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# Isotonic calculation

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### Isotonic calculation

1. 1. CALCULATION FOR SOLUTION ISOTONIC WITH BLOOD AND TEARSSome of the many ways; 1. Freezing point depression method 2. NaCl equivalent method 3. White Vincent method 4. Sprowl method 5. Molecules concentration method 6. Graphical method on vapor pressure and freezing point determination 1. FREEZING POINT DEPRESSION METHOD • The plasma and blood freezing point temperature = -0.52°C • The dissolved substances in plasma or tear depress the solution freezing point below 0.52°C • Any solution that freeze at T=-0.52°C is isotonic with blood and tear • weight of substance that need to be adjusted to make it hypotonic given by the following formula; 0.52 − = Where, W=the weight, in g, of the added substance in 100ml of the final solution a =the depression of the freezing point of water produced by the medicament already in the solution (calculated by multiplying the value for b for the medicament by the strength of the solution express as % w/v) b=the depression of freezing of water produced by 1 per cent w/v of the added substance ~mlk~ | 1Freezing Point Depression Method
2. 2. Example 1.1 Rx ∆Tf 1% NaCl 900mg 0.576 Dextrose q.s (20.5g) 0.101 Ft. isotonic soulution500ml Or; 0.9% ⁄ = Solution; 500 100 . . . ( × . ) = 0.18% ⁄ = = . = 4.12 100 For 500ml; 500 = (4.12 ) × 100 = 20.6Example 1.2 Rx Ephedrine HCl 1g Chlorobutol 0.5g NaCl (1.4g) q.s Distilled water q.s ad 200ml Solution; 0.52 − 0.52 − [(0.5 × 0.165) − (0.25 × 0.138)] = = 0.576 = 0.7 100 200 ; 200 = (0.7 ) × 100 = 1.4 ~mlk~ | 2Freezing Point Depression Method
3. 3. EXercise 1.3Prepare 500ml NaHCO3 (∆T.f 1%=0.38) so that when it is dilute with the same amount of water, it wouldbe isotonic.Solution; Rx ∆T.f 1% NaHCO3 Xg(13.7) 0.38 Water qs ad 500mlThe solution is hypertonic When diluted NaHCO3 Xg Water qs ad 1000mlThis solution is isotonic 0.52 − 0.52 − 0 = = = 1.37 ℎ ℎ 0.38NaHCO3 required to make the solution isotonic upon dilution is;1.37 χ =100 10001.37 X 10 = 13.7g (to be dissolved to 500ml solution) ~mlk~ | 3Freezing Point Depression Method
4. 4. 2. NaCl EQUIVALENT METHOD (E) • Based on the factor called the sodium chloride equivalent which can be used to convert a specified concentration of medicament to the concentration of medicament to the concentration of sodium chloride that will produce the same osmotic effect. • Standard → 0.9 % w/v NaCl at isotonic • Method to calculate % at isotonic, ∆Tf 1% =0.576 ~ NaCl 0.52 − 0.52 − 0 = = = 0.9 0.9% ⁄ 0.576 Known as normal saline • Sodium chloride equivalents (E1%) can be calculated from the following formula; ( , . 927) = 0.576 ( ℎ ℎ )Formula & Method 1 •Look up in the table the sodium chloride equivalent for the srenght of solution nearest to the strenght = 0.9% - ([E1%X %w/v] + …….) of medicament in the preparation. 2 •Multiply this by the strenght of the medicament. 3 •substract the result from 0.9 per cent; the difference is strenght of sodium chloride necessary to adjust the solution to iso-osmoticity. ~mlk~ | 4 aCl Equivalent Method
