Buffered Isotonic Solutions PDF

## BUFFERED ISOTONIC SOLUTIONS 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. For example, Isotonic sodium chloride. - Osmolality and osmolarity are c...

## BUFFERED ISOTONIC SOLUTIONS 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. For example, Isotonic sodium chloride. - 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. - 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. 1. 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. 2. Isotonic solution when the concentration of impermeable solutes in the solution and cytosol are equal and the RBCs remain unchanged, so there is no net fluid flow. 3. 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. | Type | Description | |---|---| | Hypertonic | Higher concentration of impermeable solutes | | Isotonic | Equal concentration of impermeable solutes | | Hypotonic | Lower concentration of impermeable solutes | Therefore, a solution containing a quantity of drug calculated to be isosmotic with blood is isotonic only when the blood cells are impermeable to the solute molecules and permeable to the solvent, water. Although all solutions having an isotonicity value of 0.9 g of NaCl per 100 mL of solution need not necessarily be isotonic with respect to the living membranes concerned, many of them are roughly isotonic in this sense, and all may be considered isotonic across an ideal membrane. ## 11- Measurement of Tonicity The tonicity of solutions can be determined by one of two methods: 1. The hemolytic method, the effect of various solutions of the drug is observed on the appearance of red blood cells suspended in the solutions. 2. Methods that determine colligative properties. This method is based on a measurement of the slight temperature differences arising from differences in the vapor pressure of thermally insulated samples contained in constant-humidity chambers. ### Calculating Tonicity Using Liso Values The L value can 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 written as Liso. The Liso value for a 0.90% (0.154 M) solution of sodium chloride, which has a freezing point depression of 0.52°C and is thus isotonic with body fluids, is 3.4. Liso = 0.52°C / 0.154 = 3.4 | Type | Liso Value | | ----------- | ----------- | | Nonelectrolyte | 1.9 (sucrose) | | Weak electrolyte | 2 (zinc sulfate) | | Uni-univalent electrolyte | 3.4 (NaCl) | | Uni-divalent electrolyte | 4.3 (Na₂SO₄) | | Di-divalent electrolyte | 4.8 (MgSO₄) | The Liso values can also be used for calculating sodium chloride equivalents and Sprowls V values. ## EXAMPLE 8-11 #### Freezing Point Lowering 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 0.104. We have ΔΤ = 3.4 × 0.104 = 0.35°C ## 12-Methods of Adjusting Tonicity and PH 1. 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 the sodium chloride equivalent method. 2. Class II methods, water is added to the drug in a sufficient amount to form an isotonic solution. Included in this class are the White-Vincent method and the Sprowls method. **a.freezing point lowering** ## Isotonicity How much sodium chloride is required to render 100 mL of a 1% solution of apomorphine hydrochloride isotonic with blood serum? From Table 8–4 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 - 0.08°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 = 0.58° / 0.440 ; 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. ## B.Sodium Chloride Equivalent Method( E value *Also called "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. Calculations for determining the amount of sodium chloride to render a solution isotonic simply involve multiplying the quantity of each drug in the prescription by its sodium chloride equivalent and subtracting this value from the concentration of sodium chloride that is isotonic with body fluids, namely, 0.9 g/100 mL. **Alternatives** Mannitol, propylene glycol, or glycerin could serve as alternatives to sodium chloride. The concentration of these agents for isotonicity is readily calculated by use of the equation X = Y(Additional amount of NaCl for isotonicity) / E(Grams of NaCl equivalent to 1 g of the isotonic agent) - Where X is the grams of isotonic agent required to adjust the tonicity. - Y is the additional amount of NaCl for isotonicity over and above the osmotic equivalence of NaCl provided by the drugs in the solution. E is the sodium chloride equivalence of the isotonic agent E = 17 * Liso / MW ## EXAMPLE 8-13 ### Sodium Chloride Equivalents Calculate the approximate E value for a new amphetamine hydrochloride derivative (molecular weight 187). Because this drug is a uni-univalent salt, it has an Liso value of 3.4. Its E value is calculated from equation (8-45): E = 17 * 3.4 / 187= 0.31 ## EXAMPLE 8-14 ### Tonicity Adjustment  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 x 0.23 = 0.23 g The ephedrine sulfate has contributed a weight of material osmotically equivalent to 0.23 g of sodium chloride. Because a total From equation (8–45) we calculate the E value of propylene glycol, a nonelectrolyte with an Liso value of 1.9 and a molecular weight of 76.09 g/mole: E = 17 * 1.9 / 76.09 = 0.42 Using equation (8–46), X = Y/E, we obtain X =1.794 / 0.42 = 4.3 g where X = 4.3 g is the amount of propylene glycol required to adjust the 200-mL solution of thimerosal to isotonicity. ## Isotonic Solutions Let us prepare 200 mL of an isotonic aqueous solution of thimerosal, molecular weight 404.84 g/mole. The concentration of this anti-infective drug is 1:5000, or 0.2 g/1000 mL. The Liso for such a compound, a salt of a weak acid and a strong base (a 1:1 electrolyte), is 3.4, and the sodium chloride equivalent E is ## White-Vincent Method The class II methods of computing tonicity involve the addition of water to the drugs to make an isotonic solution V=wxE × 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. - E is the sodium chloride equivalent. ## EXAMPLE 8-16 ### Isotonic Solutions Make the following solution isotonic with respect to an ideal membrane: - Phenacaine hydrochloride 0.06 g - Boric acid 0.30 g - Sterilized distilled water, enough to make 100.0 mL V = [(0.06 x 0.20) + (0.3 × 0.50)] × 111.1 V = 18 mL 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.