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Concentration Expression

Concentration Expression. Percentage Weight-In-Volume. The grams of solute or constituent in 100 ml of solution. The volume, in milliliter, represents the weight in grams of solution or liquid preparation as if it were pure water.

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Concentration Expression

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  1. Concentration Expression

  2. Percentage Weight-In-Volume • The grams of solute or constituent in 100 ml of solution. • The volume, in milliliter, represents the weight in grams of solution or liquid preparation as if it were pure water. • Volume of solution (ml)( representing grams ) X % ( expressed as a decimal ) = g of solute or constituent. Weight of solute ( g ) • percentage Weight-In-Volume = ---------------------------------- X 100 Volume of solution ( ml )

  3. Example I How many grams of dextrose are required to prepared 4000 ml of a 5 % solution ? 4000 ml represent 4000 gm of solution. 5 % = 0.05 g of solute or constituent = volume (ml) X % ( expressed as a decimal ). = 4000 g X 0.05 = 200 g

  4. Example II How many grams of potassium permanganate should be used in compounding the following prescription? Rx Potassium Permanganate 0.02 % Purified Water ad 250 ml Sig. As directed 250 ml represent 250 gm of solution. 0.02 % = 0.0002 g of solute or constituent = volume (ml) X % ( expressed as a decimal ). = 250 g X 0.0002 = 0.05 g

  5. Weight of solute ( g ) percentage Weight-In-Volume = ---------------------------------- X 100 Volume of solution ( ml ) Example What is the percentage strength ( w/v ) of a solution of urea, if 80 ml contain 12 g ? 12 % w/v = -------- X 100 = 15 % 80

  6. Weight of solute ( g ) percentage Weight-In-Volume = ------------------------------------ X 100 Volume of solution ( ml ) Example How many milliliters of a 3 % w/v solution can be made from 27 g of ephedrine sulfate ? 27 3 = -------- X 100 x 27 x = --------- X 100 = 900 ml 3 Volume ( in ml ) = 900 ml

  7. Percentage Volume-In-Volume • The milliliters of solute or constituent in 100 ml of solution. • Volume of solution (ml) X % ( expressed as a decimal ) = ml of solute or constituent. Volume of solute ( ml ) • percentage Volume-In-Volume = -------------------------------- X 100 Volume of solution ( ml )

  8. Example I How many milliliters of liquide phenol be used in compounding the following prescription? Rx liquide phenol 2.5 % Calamine Lotion ad 240 ml Sig. For external use. Volume (ml) X % ( expressed as a decimal ) = ml of solute or constituent. 240 X 0.025 = 6 ml

  9. Volume of solute ( ml ) percentage Volume-In-Volume = ---------------------------------- X 100 Volume of solution ( ml ) Example I In preparing 240 ml of a certain lotion, a pharmacist used 4 ml of liquefied phenol. What was the percentage ( v/v ) of liquefied phenol in the lotion ? 4 % v/v = -------- X 100 = 1.67 % 240

  10. Volume of solute ( ml ) percentage Volume-In-Volume = ---------------------------------- X 100 Volume of solution ( ml ) Example II What is the percentage strength ( v/v ) of solution of 800 g of liquid with a specific gravity of 0.800 in enough water to make 4000 ml ? Volume = Weight / Specific Gravity. Volume = 800 / 0.8 Volume = 1000 ml 1000 % v/v = -------- X 100 = 25 % 4000

  11. Volume of solute ( ml ) percentage Volume-In-Volume = ---------------------------------- X 100 Volume of solution ( ml ) Example III Peppermint spirit contains 10 % of ( v/v ) of peppermint oil. What volume of the spirit will contain 75 ml of peppermint oil ? 75 10 = -------- X 100 x 75 x = -------- X 100 = 750 ml 10

  12. Percentage Weight-In-Weight • The grams of solute or constituent in 100 grams of solution. • Weight of solution (g) X % ( expressed as a decimal ) = g of solute or constituent. Weight of solute ( g ) • percentage Weight-In- Weight = ------------------------------- X 100 Weight of solution ( g )

  13. Example I How many grams of phenol should be used to prepare 240 g of 5 % ( w/w ) solution in water ? Weight of solution (g) X % ( expressed as a decimal ) = g of solute or constituent. 240 X 0.05 = 12 g

  14. Example II How many milligrams of hydrocortisone should be used in compounding the following prescription? Rx hydrocortisone 0.125 % Hydrophilic Ointment ad 10 g Sig. Apply. Weight of solution (g) X % ( expressed as a decimal ) = g of solute or constituent. 10 X 0.00125 = 0.0125 g = 12.5 mg

  15. Example III How many grams of a drug substance are required to make 120 ml of a 20 % ( w/w ) solution having a specific gravity of 1.15 ? Volume = Weight / Specific Gravity. Weight = Volume X Specific Gravity. Weight = 120 X 1.15 Weight = 138 g ( weight of 120 ml of solution ) Weight of solution (g) X % ( expressed as a decimal ) = g of solute or constituent. 138 X 0.2 = g of solute or constituent. = 27.6 g plus enough water to make 120 ml.