5. 5. Example 2.1 ∆Tf 1% ascorbic acid =0.105°C ∆Tf 1% NaCl = 0.576°CWhat is the E for 1% ascorbic acid?Solution . ° E 1% ascorbic acid= . ° = 0.18Example 2.2 Rx NaCl 0.2%w/v Dextrose q.s Ft. isotonic solution 500mlSolution; ∆Tf 1% NaCl = 0.576°C ∆Tf 1% dextrose = 0.101 °C . °For dextrose 1 % (E1%) = . ° = 0.18Conclusion → 1% dextrose has osmotic pressure equals to 0.18% NaClNaCl in the solution = 0.2 %∴Remaining amount of NaCl to be added for isotonic 0.9% - 0.2% =0.7%1% dextrose equivalent to 0.18% NaCl Convert amount of NaCl requiredχ% dextrose equivalent to 0.7% NaCl to amount of dextrose required (0.7%)(1%)χ= = 3.89% 0.18%For 500ml;3.89 χ =100 500χ = 19.45 g ~mlk~ | 5 aCl Equivalent Method
6. 6. Example 2.3 Rx E 1% Ephedrine HC 1g 0.3 Chlorobutol 0.5g 0.24 NaCl q.s Distilled water 200ml isotonic Solution; For ephedrine HCl × 100 % × 0.3 = 0.15% . For chlorobutol × 100 % × 0.24 = 0.06% Amount of NaCl for isotonic; 0.9% - (0.15% + 0.06%) = 0.69% So for 200ml……………….= 1.38 Example 2.4 Rx E 1% Let adjust isotonicity using KCl instead Ephedrine HCl 1g 0.3 of NaCl Chlorobutol 0.5g 0.24 KCl q.s 0.4 Distilled water 200ml isotonic Solution; 1% KCl equivalent to 0.4% NaCl Convert amount of NaCl required to χ% ← 0.69% NaCl (refer to example 3) amount of KCl required 0.69 χ= = 1.725% 0.4 So for 200ml …..= 3.45g KCl ~mlk~ | 6 aCl Equivalent Method
7. 7. Example 2.5 Calculate the percentage of sodium chloride required to render a 0.5 per cent solution of potassium chloride iso-osmotic with blood plasma. Solution Sodium chloride equivalent of 0.5 per cent potassium chloride = 0.76 ∴ Percentage of sodium chloride for adjustment = 0.9 − (0.5 Χ 0.76) = 0.9 − 0.38 = 0.52 Example 2.6 Calculate the percentage of anhydrous dextrose required to render a 1 per cent solution of ephedrine hydrochloride iso-osmotic with body fluid. ∴ Percentage of sodium chloride for adjustment = 0.9 − (1 Χ 0.3) = 0.6 Equivalent percentage of anhydrous dextrose = 0.6/0.18 = 3.33 Example 2.7 Select a suitable substance for an eye lotion 0.5 per cent of silver nitrate and calculate the percentage required to render the lotion iso-osmotic with lachrymal secretion. Sodium chloride is unsuitable because silver nitrate is incompatible with chloride. Potassium nitrate will be used Sodium chloride equivalent of 0.5 per cent silver nitrate = 0.33 = 0.9 − (0.5 Χ 0.33) = 0.9 − 0.165 = 0.753 . Equivalent percentage of potassium nitrate = = 1.3 . ~mlk~ | 7aCl Equivalent Method
8. 8. 3. WHITE VINCENT METHOD Principle: A Isotonic solution + B isotonic solution = C isotonic solution→This method involves the addition of water to medicament to obtain an isotonic solution. Thisfollowed by the addition of isotonic buffer solution or preservatives isotonic solution to the requiredvolume. V = W X E1% X 111.1Where;V= V (ml) of isotonic solution that could be obtain in wg of drug in water (the amount of water to addedto form isotonic solution)W= amount of drug in the formulaE1%=NaCl equivalent of the drug111.1= constant that could be find from volume for 1% isotonic NaCl 0.9% NaCl → 1% NaCl 0.9g NaCl → 100 ml 1 g NaCl → 111.1 ml ~mlk~ | 8White Vincent & Sprowl Method
9. 9. Example 3.1 Rx E 1% Ephedrine HCl 1g 0.3 Chlorobutol 0.5g 0.24 NaCl q.s Distilled water 200ml isotonicSolution; This solution is isotonic (A)For ephedrine HCl, V=1 X 0.3 X 111.1 = 33.33ml (final volume) So the overallFor chlorobutol, V=0.5 X 0.24 X 111.1 = 13.33ml (final volume) This solution is isotonic (B) solution is isotonic200ml – (33.33 + 13.33) ml = 153.34ml∴ Amount of NaCl required to adjust 153.34ml to isotonic; This solution is isotonic (C) 0.9= × 153.34 = 1.38 100 4. SPROWL METHOD • Using white Vincent method but w is set to constant, 0.3 V = W X E1% X 111.1 or V = 33.33E1% ~mlk~ | 9White Vincent & Sprowl Method
