Author name: contact@devinevisiontech.com

NTA Level 5 • Semester 2 • PST05208 Calculations Involving Osmolarity Pharmaceutics Theory and Compounding • Source Session/Topic 18 Full source-text version: all educational wording from the extracted learning source is retained; only presenter/tutor metadata and web-layout noise are removed, while formatting is improved for readability. Session 18: Calculations Involving Osmolarity Total Session Time: 120 minutes + 6 hours of Practices Prerequisites • None Learning Tasks By the end of this session students are expected to be able to: • Give introduction to calculation involving Osmolarity • Concept on osmolarity vs. osmolality • Perform calculations Involving Osmolarity Resources Needed: • Flip charts, marker pens, and masking tape • Black/white board, chalk and whiteboard markers • LCD projector and compute SESSION OVERVIEW Activity/ Step Time Content Method 1 05 Minutes Presentation Introduction, Learning Tasks 2 40 Minutes Presentation Introduction to Calculation Involving Osmolarity 25 Minutes Presentation Concept on Osmolarity vs. Osmolality 40 Minutes Presentation Performing Calculation Involving Osmolarity 3 Demonstration 4 05 Minutes Presentation Key Points 5 05 Minutes Presentation Evaluation 125 SESSION CONTENTS STEP1: Presentation of Session Title and Learning Tasks (5 minutes) READ or ASK students to read the learning tasks and clarify ASK students if they have any questions before continuing STEP 2: Introduction to Calculation Involving Osmolarity (40 Minutes) Osmolarity Osmolarity is defined as the number of osmoles of solute per liter (L) of solution. It is expressed in terms of osmol/L or Osm/L. Osmolarity depends on the number of particles in a chemical solution, but not on the identity of those molecules or ions. Sample Osmolarity Calculations A 1 mol/L NaCl solution has an osmolarity of 2 osmol/L. A mole of NaCl dissociates fully in water to yield two moles of particles: Na+ ions and Cl- ions. Each mole of NaCl becomes two osmoles in solution. A 1 M solution of sodium sulfate, Na2SO4, dissociates into 2 sodium ions and 1 sulfate anion, so each mole of sodium sulfate becomes 3 osmoles in solution (3 Osm). To find the osmolarity of a 0.3% NaCl solution, you first calculate the molarity of the salt solution and then convert the molarity to osmolarity. Convert percent to molarity: 0.03 % = 3 grams 100 ml = 3 grams / 0.1 L = 30 g/L molarity NaCl = moles / liter = (30 g/L) x (1 mol / molecular weight of NaCl) Look up the atomic weights of Na and Cl on the periodic table and add the together to get the molecular weight. Na is 22.99 g and Cl is 35.45 g, so the molecular weight of NaCl is 22.99 + 35.45, which is 58.44 grams per mole. Plugging this in: Molarity of the 3% salt solution = (30 g/L) / (58.44 g/mol) molarity = 0.51 M You know there are 2 osmoles of NaCl per mole, so: Osmolarity of 3% NaCl = molarity x 2 Osmolarity = 0.51 x 2 Osmolarity = 1.03 Osm 126 STEP 3: Concept on Osmolarity vs. Osmolality (25 Minutes) • Osmolality and osmolarity are units of measurement. Osmolality is the number of osmoles of solute in a kilogram of solvent, while osmolarity is the number of osmoles of solute in a litre of solution. Osmolarity is the concentration of an osmotic solution. • Osmolality is convenient to use because the amount of solvent remains constant, regardless of changes in temperature and pressure. • While osmolarity is easy to calculate, it's less difficult to determine because the volume of solution changes according to temperature and pressure. Osmolarity is most commonly used when all measurements are made at a constant temperature and pressure. • If osmolality is the number of osmoles of solute in a kilogram of solvent, then osmolarity is the number of osmoles of solute in a litre of solution. • Osmolarity deals with the concentration of an osmotic solution, while osmolality deals with the concentration of particles in a fluid. • It is easier to determine the osmolality than the osmolarity. Osmolarity is expressed as Osm/L, and osmolality is expressed as Osm/Kg. • Osmolality is used to determine medical conditions like diabetes, shock and dehydration, while osmolarity is used for the detection of the concentration of dissolved particles in urine.Osmolality is the commonly used method of measurement in Osmometry. • When the concentration of solutes is very low, the osmolality and osmolarity are similar. 127 STEP 3: Performing Calculations Involving Osmolarity (40 Minutes) Activity: Small Group Discussion ( 20 minutes) DIVIDE students in small manageable groups ASK students to discuss in groups on the following questions • How many mOsmol are represented in a liter of a 0.9% sodium chloride solution REFER Students to Pharmaceutical Calculation. 13th Edition by HOWARD C. ANSEL: Chapter 11, for reference ALLOW students to discuss for 20 minutes ALLOW each groups to present for 5 minutes CLARIFY and SUMMARIZE by using the contents below Formula weight of NaCl = 58.5 1 mmol of NaCl (= 58.5mg) = 2 mOsmol 900mg of NaCl per 100mL = 9000mg NaCl per liter Then by using proportion: 58.5mg = 2mOsmol x = 9000 x 2 9000mg x mOsmol 58.5 X = 307.7 = 308mOsmol/L STEP 4: Key Points (5 Minutes) • Milliosmoles (mOsmol) is an expression of the osmotic activity of 1 Millimole • A distinction also should be made between the terms osmolarity and osmolality. Whereas Osmolarity is the milliosmoles of solute per liter of solution while osmolality is the milliosmoles of solute per kilogram of solvent. For dilute aqueous solutions, osmolarity and osmolality are nearly identical. • For more concentrated solutions, however, the two values may be quite dissimilar. The Pharmaceutical personnel should pay particular attention to a product‟s label statement regarding molarity versus molality. 128 STEP 5: Evaluation (5 Minutes) • What is Milliosmoles? • What are the difference between osmolarity and osmolality? STEP 6: Take Home Assignment (15 minutes) Activity: Take home Assignment (15 minutes) ASK each individual student to do the following assignment • A solution contains 156mg of k+ ions per 100 mL. How

NTA Level 5 • Semester 2 • PST05208 Calculations Involving Milliosmoles (mOsmol) Pharmaceutics Theory and Compounding • Source Session/Topic 17 Full source-text version: all educational wording from the extracted learning source is retained; only presenter/tutor metadata and web-layout noise are removed, while formatting is improved for readability. Session 17: Calculations Involving Milliosmoles (mOsmol) Total Session Time: 120 minutes + 6 hours of Practices Prerequisites • None Learning Tasks By the end of this session students are expected to be able to: • Give introduction to calculations involving milliosmoles • Perform calculations Involving Milliosmoles Resources Needed: • Flip charts, marker pens, and masking tape • Black/white board, chalk and whiteboard markers • LCD projector and compute SESSION OVERVIEW Activity/ Step Time Content Method 1 05 Minutes Presentation Introduction, Learning Tasks 45 Minutes Introduction to Calculations Involving 2 Presentation Milliosmoles 60 Minutes Presentation Performing Calculation Involving Milliosmoles 3 Demonstration 4 05 Minutes Presentation Key Points 5 05 Minutes Presentation Evaluation 120 SESSION CONTENTS STEP1: Presentation of Session Title and Learning Tasks (5 minutes) READ or ASK students to read the learning tasks and clarify ASK students