Eq)/liter of a strong base added to the buffer solution to produce a pH change of ΔpH. # Van Slyke's Equation for Buffer Capacity $β = 2.3C * Ka[H_3O+]/[(Ka+[H_3O+])]²$ Where C is the total buffer concentration, i.e., the sum of the molar concentration of the acid and the salt. ## Numerical Example (3) At a hydrogen ion concentration of $1.75 \times 10^-5$ (pH = 4.76), what is the capacity of a buffer containing 0.10 mole each of acetic acid and sodium acetate per liter of solution? The total concentration, C = [Acid] + [Salt], is 0.20 mole/liter, and the dissociation constant is $1.75 \times 10^-5$. Using the Van Slyke's equation: $β = 2.3C * Ka[H_3O+]/[(Ka+[H_3O+])]²$ Substituting the values we get: $β = 2.3 \times 0.20 \times (1.75 \times 10^{-5}) \times (1.75 \times 10^{-5})/[ (1.75 \times 10^{-5}) + (1.75 \times 10^{-5})]²$ $β = 0.115$ ## Maximum Buffer Capacity: The maximum buffer capacity, $β_{max}$, is obtained when the concentration of acid equals the concentration of its conjugate base. In this case, the Henderson-Hasselbalch equation shows that pH = pKa. Under these conditions, the maximum buffer capacity is: $β_{max} = 0.576C$ where C is the total buffer concentration (Acid + salt). ## Numerical Example (4) Calculate the maximum buffer capacity for the buffer described in numerical example (3). The maximum buffer capacity can be calculated using the formula: $β_{max} = 0.576C$ Substituting the value of C = 0.20 mole/liter, we get: $β_{max} = 0.576 \times 0.20$ $β_{max} = 0.1152$ # Buffers in Pharmaceuticals ### (1) Buffers in Biologic Systems: In Vivo Biologic Buffer Systems * **Blood is maintained at a pH of about 7.4 by the so-called primary buffers in the plasma and the secondary buffers in the erythrocytes.** * **The plasma contains carbonic acid/bicarbonate and acid/alkali sodium salts of phosphoric acid as buffers.** The $pK_1$ for the first ionization stage of carbonic acid (to give bicarbonate) in the plasma at body temperature and an ionic strength of 0.16 in pH 7.4 is about 6.1. The ratio of bicarbonate to carbonic acid can be calculated: $pH = 6.1 + log [HCO_3^-]/[H_2CO_3]$ $log [HCO_3^-]/[H_2CO_3] = 7.4 - 6.1 = 1.3$ $[HCO_3^-]/[H_2CO_3] = 20:1$ * **Plasma proteins, which behave as acids in blood, can combine with bases and do act as buffers.** * **In the erythrocytes, the two buffer systems consist of hemoglobin/oxyhemoglobin and acid/alkali potassium salts of phosphoric acid.** Ellison et al. obtained a buffer capacity of about 0.039 for whole blood, of which 0.031 was contributed by the cells and 0.008 by the plasma. * **Usually when the pH of the blood goes below 7.0 or above 7.8, life in serious danger. The pH of the blood in diabetic coma is alleged to drop as low as about 6.8.** ### (2) Pharmaceutical Buffers ## Influence of Buffer Capacity and pH on Tissue Irritation: Solutions to be applied to tissues or administered parenterally are liable to cause irritation if their pH is greatly removed 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, will be minimal if: - The lower the buffer capacity of the solution - The smaller the volume used - The larger the volume and buffer capacity of the physiologic fluid. Parenteral solutions for injection into the blood are usually not buffered, or they are buffered to a low capacity so that the buffers of the blood may readily bring them within the physiologic pH range. ## Drug Stability * **In aqueous solution, instability may arise through the catalytic effect of acids or bases. So pH adjustment of solutions that require sterilization is very important.** * **Morphine solutions are not decomposed during 50-min exposure at a temperature of 100°C if the pH is less than 5.5. Neutral and alkaline solutions, however, are highly unstable.** * **Solution of Cocaine, a minimum hydrolytic decomposition occurs in the range of pH of 2 and 5. In one study, a solution of cocaine hydrochloride initially at a pH of 5.7 remained stable during 2 months, while another solution buffered to about pH 6 underwent approximately 30% hydrolysis in the same time.** ## Drug Activity Drugs that are weak acids or weak bases may exist in ionized or non-ionized form (or a mixture of both). They may be active in one form but inactive in the other. Often such drugs have an optimum pH range for maximum activity. For example, mandelic acid, benzoic acid, and salicylic acid have pronounced antibacterial activity in nonionized form but have practically no such activity in ionized form. Accordingly, these substances require an acid environment to function effectively as antibacterial agents. ## Drug Absorption * **The degree of ionization and lipoid solubility of a drug are two important factors that determine the rate of absorption of drugs from the gastrointestinal tract and indeed, their passage through cellular membranes generally.