10. 10. 5. MILIEQUIVALENT (mEq) • Definition; the gram equivalent weight of an ion is the ionic weight (the sum of atomic weights of the element in an ion) in gram divided by the valence of that ion. • A mililequivalent is one thousandth part of the gram equivalent weight, the same figure expressed in milligram. Table 5.1 makes this clear Gram equivalent weight Ion Ionic weight ℎ Weight of 1mEq (mg) ( ) Sodium Na+ 23 23 23 Potassium K+ 39.1 39 39.1 Calcium Ca2+ 40 20 20 Chloride Cl− 35.5 35.5 35.5 Bicarbonate HCO3− 61 61 61 Phosphate HPO4− 96 48 48 • ∴ Gram equivalent weight i) For ion; ℎ ℎ = Example 1 Eq sodium → 23/1 = 23g 1Eq chloride → 35.5/1 =35.5 • The weight of a salt containing 1mEq of a particular ion is obtained by dividing the molecular weight of the salt by the valency of that ion multiply by the number of such ion in the molecules ℎ , , 1 ℎ = × ℎ ℎ ~mlk~ | 10Miliequivalent (mEq)
11. 11. Table 5.2 makes this clear Weight of salt M. of Valency of Number of containing 1mEq Ion Salt used the salt the ion ion in salt of the ion Na+ Sodium chloride 58.5 1 1 58.5mg Na+ Sodium phosphate 358† 1 2 179.0mg Ca2+ Calcium chloride 147† 2 1 73.5mg Cl- Calcium chloride 147 1 2 73.5mg († N.B. the water of crystallization must not be ignored) ii) For salt; ℎ ℎ ℎ = Example (monovalent) 1 Eq → 58.5/1= 58.5g Example (not monovalent) 1Eq CaCl2.H2O → 147/2 =73.5Miliequivalent (mEq) → 1/1000 Gram Eq weightExample1mEq sodium →23mg1mEq chloride → 35.5g1mEq NaCl → 58.5 mg1mEq CaCl2.H2O →73.5 mg ~mlk~ | 11Miliequivalent (mEq)
12. 12. Conversion Equations I. To convert mEq/litre to mg/litre; × II. To convert mEq/litre to g/litre; × 1000 III. To convert mEq/litre to %w/v; × 10000• The BPC includes table showing the weight s of salt that contain 1 mEq of specified ion; Mg of salt containing 1 Ion Miliequivalent (mEq) mg Salt mEq of specified ion Na+ 23.0 Sodium chloride 58.5 Sodium bicarbonate 84 K+ 39.1 Potassium chloride 74.5 Ca2+ 20.0 Calcium chloride 73.5 Mg2+ 12.5 Magnesium sulphate 123 Magnesium chloride 101.5 Cl- 35.5 Sodium chloride 58.5 HCO3- 61.0 Sodium bicarbonate 84 HPO4- 48 Sodium phosphate 179 ~mlk~ | 12Miliequivalent (mEq)
13. 13. Example 5.1 Calculate the quantities of salt required for the following electrolyte solution; Sodium ion 20 mEq Potassium ion 30 mEq Magnesium ion 5 mEq Phosphate ion (HPO4-) 10 mEq Chloride ion 45 mEq Water for injection, to 1 litre From table 5.3:- 1 mEq of both potassium and chloride ion are contained in 74.5 mg of potassium chloride. 1 mEq of both sodium and phosphate ion are contained in 179 mg of sodium phosphate. 1 mEq of both magnesium and chloride ion are contained in 101.5 mg of magnesium chloride. 1 mEq of both sodium and chloride ion are contained in 58.5 mg of sodium chloride. Therefore; 30 mEq of potassium ion is provided by 30 X 74.5 mg of potassium chloride which will also supply 30 mEq of chloride ion. 10 mEq of phosphate ion is provided by 10 X 179 mg of sodium phosphate which will also supply 10 mEq of sodium ion. 5 mEq of magnesium ion is provided by 5 X 101.5 mg of magnesium chloride which will also supply 5 mEq of chloride ion. There remains a deficiency of 10mEq of each sodium and chloride ions which cn be provided by 10 X 58.5 mg of sodium chloride. The formula becomes; ~mlk~ | 13Miliequivalent (mEq)