if they have any questions before continuing STEP 2: Introduction to Milliosmoles (45 minutes) • The osmotic activity of a solution may be stated in terms of Milliosmoles which is an expression of the osmotic activity of 1 Millimole • The term Osmolarity is used to express the strength in mOsmol per liter and indicates the total ionic concentration of the solution • Labels of some pharmacopoeia solutions that provide intravenous replenishment of fluids, nutrients, electrolytes and the osmotic diuretic mannitol are required to state osmolar concentration. • This information indicates to the practitioner whether the solution is hypotonic, isotonic or hypertonic with regard to biologic fluids and membranes • Since the unit that is used to measure osmotic concentration is Milliosmoles then; o For non – electrolytes e.g. dextrose 1 mmol (formulary weight in mg) represents 1 mOsmol o For electrolytes, the total number of particles in solution depends on the degree of dissociation of the substance in question, assuming complete dissociation • 1 mmol of NaCl represents 2 mOsmol ( Na+ and Cl-) of the total particles • 1 mmol of CaCl2 represents 3 mOsmol ( Ca2+ and 2Cl-) of the total particles • 1 mmol of Sodium citrate (Na3C6H5O7) represents 4 mOsmol ( 3Na+ -C6H5O7) of the total particles • The milliosmolar value of separate ion of an electrolyte may be obtained by dividing the concentration in mg per liter of an ion by its atomic weight • The milliosmolar value of the whole electrolyte in solution is equal to the sum of the milliosmolar value of separate ions • According to USP, the ideal osmolar concentration may be calculated using the following equation: • mOsmol/L = Weight of the substance(in g/L) x Number of specie x 1000 Molecular weight (in g) 121 • Examples o A solution contains 5% anhydrous dextrose in water for injection. How many mOsmol per liter are represented by this concentration? o Formula weight of anhydrous dextrose = 180 o 1mmol of anhydrous dextrose (= 180 mg) = 1 mosmol o 5% solution contains 50g or 50,000mg/L o Therefore:  mOsmol/L = 50,000 = 278mOsmol/ 180  OR Using USP equation:  mOsmol/L = 50 x 1000 = 278 mOsmol/ L o 180 STEP 3: Performing Calculations Involving Milliosmoles (60 Minutes) Activity: Small Group Discussion ( 40 minutes) DIVIDE students in small manageable groups ASK students to discuss in groups on the following questions • How many mOsmol are represented in a liter of a 0.9% sodium chloride solution REFER Students to Pharmaceutical Calculation. 13th Edition by HOWARD C. ANSEL: Chapter 11, for reference ALLOW students to discuss for 20 minutes ALLOW each groups to present for 5 minutes CLARIFY and SUMMARIZE by using the contents below Formula weight of NaCl = 58.5 1 mmol of NaCl (= 58.5mg) = 2 mOsmol 900mg of NaCl per 100mL = 9000mg NaCl per liter Then by using proportion: 58.5mg = 2mOsmol x = 9000 x 2 9000mg x mOsmol 58.5 X = 307.7 = 308mOsmol/ 122 STEP 4: Key Points (5 minutes) • Milliosmoles (mOsmol) is an expression of the osmotic activity of 1 Millimole • The osmotic activity of a solution may be stated in terms of Milliosmoles which is an expression of the osmotic activity of 1 Millimole • The term Osmolarity is used to express the strength in mOsmol per liter and indicates the total ionic concentration of the solution • This information indicates to the practitioner whether the solution is hypotonic, isotonic or hypertonic with regard to biologic fluids and membranes STEP 5: Evaluation (5 minutes) • What is Milliosmoles? • What are the uses of Osmolarity? STEP 6: Take Home Assignment (15 Minutes) Activity: Take home Assignment (15 minutes) ASK each individual student to do the following assignment • A solution contains 156mg of k+ ions per 100 mL. How many mOsmol are represented in a liter of the solution? ALLOCATE time for students to do the assignment and submit REFER students to recommended references 123 References Ansel, H. C &Stocklosa, M. J. (2001). Pharmaceutical Calculations (11th ed.). Philadelphia, United States: LIPPINCOTT WILLIAMS & WILKINS Ansel, H. C (2010) Pharmaceutical Calculations (13rd ed.). Philadelphia, United States: LIPPINCOTT WILLIAMS & WILKINS Senya, S. S, Mwasha, C.Y, Muyinga, A. M, Amiri,R. I. and Mauga E.A.S.K. (2011) Tanzania Pharmaceutical Handbook (2nd ed.). Dar eS Salaam, Tanzania: School of Pharmaceutical Sciences. Zatz, J.L and Teixeira, M.G. (2005). Pharmaceutical Calculation (4th ed.). New Jersey: John Wiley & Sons, Inc 124 ← Previous TopicNext Topic →View all Pharmaceutics Theory and Compounding topicsOpen Complete Full Notes PDF / OFFLINE NOTES Unataka kutumiwa notes hizi kupitia WhatsApp?Kwa notes zilizopangiliwa vizuri kwa kusoma offline au PDF, bonyeza kitufe hapa chini. Ujumbe wenye Level, Semester, Module na Topic utaandaliwa moja kwa moja.TUMIWA NOTES WHATSAPP WhatsApp: 255620339260

NTA Level 5 • Semester 2 • PST05208 Calculations Involving Millimoles Pharmaceutics Theory and Compounding • Source Session/Topic 16 Full source-text version: all educational wording from the extracted learning source is retained; only presenter/tutor metadata and web-layout noise are removed, while formatting is improved for readability. Session 16: Calculations Involving Millimoles Total Session Time: 120 minutes + 6 hours of Practices Prerequisites • None Learning Tasks By the end of this session students are expected to be able to: • Give overview of calculations involving Millimoles • Perform calculations involving Millimoles Resources Needed: • Flip charts, marker pens, and masking tape • Black/white board, chalk and whiteboard markers • LCD projector and computer • Handout 11.1 Powders and granules for oral administration SESSION OVERVIEW Activity/ Step Time Content Method 1 05 minutes Presentation Introduction, Learning Tasks 2 45minutes Presentation Introduction to Calculations Involving Millimoles 60 minutes Presentation Performing Calculations Involving Millimoles 3 Demonstration 4 05 minutes Presentation Key Points 5 05 minutes Presentation Evaluation 115 SESSION CONTENTS STEP1: Presentation of Session Title and Learning Tasks (5 minutes) READ or ASK students to read the learning tasks and clarify ASK students if they have any questions before continuing STEP 2: Millimoles (45 minutes) • A Millimole (mmol) of an element, a compound, or an ion is its atomic, molecular, or ionic weight respectively e.g. the ionic weight of Ca2+ is 40, therefore, 1 mmol Ca2+ = 40mg calcium • The number of mmol of each ion obtained from a salt in solution depends on the number of each ion in the molecule of the salt. e.g. Sodium chloride (NaCl, mol wt.: 58.5) has 1 Na and 1 Cl in each molecule. Hence 1 mmol NaCl ( 58.5mg) provides 1 mmol Na (23mg) and 1 mmol Cl (35.5mg) • Calcium chloride (CaCl2.2H2O, mol wt.: 147) has 1 Ca2+ and 2Cl in each molecule. Hence 1mmolCaCl2.2H2O (147mg) provides 1 mmol Ca2+ ( 40mg) and 2mmol Cl(71mg) • Therefore the amount of mg salt containing 1 mmol of a specified ion is calculated by the following equation: mg salt containing 1 mmol of specified ion = molecular weight of the salt Number of specified ions in the molecule Example: • How many mg calcium chloride are needed to provide 1 mmol of Ca2+ and 1 mmol of Cl- • mgCaCl2.2H2O containing 1 mmol of Ca2+= 147 = 147 mg 1 • 147mg CaCl2. 2H2O will provide 1 mmol Ca2+ • mgCaCl2.2H2O containing 1 mmol of Cl- = 147 = 73.5mg 2 • 73.5 mg CaCl2. 