** * **Drugs that are weak organic acids or bases, and which in nonionized form are soluble in lipids are absorbed through cellular membranes.** * **Completely ionized drugs on the other hand are absorbed poorly.** * **Rates of absorption of a variety of drugs are related to their ionization constants (pKa) and in many cases may be quantitatively predicted on the basis of this relationship.** * **The pH of the physiological medium (gastric or intestinal fluid, plasma, cerebrospinal fluid, etc), in which a drug is dissolved or dispersed, determines the extent to which the drug will be converted ionic or nonionic form, so it is one of the important parameters of drug absorption.** ## Stability vs. Optimum Therapeutic Response 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 a compromise between that of optimum stability and the pH of maximum therapeutic action. When the solution of pilocarpine is instilled in the eye, the tears (pH 7.4) participate in the gradual neutralization of the solution, and conversion of the drug occurs from the physiologically inactive form to the undissociated base. The base can then readily penetrate the lipoidal membrane. ## pH and Solubility 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. When the amount of base exceeds the limited water solubility of this form, free base precipitates from solution. Therefore, the solution should be buffered at a sufficiently low pH so that the concentration of alkaloidal base in equilibrium with its salt is calculated *to be less than* the solubility of the free base at the storage temperature. # Buffered Isotonic Solutions ## Isotonic Solutions (0.9 g NaCl per 100 ml): Isotonic solutions neither cause swelling nor contraction of tissues with which they come in contact. An isotonic solution is generally one that produces no discomfort when instilled in the eye, nasal tract, blood or other body tissues. ## Hypertonic Solution (2.0% NaCl solution): If red blood cells are suspended in a 2.0% NaCl solution, 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 erythrocyte membrane are identical. This outward passage of water causes the cells to shrink and become wrinkled. ## Hypotonic Solution (0.2% NaCl): If the blood is mixed with a 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. ## Distinction Between Iso-Osmotic and Isotonic The terms of iso-osmotic and isotonic are not equivalent. If a plant or animal cell is in contact with a solution that has the same osmotic pressure as the cell contents, there will be no gain or loss of water by either solution provided. The cell membrane is impermeable to all solutes that are present, so the volume of the cell contents remains unchanged, and therefore the tone or normal state of the cell may be described not only as being iso-osmotic with the solution in the cell, but also as being isotonic with it. It is possible that some substances used in an injection dosage form can cause hemolysis of red blood cells, even when their concentrations are such as to produce solutions theoretically iso-osmotic with the cells. This is because the solutes diffuse through the membrane of the cells. For example, a 0.8% solution of urea has the same osmotic pressure as a 0.9% solution of sodium chloride, but the former solution produced hemolysis of red blood cells. This means that the urea solution is not isotonic. ## Measurement of Tonicity The tonicity of solutions can be measured using one of two methods: - **Hemolytic method**: This method is based on the fact that a hypotonic solution liberates oxyhemoglobin in direct proportion to the number of cells hemolyzed. - **Colligative properties method**: This method is based on any of the methods that determine colligative properties. For example, -0.52° is the freezing point of both human blood and lacrimal fluid. This temperature corresponds to the freezing point of a 0.9% NaCl solution. Thus, 0.9% NaCl solution is considered to be isotonic with both blood and lacrimal fluid. ## Calculating Tonicity Using Liso Values - Since the freezing point depressions for solutions of electrolytes of both the weak and strong types are always greater than those calculated from the equation ΔTf = iKC, a new factor L = iKf is introduced to overcome this difficulty. - ΔTf = LC - The L value may be obtained from the freezing point lowering of solutions of representative compounds of a given ionic type at a