14. 14. mg g mEq × × K+ Na+ Mg2+ Cl- HPO4- 1000 Potassium chloride 74.5 X 30 2.235 30 30 Sodium phosphate 179.0 X 10 1.790 10 10 Magnesium chloride 101.5 X 5 0.508 5 5 Sodium chloride 58.5 X 10 0.585 10 10 Water for injection, to 1 litre to 1 litre 30 20 5 45 10 55 55 The fact that cation and anion balance confirm that the formula has been worked out correctly Example 5.2 Express the following formula as percentage w/v. Sodium ion 147 mEq Potassium -ion 4 mEq Calcium ion 4 mEq Chloride ion 155 mEq Water for Injections, to 1 liter. From Table 20.6 and the appropriate conversion equation, the required percentages are— Sodium Chloride 5.85 X 147 ÷ 10 000 = 0.860% W/v Potassium Chloride 74.5 X 4 ÷ 10 000 = 0.030% w/v Calcium Chloride 73.5 X 4 ÷ 10 000 = 0.029% w/v Water for injections, to 1 liter ~mlk~ | 14Miliequivalent (mEq)
15. 15. Example 5.3 Prepare 500 ml of an Intravenous solution containing 70 mEq of sodium, 2 mEq of potassium, 4 mEq of calcium and 76 mEq of chloride. The number of milligrams of the various chlorides which contain I mEq of the required ions is obtained from Table 20.6 and the formula becomes— Sodium Chloride 70 X 58.5 ÷ 1000 = 4.0Q5 g Potassium Chloride 2 X 74.5 ÷ 1000 = 0.149 g Calcium Chloride 4 X 73.5 ÷ 1000 = 0.294 g Water for Injections, to 500 ml It is not unusual for students to miscalculate the amounts for this type of formula due to failing to appreciate that mEq is a unit of weight and not an abbreviation for mEq per litre. This leads to an incorrect halving of the final quantities in the above example. Example 5.4 Express 0.9 per cent sodium chloride solution to mililequivalent per litre. 0.9 × 10000 = 154 / 58.5 To Convert Percentage w/v to mEq The number of grams (C) per 100 milliliters is converted to mg/liter by multiplying by 10 000. This, divided by the weight (W) in mg of salt containing 1mEq, will give the number of mEq/liter. × 10000 = / ~mlk~ | 15Miliequivalent (mEq)
16. 16. Adjustment to Iso-osmoticity with Blood Plasma Based on Miliequivalent The equation used is:- = 310 − Where; a = number of mililequivalent per liter of medicament present and b = number of mililequivalent per liter of adjusting substance required Example 5.5 Calculate the amount of sodium chloride required to adjust a solution containing 40 mEq of each potassium and chloride ion to iso-osmoticity with blood plasma. A solution containing 40 mEq of chloride ion provide a total of 80 mEq of anion and cation. ∴ = 310 − 80 = 230 230 mEq will be provided by 115 mEq of sodium ions and 115 mEq chloride ions. The formula of the solution will be:- Potassium chloride 74.5 mg X 40 = 2.98 g Sodium chloride 58.5 mg X 115 = 6.73 g Water for injection, to 1 litre ~mlk~ | 16Miliequivalent (mEq)
17. 17. 6. MILIMOLES In the SI, the unit for chemical quantity is mole and the term equivalent and mililequivalent becomes absolute. Consequently the method of expressing the composition of body and infusion fluid is changing from the mililequivalent to mole notation. By analogy with atoms and molecules, a mole of an ion is its ionic weight in grams but the number of moles of each of the ions of a salt in solution depends on the number of each ion in the molecule of the salt. It follows that the quantity of salt, in mg, containing 1 mmol of a particular ion can be found by dividing the molecular weight of the salt by the number of that ion contained in the salt. For example— Salt Ion Quantity of salt (In mg) containing 1mmol NaCl Na+ M.Wt /1 = 58.5 CI- M.Wt / 1 = 58.5 CaCl2 Ca2+ M.Wt / 1 = 147 Cl- M.Wt /2 = 73.5 Na2HPO4 Na+ M.Wt /2 = 179 HP042- M.Wt /1 = 258 NaH2PO4 Na+ M.Wt /1 = 156 H2P04- M.Wt /1 = 156 Conversion Equations To convert quantities expressed in mmol per litre into weighable amounts. The following formulae may be used:- = × = × ÷ 1000 ⁄ = × ÷ 10000 Where W is the number of mg of salt containing 1 mmol of the required ion and M is the number of mmol per litre ~mlk~ | 17Milimole(mmol)