2H2O will provide 1 mmol Cl- • When g or mg salt are stated the number of mmol can be calculated by simple proportion 116 Example: • How many mmol Cl are obtained from234mg sodium chloride? • 58.5mg NaCl provide 1 mmol Cl-, therefore; 1 == x x = 1 x 234 = 4mmol 58.5 234 58.5 234mg NaCl provide 4 mmol Cl- STEP 3: Calculating Millimoles (60 minutes) Activity: Small Group Discussion ( 40 minutes) DIVIDE students in small manageable groups ASK students to discuss in groups on the following questions • 367mg calcium chloride provide how many mmol Ca2+ and how many mmol Cl-? REFER Students to Pharmaceutical Calculation. 13th Edition by HOWARD C. ANSEL: Chapter 11, for reference ALLOW students to discuss for 20 minutes ALLOW each groups to present for 5 minutes CLARIFY and SUMMARIZE by using the contents below 1mmol Ca2+ = 147 == 147mg 1 Therefore: 1 = x x = 1 x 367 = 2.49 = 2.5 mmol Ca2+ 147 367 147 – 1mmol Cl = 147mg = 73.5mg 2 117 Therefore: 1 = x x = 1 x 367 = 4.99 = 5mmol Cl-73.5 367 73.5 367mg calcium chloride provide 2.5mmol Ca2+ and 5 mmol Cl- STEP 4: Key Points (5 minutes) • A Millimole (mmol) of an element, a compound, or an ion is its atomic, molecular, or ionic weight respectively e.g. the ionic weight of Ca2+ is 40, therefore, 1 mmol Ca2+ = 40mg calcium • The number of mmol of each ion obtained from a salt in solution depends on the number of each ion in the molecule of the salt. e.g. Sodium chloride (NaCl, mol wt.: 58.5) has 1 Na and 1 Cl in each molecule. Hence 1 mmol NaCl ( 58.5mg) provides 1 mmol Na (23mg) and 1 mmol Cl (35.5mg) STEP 5: Evaluation (5 minutes) • What is Millimole (mmol) of an element? STEP 6: Take Home Assignment (15 minutes) Activity: Take home Assignment (15 minutes) ASK each individual student to do the following assignment • Calculate the strength of sodium chloride solution which is iso – osmotic with blood serum and tears ALLOCATE time for students to do the assignment and submit REFER students to recommended references 118 References Ansel, H. C &Stocklosa, M. J. (2001). Pharmaceutical Calculations (11th ed.). Philadelphia, United States: LIPPINCOTT WILLIAMS & WILKINS Ansel, H. C (2010) Pharmaceutical Calculations (13rd ed.). Philadelphia, United States: LIPPINCOTT WILLIAMS & WILKINS Senya, S. S, Mwasha, C.Y, Muyinga, A. M, Amiri,R. I. and Mauga E.A.S.K. (2011) Tanzania Pharmaceutical Handbook (2nd ed.). Dar eS Salaam, Tanzania: School of Pharmaceutical Sciences. Zatz, J.L and Teixeira, M.G. (2005). Pharmaceutical Calculation (4th ed.). New Jersey: John Wiley & Sons, Inc 119 ← Previous TopicNext Topic →View all Pharmaceutics Theory and Compounding topicsOpen Complete Full Notes PDF / OFFLINE NOTES Unataka kutumiwa notes hizi kupitia WhatsApp?Kwa notes zilizopangiliwa vizuri kwa kusoma offline au PDF, bonyeza kitufe hapa chini. Ujumbe wenye Level, Semester, Module na Topic utaandaliwa moja kwa moja.TUMIWA NOTES WHATSAPP WhatsApp: 255620339260

NTA Level 5 • Semester 2 • PST05208 Calculations Involving Milliequivalent Pharmaceutics Theory and Compounding • Source Session/Topic 15 Full source-text version: all educational wording from the extracted learning source is retained; only presenter/tutor metadata and web-layout noise are removed, while formatting is improved for readability. Session 15: Calculations Involving Milliequivalent Total Session Time: 120 minutes + 6 hours of Practices Prerequisites • None Learning Tasks By the end of this session students are expected to be able to: • Explain Milliequivalent • Calculate Milliequivalent Resources Needed: • Flip charts, marker pens, and masking tape • Black/white board, chalk and whiteboard markers • LCD projector and computer • Handout 15.1: Values for some important ions SESSION OVERVIEW Activity/ Step Time Content Method 1 05 minutes Presentation Introduction, Learning Tasks 2 45minutes Presentation Milliequivalent 60 minutes Presentation Calculating Milliequivalent 3 Demonstration 4 05 minutes Presentation Key Points 5 05 minutes Presentation Evaluation 108 SESSION CONTENTS STEP1: Presentation of Session Title and Learning Tasks (05 Minutes) READ or ASK students to read the learning tasks and clarify ASK students if they have any questions before continuing STEP 2: Milliequivalent (45 Minutes) • The equivalent weight of an element is the gram atomic weight divided by its valency OR • A Milliequivalent of an ion is the ionic weight in mg divided by the valence of that ion • Divide the Equivalent by 1000, and you get a • For example, Na has an atomic weight of 23. So 23 mg of Na+ in solution means 1 mmol of Na+ is in solution. We could also say 1 mEq of Na+ is in solution. • Ca has an atomic weight of 40. So 40 mg of Ca2+ in solution means 1 mmol of Ca2+ is in solution. In this case, 2 mEq of Ca is in solution. 1mEq = ionic weight in mg e.g. 1mEqCa2+ = 40 = 20mg Valency 2 2+ 1mEqCa is equivalent to 20mg calcium The number of mEq of each ion obtained from a salt in solution therefore depends on the valency of the ion e. g. Sodium chloride has 1 Na+ and Cl- in each molecule and both have the valency of one 1mEq Na+ = 23 = 23mg sodium 1 + 1mEq Cl = 35. 5 = 35.5mg sodium 1 Therefore, 58.5mg sodium chloride provide 1 mEq Na+ and 1 mEq Cl- • In this case mmol and mEq give numerically the same results, because both ions have the valency of one CaCl2 2H2O provides 1 Ca2+ which has the valency of two and Cl- with the valency of one 1mEqCa2+ = 40 = 20mg calcium 2 – 1mEqCl = 35 = 35.5mg chloride 1 109 Hence 147mg CaCl2.2H2O will provide 2mEq Ca2+ and 2 mEq Cl; (20 x 2) + (35.5 x 2) + (18 x 2) = 147 N.B: The molecular weight of H2O is 18 • Therefore, the amount of salt containing 1 mEq of specified ion is calculated by the following equation: • Mg salt containing 1 mEq of specified ion = molecular weight of salt Valency of specified ion x number of specified ions in the molecule • E.g. How many mg of calcium chloride are needed to provide 1mEq of Ca2+ and 1 mEq Cl? • MgCaCl2.2H2O containing 1mEq of Ca2+ = 147 = 73.5mg 2×1 • MgCaCl2.2H2O containing 1mEq of Cl = 147 = 73.5mg – 2×1 • 73.5mg CaCl2 .2H2O provide 1mEq Ca and 1 mEq Cl- 2+ • When g or mg salt are stated the number of mEq can be calculated by simple proportion e.g. How many mEq Na+ are contained in 351mg NaCl? 1 = x x = 1 x 351 = 6mEq 58.5 351 58.5 • The number of mEq of anions cations in any amount of salt is always the same, whereas the number of mmol of anions and cations differs with certain salts, depending on the number of ions in the molecule • Conversion of mmol to mEq and vice versa can be done by the following equation: Mmol == mEq Valency Examples • Molecular weights can be obtained from the table below 1. How many mg of sodium phosphate contain 1mEq HPO42- Mg Na2HPO4.12H2O containing 1mEq HPO42- = 358 = 179mg 2X1 2- 179mg ofNa2HPO4.12H2O provide 1mEq HPO4 • According to the note above, 179mg of the salt will provide as well 1mEq Na+ 2. A solution contains 90 mEq Na+, 60mEq K+ and 150 mEq Cl- per litre. Convert to g/L • A convenient way to solve the question is first to arrange the mEq in a table, so that the composition of the salt is obvious 110 Cations Anions Na+ K + Cl- 90 90 60 60 150 150 NaCl: As 1 mEq Na+ or Cl- is provided by 58.5mg NaCl, then 90mEq will be provided by 58.5 x 90 58.5x 90 = 5265mg KCL: The amount is found respectively 74.5 x 60 = 4470mg Therefore the solution contains 5.265g NaCl and 4.47g KCL per litre STEP 3: Calculating Milliequivalent (60 minutes) Activity: Small Group Discussion ( 30 minutes) DIVIDE students in small manageable groups ASK students to discuss in groups on the following questions • 367mg calcium chloride provide how many mmol Ca2+ and how many mmol Cl-? REFER Students to Pharmaceutical Calculation. 13th Edition by HOWARD C. ANSEL: Chapter 11,for reference ALLOW students to discuss for 20 minutes ALLOW each groups to present for 5 minutes CLARIFY and SUMMARIZE by using the contents below 1mmol Ca2+ = 147 == 147mg 1 111 Therefore: 1 = x x = 1 x 367 = 2.49 = 2.5 mmol Ca2+ 147 367, 147 1mmol Cl- = 147mg = 73.5mg 2 Therefore: 1 = x x = 1 x 367 = 4.99 = 5mmol Cl- 73.5 367 73.5 367mg calcium chloride provide 2.5mmol Ca2+ and 5 mmol Cl- Handout 15.1: Values for some important ions STEP 4: Key Points (5 minutes) • A Milliequivalent of an ion is the ionic weight