concentration c that is isotonic with body fluids. This specific value of L is symbolized as Liso. - The $L_{iso}$ value for a 0.90% (0.154 M) solution of sodium chloride, which has a freezing point depression of 0.52° and is thus isotonic with body fluids, is 3.4. - $L_{iso} = ΔTf /C$ $L_{iso} = 0.52°/0.154 = 3.4$ ## Methods of Adjusting Tonicity and pH Several methods may be used to calculate the quantity of sodium chloride, dextrose, and other substances that may be added to solutions of drugs to render them isotonic. The methods are divided into two classes: ### Class I Methods 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° and thus make it isotonic with body fluids. - Under this class are included the Cryoscopic method and the Sodium Chloride Equivalent method. ### Class II Methods 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. ## Class I Methods: Cryoscopic Method The calculations involved in the cryoscopic method are best explained by an example. **Example:** How much sodium chloride is required to render 100 ml of a 1% solution of apomorphine hydrochloride isotonic with blood serum? * It is found that a 1% solution of the drug has a freezing point lowering of 0.08°. To make this solution isotonic with blood, sufficient sodium chloride must be added to reduce the freezing point by an additional 0.44° (0.52°-0.08°). * In the freezing point table, it is also observed that a 1% solution of sodium chloride has a freezing point lowering of 0.58°. * By the method of proportion: $1\%/X = 0.58°/0.44 ; X = 0.76\%$ * The required sodium chloride = 0.9 - 0.76 = 0.14 g for 100 ml. ## Class I Methods: Sodium Chloride Equivalent Method The tonic equivalent of a drug is the amount of sodium chloride that is equivalent to (i.e., has the same osmotic effect as) 1 gram, or other weight unit, of the drug. The sodium chloride equivalents E for a number of drugs are calculated. $E = 17 * L_{iso}/M$ (M molecular weight) ## Class II Methods: White-Vincent Method The Class II methods of computing tonicity involve adding 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. The equation is best shown as follows: $V = W \times E \times 111.1$ where: * V is the volume in ml of isotonic solution * W is the weight in grams of the drug * E is the sodium chloride equivalent * 111.1 is a constant representing the volume in ml of isotonic solution obtained by dissolving 1 gram of sodium chloride in water. The value of the ratio 100/0.9 is 111.1. When more than one ingredient is contained in an isotonic preparation, the volumes of isotonic solution, obtained by mixing each drug with water, are *additive*. The same is applied to all methods of calculation. **Example:** Make the following solution isotonic with respect to an ideal membrane. | | | | :--------------- | :------- | | Phenacetine hydrochloride | 0.06g | | Boric acid | 0.30g | | Sterilized distilled water (enough to make) | 100.0ml | Using the White-Vincent equation: $V = [(0.06 \times 0.16) + (0.3 \times 0.50)] \times 111.1$ $V = 17.7 ml$ ## The Sprowls method * A further simplification of the method of White and Vincent was introduced by Sprowls. * Sprowls constructed a table of values of V (of water) when the weight of the drug, w, was arbitrarily fixed. Sprowls chose as the weight of drug 0.3 g, the quantity for 1 fluid ounce of a 1% solution. * The volume, V, of isotonic solution that can be prepared by mixing 0.3 g of a drug with sufficient water can be computed for drugs commonly used in ophthalmic and parenteral solutions. ## Answering the following problem: A solution contains 1 gm of ephedrine sulfate in a volume of 100 ml. * NaCl equivalent of Ephedrine sulfate = 0.23 * NaCl equivalent of dextrose = 0.16 * (0.9% NaCl is isotonic) 1. **What quantity of NaCl must be added to make the solution isotonic?** 2. **How much dextrose would be required for this purpose?** To solve this problem, we can use the Sodium Chloride Equivalent method. The NaCl equivalent for ephedrine sulfate is 0.23, which means that 1g of ephedrine sulfate has the same osmotic effect as 0.23 g of NaCl. 1. **To make the solution isotonic, we need to add enough NaCl to make the total NaCl equivalent equal to 0.9 g/100 ml.** Since the solution already contains 1 g of ephedrine sulfate with an NaCl equivalent of 0.23, we need to add: 0.9 - 0.23 = 0.67 g of NaCl per 100 ml. 2. **To calculate the amount of dextrose required, we use the NaCl equivalent of dextrose.** We know that 1 g of dextrose has the same osmotic effect as 0.16 g of NaCl. To make the solution isotonic using dextrose, we would need: 0.9/0.16 = 5.625 g of dextrose per 100 ml.