18. 18. Example 6.1 Calculate the quantities of salts required for the following electrolyte solution. Sodium 60 mmol Potassium 5 mmol Magnesium 4 mmol Calcium 4mmol Chloride 81 mmol Water for injection, to 1 litre From Table 20.8— 5 mmol of potassium ion is provided by 5 x 745 mg of potassium chloride which also provides 5 mmol of chloride ion. 4 mmol of magnesium ion is provided by 4 x 203 mg of magnesium chloride which also provides 2 x 4 mmol = 8 mmol of chloride ion, since there are two chloride ions in the molecule. 4 mmol of calcium ion are provided by 4 x 147 mg of calcium chloride which, like magnesium chloride, also provides 8 mmol of chloride ion. 60 mmol of sodium ion is provided by 60 x 58.5 mg of sodium chloride which provides a further 60 mmol of chloride. The formula becomes— mmol WXM K+ Mg2+ Ca2+ Na+ Cl- Potassium chloride 5 x 74.5 = 0.373 g 5 5 Magnesium chloride 4 x 203 = 0.812 g 4 8 Calcium chloride 4 x 147 = 0.588 g 4 8 Sodium chloride 60 x 58.5 = 3.510 g 60 60 Water for injection, to 1 litre 73 31Although there appears to be inequality between the anions and cations, the charges are equallybalanced. ~mlk~ | 18Milimole(mmol)
19. 19. Example 6.2 Calculate the number of millimoles of (a) dextrose and (b) sodium ions in 1 litre of Sodium Chloride and Dextrose Injection containing 4.3 per cent w/v of anhydrous dextrose and 0.18 per cent w/v of sodium chloride. Use the conversion equation— ⁄ = × ÷ 10000 ⁄ × 10000 = a) For dextrose Since dextrose is non-electrolyte, W = M.Wt Hence, 4.3 × 10000 = = 239 180.2 b) For sodium chloride 0.18 × 10000 = = 31 58.5 Since 1 mmol sodium chloride provides 1 mmol of sodium ion and 1 mmol of chloride ion, 1 litre of the solution will contain 31 mmol of sodium ion (and 31 mmol of chloride ion ). Example 6.3 Calculate the number of milimoles of calcium and chloride ion in a litre of a 0.029 per cent solution of calcium chloride. 0,029 × 10000 = =2 147 But, each mole of calcium chloride provides 1 mol of calcium ions and 2 moles of chloride ions. Therefore, 1 litre of solution contains 2mmol of calcium ion and 4 mmol of chloride ion. ~mlk~ | 19Milimole(mmol)
20. 20. ppm calculationEXamplePrepare 90ml NaF solution so that 5ml of the solution when diluted with water to 1cupfull(240ml), a3ppm solution obtained. Solution; 5ml solution → 240ml final volume 90ml solution → ? ml final volume (240 )(90 ) = = 4329 5 3g Na → 1000000ml ? ml ←4320ml ( )( ) = = 0.01296 ∴ 0.01296mg needed ≈ 13mg NaFFormula NaF 13mg Distilled water q.s ad. 90ml ~mlk~ | 20
21. 21. Calculation involving density factorEXampleGiven acetic acid BPC (33%w/w, d=1.04g/ml).What is the concentration of acetic acid in %w/wSolution; Acetic acid 33%w/w= 33g acetic acid in 100g solution d= 1.04g/ml ? w/v33g acetic acid →100g acid solution33g acetic acid → ?ml solutionFrom density factor information; 104g acid → 100ml acid solution 100g acid → ?ml acid solution ( )( ) = = 96.15 ? gram acetic acid → 100ml solution 33 = × 100 = 34% ⁄ 96.15 ~mlk~ | 21