NTA Level 5 • Semester 2 • PST05208 Determination of Isotonicity by Molecular Pharmaceutics Theory and Compounding • Source Session/Topic 14 Full source-text version: all educational wording from the extracted learning source is retained; only presenter/tutor metadata and web-layout noise are removed, while formatting is improved for readability. Session 14: Determination of Isotonicity by Molecular Concentration Method Total Session Time: 120 minutes + 6 hours of Practices Prerequisites • None Learning Tasks By the end of this session students are expected to be able to: • Explain isotonicity by molecular concentration method • Determine the isotonicity by molecular method Resources Needed: • Flip charts, marker pens, and masking tape • Black/white board, chalk and whiteboard markers • LCD projector and computer SESSION OVERVIEW Activity/ Step Time Content Method 1 05 minutes Presentation Introduction, Learning Tasks 45 minutes Presentation Isotonicity by Molecular Method 2 Brainstorming 60minutes Presentation Determining the Isotonicity by Molecular 3 Concentration Method Demonstration 4 05 minutes Presentation Key Points 5 05 minutes Presentation Evaluation 104 SESSION CONTENTS STEP1: Presentation of Session Title and Learning Tasks (5 minutes) READ or ASK students to read the learning tasks and clarify ASK students if they have any questions before continuing STEP 2: Molecular Concentration Method (45 minutes) • Molecular concentration: number unit i.e. molecules or ions or both presents in a solution • A solution containing 1g molecule of a non – ionizing solute in 22.4 liters at normal temperature and pressure (NTP) has an atmospheric pressure of one atmosphere • Therefore a solution containing on gram molecule in 1 litre (a mole solution) will have osmotic pressure of 22.4 atmosphere • The molarity or molar concentration of a solute is defined as the number of moles of solute per liter of solution (not per liter of solvent!): • M=N/L Where M=molarity, N = number of mole and L = 1 litre of a solution • N =M/MW where M = mass or weight and MW = molecular weight of compound STEP 3: Determining the Isotonicity by Molecular Concentration Method (60 minutes) Activity: Small Group Discussion ( 30 minutes) DIVIDE students in small manageable groups ASK students to discuss in groups on the following questions • Calculate MW of compound x which contains 9.06g isotonic solution given dissociation factor is 1.8. REFER Students to Pharmaceutical Calculation. 13th Edition by HOWARD C. ANSEL: Chapter 11, for reference ALLOW students to discuss for 20 minutes ALLOW each groups to present for 5 minutes CLARIFY and SUMMARIZE by using the contents below 105 From freezing point = 1.86Ni Where N = Number of mole and i dissociation factor N = m/ mw and i = (% x p) + (100% – %) where p = number of ions after dissociation Now 0.52 = 1.86 x N x 1.8 N = 0.52/ 1.86 x 1.8 N = 0.1553166 But N = M/MW MW = m/N therefore 9.06g/0.1553166 Mw =58.33 STEP 4: Key Point (5 minutes) • Molecular concentration is the number of units i.e. molecules or ions or both present in a solution • A solution containing 1g molecule of a non – ionizing solute in 22.4 liters at normal temperature and pressure (NTP) has an atmospheric pressure of one atmosphere • Therefore a solution containing one gram molecule in 1 litre (a mole solution) will have osmotic pressure of 22.4 atmosphere STEP 5: Evaluations (5 minutes) • What is molecular concentration STEP 6: Take Home Assignment (5 minutes) Activity: Take home Assignment (15 minutes) ASK each individual student to do the following assignment • Compound x contains 9.07g dissociate by 80% into two ions. Calculate molecular weight of isotonic solution ALLOCATE time for students to do the assignment and submit REFER students to recommended references 106 References Ansel, H. C & Stocklosa, M. J. (2001). Pharmaceutical Calculations (11th ed.). Philadelphia, United States: LIPPINCOTT WILLIAMS & WILKINS Ansel, H. C (2010) Pharmaceutical Calculations (13rd ed.). Philadelphia, United States: LIPPINCOTT WILLIAMS & WILKINS Senya, S. S, Mwasha, C.Y, Muyinga, A. M, Amiri,R. I. and Mauga E.A.S.K. (2011) Tanzania Pharmaceutical Handbook (2nd ed.). Dar eS Salaam, Tanzania: School of Pharmaceutical Sciences. Zatz, J.L and Teixeira, M.G. (2005). Pharmaceutical Calculation (4th ed.). New Jersey: John Wiley & Sons, 107 ← Previous TopicNext Topic →View all Pharmaceutics Theory and Compounding topicsOpen Complete Full Notes PDF / OFFLINE NOTES Unataka kutumiwa notes hizi kupitia WhatsApp?Kwa notes zilizopangiliwa vizuri kwa kusoma offline au PDF, bonyeza kitufe hapa chini. Ujumbe wenye Level, Semester, Module na Topic utaandaliwa moja kwa moja.TUMIWA NOTES WHATSAPP WhatsApp: 255620339260

NTA Level 5 • Semester 2 • PST05208 Determination of Isotonicity by Sodium Chloride Pharmaceutics Theory and Compounding • Source Session/Topic 13 Full source-text version: all educational wording from the extracted learning source is retained; only presenter/tutor metadata and web-layout noise are removed, while formatting is improved for readability. Session 13: Determination of Isotonicity by Sodium Chloride Equivalent Method Total Session Time: 120 minutes + 6 hours of Practices Prerequisites • None Learning Tasks By the end of this session students are expected to be able to: • Give overview of sodium chloride equivalent method • Determine the isotonicity by sodium chloride equivalent method Resources Needed: • Flip charts, marker pens, and masking tape • Black/white board, chalk and whiteboard markers • LCD projector and computer SESSION OVERVIEW Activity/ Step Time Content Method 1 05 Minutes Presentation Introduction, Learning Tasks 45 Minutes Presentation Introduction to Sodium Chloride Equivalent 2 Method Brainstorming 60 Minutes Presentation Determining the Isotonicity by Sodium Chloride 3 Method Demonstration 4 05 Minutes Presentation Key Points 5 05 Minutes Presentation Evaluation 98 SESSION CONTENTS STEP1: Presentation of Session Title and Learning Tasks (5 minutes) READ or ASK students to read the learning tasks and clarify ASK students if they have any questions before continuing STEP 2: Sodium Chloride Equivalent Method (45 minutes) • The Sodium Chloride equivalent method (E) is the amount of sodium which has the same osmotic effect (based on number of particles) as 1g of the drug sample calculation • Percentage of sodium chloride for adjustment to isotonicity = 0.9 – ( % of medicament in solution x sodium chloride equivalent of medicament) • When adjustment is to be made with a substance other than sodium chloride first calculate the required percentage of sodium chloride and then