Calculations Involving Milliequivalent – PST05208 Pharmaceutics Theory and Compounding
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