divide this by the sodium chloride equivalent of the chosen adjusting substance • Example o Calculate the percentage of anhydrous dextrose required to render a 1% solution of ephedrine hydrochloride iso – osmotic with body fluid o Sodium chloride equivalent  Ephedrine hydrochloride = 0.30  Anhydrous dextrose = 0,18 • Therefore percentage of sodium chloride for adjustment = 0.9 – ( 1 x 0.3) == 0.6 • Equivalent percentage of anhydrous dextrose = 0.6 = 6 0.1 STEP 3: Determining the Isotonicity by Sodium Chloride Equivalent Method (60 minutes) • Calculate the amount of NaCl required to make the following ophthalmic solution isotonic. Rx Atropine Sulfate 2% NaCl qs Aqua. dist. qs. ad. 30 ml M.ft. isotonic solution 99 1. Determine the amount of NaCl to make 30 ml of an isotonic solution 2. Calculate the contribution of atropine sulfate to the NaCl equivalent 3. Determine the amount of NaCl to add to make the solution isotonic by subtracting (2) from (1) Other substances may be used, in addition to or in place of NaCl, to render solutions isotonic. This is done by taking the process one step further and calculating the amount of the substance that is equivalent to the amount of NaCl calculated in step 3. For example, boric acid is often used to adjust isotonicity in ophthalmic solutions because of its buffering and anti-infective properties. If E for boric acid is 0.50, then the amount of boric acid needed to replace the NaCl in step 3 can be calculated: Or or, more simply: Thus, 0.38 g or 380 mg of boric acid would be required to render the previous ophthalmic solution isotonic 100 To determine amount of ingredients required making isotonic solution Activity: Small Group Discussion (20 minutes) DIVIDE students in small manageable groups ASK students to discuss in groups on the following questions • Calculate the percentage of sodium chloride needed to make an injection containing morphine sulphate 1%, hyoscine hydrobromide 0.04% and sodium metabisulphate 0.1% iso – osmotic with blood plasma REFER Students to Pharmaceutical Calculation. 13th Edition by HOWARD C. ANSEL: Chapter 11,for reference ALLOW students to discuss for 20 minutes ALLOW each groups to present for 5 minutes CLARIFY and SUMMARIZE by using the contents below • Sodium chloride equivalent o Ephedrine hydrochloride = 0.30 o Anhydrous dextrose = 0,18 • Therefore percentage of sodium chloride for adjustment: = 0.9 – ( 1 – 0.14) + ( 0.04 x 0.12 ) + ( 0.1 x 0.70 ) = 0.9 – (0.14 + 0.0048 + 0.07) = 0.9 – 0.2148 = 0.6852 = 0.69 • Percentage of sodium chloride for adjustment = 0.69% STEP 4: Key Points (5 minutes) • The Sodium Chloride equivalent method (E) is the amount of sodium which has the same osmotic effect (based on number of particles) as 1g of the drug sample calculation • Percentage of sodium chloride for adjustment to isotonicity = 0.9 – ( % of medicament in solution x sodium chloride equivalent of medicament) 101 • When adjustment is to be made with a substance other than sodium chloride first calculate the required percentage of sodium chloride and then divide this by the sodium chloride equivalent of the chosen adjusting substance STEP 5: Evaluation (5 minutes) • What is sodium chloride equivalent method (E) in determining isotonic? • What is the percentage of sodium chloride for adjustment to isotonicity? STEP 6: Take Home Assignment (15 minutes) Activity: Take home Assignment (15 minutes) ASK each individual student to do the following assignment • Calculate the strength of sodium chloride solution which is iso – osmotic with blood serum and tears ALLOCATE time for students to do the assignment and submit REFER students to recommended references 102 References Ansel, H. C &Stocklosa, M. J. (2001). Pharmaceutical Calculations (11th ed.). Philadelphia, United States: LIPPINCOTT WILLIAMS & WILKINS Ansel, H. C (2010) Pharmaceutical Calculations (13rd ed.). Philadelphia, United States: LIPPINCOTT WILLIAMS & WILKINS Senya, S. S, Mwasha, C.Y, Muyinga, A. M, Amiri,R. I. and Mauga E.A.S.K. (2011) Tanzania Pharmaceutical Handbook (2nd ed.). Dar eS Salaam, Tanzania: School of Pharmaceutical Sciences. Zatz, J.L and Teixeira, M.G. (2005). Pharmaceutical Calculation (4th ed.). New Jersey: John Wiley & Sons, Inc 103 ← Previous TopicNext Topic →View all

NTA Level 5 • Semester 2 • PST05208 Determination of Isotonicity by Freezing Point Pharmaceutics Theory and Compounding • Source Session/Topic 12 Full source-text version: all educational wording from the extracted learning source is retained; only presenter/tutor metadata and web-layout noise are removed, while formatting is improved for readability. Session 12: Determination of Isotonicity by Freezing Point Method Total Session Time: 120 minutes + 6 hours of Practices Prerequisites • None Learning Tasks By the end of this session students are expected to be able to: • Define freezing point depression • Determine the isotonicity by freezing point depression Resources Needed: • Flip charts, marker pens, and masking tape • Black/white board, chalk and whiteboard markers • LCD projector and computer SESSION OVERVIEW Activity/ Step Time Content Method 1 05 Minutes Presentation Introduction, Learning Tasks 2 45 Minutes Presentation Introduction to Freezing Point Depression Presentation Determining the Isotonicity by freezing Point 3 60 Minutes Small Group Depression Discussion 4 05 Minutes Presentation Key Points 5 05 Minutes Presentation Evaluation 92 SESSION CONTENTS STEP1: Presentation of Session Title and Learning Tasks (5 minutes) READ or ASK students to read the learning tasks and clarify ASK students if they have any questions before continuing STEP 2: Freezing Point Depression (45 minutes) • The freezing point of body fluid is at – 0.520C. • Any solution freezing at – 0.520C will have the same osmotic pressure as body fluids • Hypotonic solution will have a higher freezing-point and consequently need the addition of adjusting substance to reach a freezing point of – 0.520C • The amount of adjusting substance which is often sodium chloride, is calculated as shown in the following examples. o Solution A freezes at – 0.390C. How much sodium has to be added to obtain an isotonic solution? o The adjusting substance sodium chloride has to decrease the freezing point by 0.52 – 0.39 = 0.130C o A 1 % w/v sodium chloride solution freezes at – 0.5760C; so the amount needed is found by the proportion: o 1 == x x == 1 x 0.13 == 0.2%w/v 0.576 0.13 0.576 • Therefore, solution A will need the addition of 0.2%w/v sodium chloride in order to become isotonic with body fluids; e.g. 50mL of this solution require 0.1g sodium chloride. • From the above used proportion, the following equation has been derived: o W == 0.52 – a b o Where:  w== concentration %w/v of adjusting substance in the final solution  a == freezing point of unadjusted solution  b==freezing point of a 1%w/v solution of adjusting substance • The freezing points of 1%w/v aqueous solution of various substances are obtain from the table below; “a „ is calculated by multiplying the freezing depression of a 1%w/v solution by the concentration of unadjusted solution expressed as %w/v 93 • Example: o Adjust 1000mL of a 2%w/v solution of anhydrous dextrose isotonic with body fluids; adjusting substance is sodium chloride  Freezing point of 1%w/v anhydrous dextrose solution : – 0.1010C  Freezing point of unadjusted solution: – 0.101 x 2 == – 0 .2020C  Freezing point of 1%w/v sodium chloride solution = -0.5760C  The last two values are substituting “ a “ and “b” in the formula o W = 0.52 – 0.202 = 0.318 = 0.55%w/v 0.576 0.576 o Therefore, a 2%w/v anhydrous dextrose solution requires the addition of 0.55%w/v sodium chloride in order to become isotonic with blood serum and tears; for 1000mL solution 5.5g are needed STEP 3: Determining the Isotonicity by Freezing Point Depression (60 minutes) To determine amount of ingredients required to make isotonic solution Activity: Small Group Discussion ( 30 minutes) DIVIDE students in small manageable groups ASK students to discuss in groups on the following questions 1. Render the following solution iso – osmotic with blood and tears Ephedrine hydrochloride…………………………0.1g Chlorbutol…………………………………………0.1g Water for preparation……………………….to 20mL Adjusting substance is sodium chloride REFER Students to Pharmaceutical Calculation. 13th Edition by HOWARD C. ANSEL: Chapter 11, For reference ALLOW students to discuss for 20 minutes ALLOW each groups to present for 5 minutes CLARIFY and SUMMARIZE by using the contents below 94 • The freezing point depression of 1%w/v solution is obtained from the table below. • First the concentrations %w/v must be calculated for ephedrine hydrochloride and Chlorbutol by proportion: 0.1 = xx =0.1 x100 x = 0.5%w/v 20 100 20 • A 0.5%w/v solution of ephedrine hydrochloride has a freezing point depression of 0.165 x 0.5 = 0.08250C • A 0.5%w/v solution for Chlorbutol has a freezing point depression of 0.14 x 0.5 =0.070C • Now the equation can be used: W = 0.52 – 0.1525 ==0.3675 = 0.638%w/v =0.64%w/v 0.576 0.576 • Therefore, the unadjusted solution needs the addition of 0.64%w/v sodium chloride. The actual amount for 20mL solution is found by simple proportion 0.638 = x x = 0.638 x 20 =0.1276gm of sodium chloride 100 20 100 • So the formula for the adjusted solution becomes: Ephedrine hydrochloride…………………………0.1g Chlorbutol………………………………………..0.1g Sodium chloride………………………………….0.13g Water for preparation……………………….to 20mL STEP 4: Key Points (5 minutes) • Freezing-point depression is the decrease of the freezing point of a solvent on addition of a non- volatile solute • The formula for determining the Isotonicity by freezing point depression is w= 0.52 – a b STEP 5: Evaluation (5 minutes) • What is freezing point depression in determining isotonicity? • What is the formula for determining isotonicity by freezing point depression? 95 STEP 6: Take Home Assignment (15 minutes) Activity: Take home Assignment (15 minutes) ASK each individual student to do the following assignment • Calculate the strength of sodium chloride solution which is iso – osmotic with blood serum and tears ALLOCATE time for students to do the assignment and submit REFER students to recommended references 96 References Aulton, M. E., & Taylor, K. (2013). Aulton's pharmaceutics : the design and manufacture of medicines (4th ed.). Edinburgh: Churchill Livingstone/Elsevier. Pharmaceutical Society of Great Britain., & Pharmaceutical Society of Great Britain. Department of

NTA Level 5 • Semester 2 • PST05208 Introduction to Isotonicity Pharmaceutics Theory and Compounding • Source Session/Topic 11 Full source-text version: all educational wording from the extracted learning source is retained; only presenter/tutor metadata and web-layout noise are removed, while formatting is improved for readability. Session 11: Introduction to Isotonicity Total Session Time: 60 minutes Prerequisites • None Learning Tasks By the end of this session students are expected to be able to: • Define terms used in isotonicity • Explain the importance of isotonic solution. • Explain effects of administering paratonic solutions Resources Needed: Resources Needed: • Flip charts, marker pens, and masking tape • Black/white board, chalk and whiteboard markers • LCD projector and computer SESSION OVERVIEW Step Time Activity/Method Content 1 05 Minutes Presentation Introduction, Learning Tasks Presentation Definition of Terms Used in Isotonicity and 2 15 Minutes Brainstorming Electrolytes. 15 Minutes Presentation Importance of isotonic solution. 3 Small group discussion 4 15 Minutes Presentation Effects of administering paratonic solutions 5 05 Minutes Presentation Key Points 05 Minutes Presentation 6 Evaluation 84 CONTENTS STEP 1: Presentation of Session Title and Learning Tasks (5 minutes) READ or ASK students to read the learning tasks and clarify ASK students if they have any questions before continuing. STEP 2: Definition of Terms used in Isotonicity and Electrolytes (15 minutes) Activity: Buzzing (5 minutes) ASK students to pair up and buzz on the following question for 2 minutes • What is isotonicity? ALLOW few pairs to respond and let other pairs to add on points not mentioned WRITE their response on the flip chart/board CLARIFY and SUMMARIZE by using the content below • Body fluids e.g. blood plasma have certain quantity of dissolved substances e.g. salts that determine power of the blood called osmotic pressure. • Osmotic pressure is the pressure created by solvent molecules moving from a low concentrated solution to higher concentrated solution through a semi-permeable membrane, a semi-permeable membrane allow only the solvent molecules to pass. • The characteristics of a solution is determined by the number of dissolved substances in it, aqueous solutions which exert the same osmotic pressure as blood plasma are said to be isotonic with plasma. This is equivalent to 0.9%w/v sodium chloride injection (Normal saline) • An isotonic solution is a solution in which body cells can be bathed without a net flow of water across a semipermeable membrane. E.g. 0.9% normal saline (NS). • Osmolality o A unit of measure of osmotic pressure. o Blood has 300 mOsmol per liter. (MiliOsmolality) o Both Normal Saline and Dextrose 5% solutions have a similar osmolarity. 85 • Solutions, which exert a different osmotic pressure as blood plasma, are said to be paratonic with plasma. (i.e. solutions with hypotonic or hypertonic osmotic pressure) • Those solutions with a lower osmotic pressure than blood plasma are said to be hypotonic solutions and a solution of less than normal tonicity is hypotonic, which has fewer numbers of dissolved solutes than blood cells. E.g. 0.45% Normal Saline. • Solutions with a higher osmotic pressure than blood plasma are said to be hypertonic solutions, a hypertonic solution has more number of dissolved solutes than the blood cells themselves. E.g. 50% Dextrose or 3% Sodium chloride. • Aqueous solutions intended to contact with body fluid should preferably be made isotonic with speci※c bodily ‼uid in order to minimize any possible adverse effects. • Tonicity is generally classified in three types: o Hypertonicity o Hypo tonicity o Isotonicity • Hypertonic, isotonic and hypotonic solutions are defined in reference to a cell membrane by comparing the tonicity of the solution with the tonicity within the cell. • For a solution to be termed isotonic (equal tone) it must have the same osmotic pressure as a speci※c bodily ‼uid. The easiest way to calculate the osmotic pressure of a solution is to utilize the more easily measured property of the freezing point depression, as they are proportional to one another. • In pharmacy, isotonicity calculations are most often performed for parenteral and ophthalmic solutions, which must have a freezing point depression of 0.520C for them to be isotonic with blood plasma and tears. Therefore a solution is considered to be isotonic if it has a freezing point of -0.520C • Electrolytes are substances containing free ions, thus rendering the substance electrically conductive • Electrolyte preparations are used in the treatment of disturbances of electrolyte and fluid balance in the body. In clinical practice, they are provided in the form of oral solutions, and syrup as dry granules intended to be dissolved in water or juice to make an oral solution and when necessary as intravenous infusions 86 STEP 3: Importance of Isotonicity (15 minutes) Activity: Small Group Discussion ( 10 minutes) DIVIDE students into small manageable groups ASK students to discuss on the following question • What is the importance of isotonicity? REFER Students to Tanzania Pharmaceutical Handbook pg. 33-50 ALLOW students to discuss for 10 minutes ALLOW few groups to present and the rest to add points not mentioned CLARIFY and SUMMARIZE by using the contents below • It is important for a solution to be isotonic with a bodily ‼uid to prevent irritation and cell damage, and to maximize drug efficacy. Its neutrality is not affecting the osmotic pressure of a cell • This state allows for the free movement of water across the membrane without changing the concentration of solutes on either side. • Isotonic solutions are used in Oral Rehydration Therapy (ORT), which is a treatment for dehydration associated with gastroenteritis. ORT replenishes the body's lost electrolytes and promotes the uptake of glucose and fluid by the intestinal epithelial cells. • Isotonic solutions given intravenously in case of severe dehydration and is used in the treatment of hypernatremia (a condition in which serum concentration of sodium is high) in individuals who have impaired water excretion, eg 0.9% NaCl solution • It can be used as a medium for intravenous administration of drugs like propofol, which are not

NTA Level 5 • Semester 2 • PST05208 Compounding of Gels/ Jellies Pharmaceutics Theory and Compounding • Source Session/Topic 10 Full source-text version: all educational wording from the extracted learning source is retained; only presenter/tutor metadata and web-layout noise are removed, while formatting is improved for readability. Session 10: Compounding of Gels/ Jellies Total Session Time: 120 Minutes Prerequisites • None Learning Tasks By the end of this session students are expected to be able to:  Give overview of gels  Describe procedures for preparing gels  Prepare gels  Label prepared gels  Dispense prepared gels into suitable containers and closure Resources Needed: • Flip charts, marker pens, and masking tape • Black/white board, chalk and whiteboard markers • LCD projector and computer • Figure 10.1: Gels jars and tube SESSION OVERVIEW Activity/ Step Time Content Method 1 05 Minutes Presentation Introduction, Learning Tasks Presentation 2 25 Minutes Buzzing Introduction to Gels 3 10 Minutes Presentation Procedures for Preparing Gels Presentation 4 Preparing Gels 35 Minutes Demonstration Presentation 5 25 Minutes Demonstration Labeling of Prepared Gels Presentation Dispensing of Prepared Gels into Suitable Containers 6 10 Minutes Brainstorming and Closure 7 05 Minutes Presentation Key Points 8 05 Minutes Presentation Evaluation 74 SESSION CONTENTS STEP 1: Presentation of Session Title and Learning Tasks (5 minutes) READ or ASK students to read the learning tasks and clarify ASK students if they have any questions before continuing. STEP 2: Introduction to Gels (25 minutes) Activity: Buzzing (5 minutes) ASK students to pair up and buzz on the following question for 2 minutes • What are Gels? ALLOW few pairs to respond and let other pairs to add on points not mentioned WRITE their response on the flip chart/board CLARIFY and SUMMARIZE by using the content below • Gels are transparent (or translucent) semi-solid or solid preparations of either suspensions made up of small inorganic particles or large organic molecules interpenetrated by a liquid • Are semisolid systems consisting of dispersion of small or large molecules in an aqueous liquid vehicle rendered jelly-like through addition of a gelling agent • Movement of dispersing system is restricted by interlacing network of particles or solvated macromolecules of dispersed phase • Gels are made using substances (called gelling agents) that undergo a high degree of cross- linking or association when hydrated and dispersed in the dispersing medium or when dissolved in the dispersing medium • Cross-linking or association of dispersed phase alters the viscosity of the dispersing medium • Hence the movement of dispersing medium is restricted by the dispersed phase • Two-phase System Gel o Is when the gel contains a network of small discrete particles • One or Single-phase System Gel o When gels appear to have no discrete particles, 75 o Macromolecules are uniformly distributed throughout the liquid, usually organics o No apparent boundaries between dispersed macromolecules and the liquid o Two-phase system gels are thixotropic Gel mass consists of floccules of small distinct particles Usually involve inorganics Semisolid on standing but liquefy on shaking If the particles in the two-phase systems are large • The gel is called a Magma or Milk Composition of gels is: o Gelling agents o Water o Co-solvents o Preservatives o Stabilizers • One-phase gels are made from synthetic or natural organic macromolecules distributed uniformly throughout a liquid without any boundary between dispersed and dispersing phases (e.g. tragacanth, cellulose, methylcellulose) • If made from natural gum, single-phase gels are called Mucilages • Some Gelling Agents o Acacia, tragacanth, Methylcellulose, o Alginic acid, Bentonite, Carbopol®, o Carboxymethylcellulose (CMC), o Gelatin, Hydroxypropyl cellulose, o Magnesium Aluminium Silicate (Veegum®) etc • Characteristic Properties of gels are:often non-greasy, smooth, elegant, produce cooling effect on application and easily washable from skin STEP 3: Procedures for Preparing Gels (10 minutes) • Specific method depends on gelling agent o e.g. gelatin gels o By dispersing gelatin in hot water followed by cooling OR o By wetting gelatin with organic liquid e.g. Propylene glycol, followed by addition of hot water and cooling • Example of a gel formulation: • Methylcellulose and Carbomer Gel Base o Methylcellulose, 4000cps 1.0% o Carbomer 934 0.35% o 1N Sodium Hydroxide Solution qs to pH 7 76 o Propylene glycol 16.7% o Methyl paraben 0.015% o Purified water, qs 100 o Disperse the methylcellulose in a portion of hot (80-90°C) water o Cool to room temperature, and disperse the Carbomer 934 in the gel using a bladed stirrer. o Adjust the pH of the dispersion to 7.0 by adding sufficient 1N Sodium hydroxide solution. o Dissolve the methylparaben in the propylene glycol. Mix the methylcellulose, Carbopol 934 and propylene glycol fractions using caution to avoid incorporating air. STEP 4: Preparing of gels (35 minutes) Activity: Demonstration (45 minutes) DIVIDE students in small manageable groups. PREPARE equipment and materials needed for demonstration POSITION students so that everyone can see and hear REVIEW steps of preparing pastes with students REFER students to the official formula of preparing Ichthammol gel IP In the Tanzania Pharmaceutical Handbook DEMONSTRATE the procedure of preparing ointments according to the instruction under the official formula in the Tanzania Pharmaceutical Handbook (Mitte 50g) ALLOW one student from each group to do a return demonstration and let others comment on it CLARIFY and SUMMARIZE Using the content below INFORM the students that “every student will practice in the skills laboratory under supervision until is competent” • Official formula given for Ichthammol gel PCx Ichthammol……………………………………………………….. 1.0g Tragacanth, in powder……………………………………………… 2.5g Alcohol 90%……………………………………………………………… 5.0g 77 Glycerin……………………………………………………………… 1.0g Purified water q.s………………………………………………………… 50.0g Mitte 50.0g for Allen Mbogo • Thus, amount of official formula = 100g • Total amount required = 50g • Amount of each ingredient (x) =? • FACTOR= Required amount Official amount = 50g 100g = 0.5 Then, amount of each ingredient can be obtain as summarised in the table below: Formula Official amount Factors Required amount Ichthammol 1.0g 0.5g Tragacanth, in powder 2.5g 0.5 1.25g Alcohol 90% 5.0g 2.5g Glycerol 1.0g 0.5g Purified water q.s 50.0g 25g

NTA Level 5 • Semester 2 • PST05208 Compounding of Creams Pharmaceutics Theory and Compounding • Source Session/Topic 9 Full source-text version: all educational wording from the extracted learning source is retained; only presenter/tutor metadata and web-layout noise are removed, while formatting is improved for readability. Session 9: Compounding of Creams Total Session Time: 120 minutes Prerequisites • None Learning Tasks By the end of this session students are expected to be able to:  Define creams  Describe procedures for preparing creams  Prepare creams  Label prepared cream  Dispense prepared cream into suitable containers and closure Resources Needed: • Flip charts, marker pens, and masking tape • Black/white board, chalk and whiteboard markers • LCD projector and computer • Figure 9.1: Cream jars and tube SESSION OVERVIEW Step Time Activity/Method Content 1 05 Minutes Presentation Introduction, Learning Tasks 15 Minutes Presentation Definition of creams 2 Buzzing 3 15 Minutes Presentation Procedures for preparing creams Presentation 4 Preparing of Creams 30 Minutes Demonstration Presentation 5 Labeling of prepared creams 30 Minutes Demonstration 15 Minutes Presentation Dispensing of prepared creams into suitable 6 Brainstorming containers and closure 7 05 Minutes Presentation Key Points 8 05 Minutes Presentation Evaluation 65 SESSION CONTENTS STEP 1: Presentation of Session Title and Learning Tasks (5 minutes) READ or ASK students to read the learning tasks and clarify ASK students if they have any questions before continuing. STEP 2: Definition of Creams (15 minutes) Activity: Buzzing (05 minutes) ASK students to pair up and buzz on the following question for 2 minutes • What are creams? ALLOW few pairs to respond and let other pairs to add on points not mentioned WRITE their response on the flip chart/board CLARIFY and SUMMARIZE by using the content below • Creams are semi-solid emulsions intended for application to the skin or mucous membrane, they are viscous semisolid emulsion system with opaque appearance • Usually they are of two types, namely: o Aqueous creams, i.e. oil-in-water emulsions (O/W). This creams contain oil-in-water emulsifying agent o Oily creams, i.e. water-in-oil emulsions (W/O). This cream contain water-in-oil emulsifying agent. • Aqueous creams are relatively non-greasy and are most useful as water-washable bases whereas oily creams are emollient and cleansing. • Patients often prefer a w/o cream to an ointment because the cream spreads more readily, is less greasy and the evaporating water soothes the inflamed tissue. • O/w creams („vanishing‟ creams) rub into the skin; the continuous phase evaporates and increases the concentration of a water-soluble drug in the adhering film. • Consistency and rheological characters depend on weather the cream is w/o or o/w. • Creams intended for application to large open wounds should be sterile. 66 STEP 3: Procedures for Preparing Creams (15 minutes) • Emulsified creams are prepared by heating the components of the oily phase including the emulgent until molten and then cooling to 60°C. • The components of the aqueous phase are mixed in a separate vessel and also heated at 60°C. • The aqueous phase is then added to the oily phase at the same temperature, this is very important and the thermometer should be used. • The resulting emulsion should be stirred until cool. • Rapid cooling may result in separation of high melting point components and excessive aeration caused by vigorous stirring may also lead to a granular product. • If necessary the product may be homogenized after cooling. • Dilution of creams o Should be done under hygienic conditions o Stability and bacteriocidal properties of original cream should NOT be reduced on dilution o The diluent should be compatible with ingredients and the whole original preparation o Dilution may alter extent of release of medicament from creams o Information on suitability of diluent should be available, otherwise DO NOT DILUTE • Preservation & hygiene during preparation o Creams can support growth of micro-organisms, hence most need preservatives o Accidental contamination will almost certainly occur during bench scale manufacture, but good technique can keep this to a minimum o All apparatus and final containers should be thoroughly cleansed before use and purified water should be used in the preparation of products 67 STEP 4: Preparing of Creams (30 minutes) Activity: Demonstration (15 minutes) DIVIDE students in small manageable groups. PREPARE equipment and materials needed for demonstration POSITION students so that everyone can see and hear REVIEW steps of preparing pastes with students REFER students to the official formula of preparing aqueous calamine cream PCx In the Tanzania Pharmaceutical Handbook DEMONSTRATE the procedure of preparing cream according to the instruction under the official formula in the Tanzania Pharmaceutical Handbook (Mitte 50g) ALLOW one student from each group to do a return demonstration and let others comment on it CLARIFY and SUMMARIZE Using the content below INFORM the students that “every student will practice in the skills laboratory under supervision until is competent” • Official formula given for aqueous calamine cream PCx Calamine……………………………………………………………… 4.0g Zinc oxide……………………………………………………… 3.0.0g Emulsifying wax………………………………………………… 6.0g Arachis oil (0.86g/mL)…………………………………………….. 30.0g Purified water, freshly boiled and cooled ………………………… 50.0g Mitte 50.0g for Okeyo Odhiambo o Thus, amount of official formula = 100g o Total amount required = 50g o Amount of each ingredient (x) =? o FACTOR= Required amount Official amount = 50gm 100gm = 0.5 68 Then, amount of each ingredient can be obtain as summarised in the table below: Formula Official amount Factor Required amount Calamine 4.0gm 2.0g Zinc oxide 3.0gm 1.5g Emulsifying wax 6.0gm 0.5 3.0g Arachis oil (0.86g/mL) 30.0gm 15.0g Purified water, freshly 57.0gm 28.5g boiled and cooled Method of preparation • Melt oily phase (heat up to 75oC, and cool to 60oC) • Prepare aqueous phase and heat to 60oC • Add aqueous phase to oily phase with slow but constant agitation until cold • Incorporate insoluble medicinal ingredient • Then transferred to suitable container, label and ready for dispensing to a patient. STEP 5: Labeling of Prepared Creams (30 minutes) • The preparation should be labelled “For External Use Only” in red ink and