  16. Weight of solute ( g ) percentage Weight-In- Weight = ------------------------------ X 100 Weight of solution ( g ) Example IV If 1500 g of a solution contain 75 g of a drug substance, what is the percentage ( w/w ) of the solution ? 75 % w/w = ---------- X 100 = 5 % 1500

  17. Weight of solute ( g ) percentage Weight-In- Weight = ------------------------------ X 100 Weight of solution ( g ) Example V If 5 g of boric acid are add to 100 ml of water, what is the percentage strength ( w/w ) of the solution ? 100 ml of water weight 100 g 100 g + 5 g = 105 g, weight of the solution 5 % w/w = ---------- X 100 = 4.76 % 105

  18. Milligram Percent • The number of milligrams of substance in 100 ml of liquid. • It is used frequently to denote the concentration of a drug or natural substance in a biologic fluid, as in the blood. • The statement that the concentration of non-protein nitrogen in the blood is 30 % means that each 100 ml of blood contains 30 mg of non-protein nitrogen.

  19. Example I If a patient is determined to have a serum cholesterol level of 200 mg/dl (a) What is the equivalent value expressed in terms of milligrams percent? (a) 200 mg/dl = 200 mg / 100 ml = 200 mg%. (b) How many milligrams of cholesterol would be present in 10 ml sample of the patient's serum ? (b) 200 (mg)  100 ml x (mg)  10 ml x ( mg ) = 200 X 10 / 100 = 20 mg.

  20. Example II If a patient is determined to have a serum cholesterol level of 200 mg/dl, what is the equivalent value expressed in term of millimoles ( mmol ) per liter ? Molecular weight of cholesterol = 387. 1 mol cholesterol = 387 g. 1mmol cholesterol = 387 mg. 200 mg/dl = 2000 mg/L. 387 ( mg )  1 (millimoles ) 2000 ( mg )  x (millimoles ) Therefore x = ( 1 X 2000 ) / 387 = 5.17 mmol / L

  21. Part Per Million ( PPM ) & Part Per Billion ( PPB ) • The strength of very dilution solution are commonly expressed in terms of part per million ( PPM ) & part per billion ( PPB ), i.e. the number of parts of the agent per 1 million or 1 billion parts of the whole. • For example, fluoridated drinking water (used to reduce dental caries) often contains 1 part of fluoride per million parts of drinking water. • The ppm or ppb concentration of a substance may be expressed in quantitatively equivalent value of percent strength or ratio strength.

  22. Example I Express 5 ppm of ion in water in percent strength and ratio strength ? 5 ppm = 5 parts in 1,000,000 parts = 1 : 200,000 ( ratio strength ) = ( 1 / 200,000 ) X 100 = 0.0005 % ( percent strength )

  23. Example II The concentration of a drug additive in an animal feed is 12.5 ppm. How many milligrams of the drug should be used in preparing 5.2 kg of feed ? 12.5 ppm = 12.5 g ( drug ) in 1,000,000 g ( feed ) Thus 12.5 g ( drug )  1,000,000 g ( feed ) x g ( drug )  5,200 g ( feed ) x = ( 5,200 X 12.5 ) / 1,000,000 = 0.065 g = 65 mg

  24. Molarity • Concept of mole Atoms & molecules are too small, tiny to weight so the concept of mole which is theoretical value or unit expressing the formula weight of substance in grams. e.g. gm atom of O2 M.Wt is 16 = 16 gm = 1 mol gm ion of Cl- M.Wt is 35.5 = 35.5 gm = 1 mol gm molecules of H2O M.Wt is 18 = 18 gm = 1 mol

  25. Molar Concentration • Moles / Liter is one of the most useful units for describing concentration. • Molarity ( M ) is the number of moles of solute / Liter of solution ( not solvent ). • M X L = moles. • M X L = Weight in grams / molecular weight.

  26. Example I How can prepare 500 ml of 0.15 M Na2CO3 solution ? i.e. How many grams needed ? M X L = Weight in grams / molecular weight. 0.15 X 0.5 = Weight in grams / 106 Weight in grams = 0.15 X 0.5 X 106 Weight in grams = 7.95 g Weighing 7.95 g of Na2CO3 & placed in volumetric flask 500 ml & diluted to 500 ml .

  27. Example II What is the molarity of NaCl solution 10 g / 100 ml ? M X L = Weight in grams / molecular weight. M X 0.1 = 10 / 58.5. M = 1.71 M Example II What is the molar concentration of 1 % (w/v) NaCl solution? 1 % = 1 g of NaCl in 100 ml M X L = Weight in grams / molecular weight. M X 0.1 = 1 / 58.5. M = 0.171 M

  28. Example IV Describe the preparation of 2 L of 0.2 M HCl solution starting with a concentrated HCl solution ( 28 % w/w, specific gravity = 1.15 ) ? M X L = Weight in grams / molecular weight. 0.2 X 2 = x / 36.5. Weight in grams = 0.2 X 2 X 36.5 Weight in grams = 14.6 g ( need to prepare the solution ) The stock of HCl is not pure only 28 % HCl : 28 g HCl  100 g of solution. 14.6 g HCl  x g of solution. x = ( 14.6 X 100 ) / 28 x = 52.143 g 52.143 g should be taken from a solution to take it in ml instead of grams ( specific gravity = weight / volume ) Volume = weight / specific gravity = 52.143 / 1.15 = 45.34 ml 45.34 ml placed in a volumetric flask and completed to 2 liters.

  29. Molality ( m ) • number of moles of solute / kg of solvent ( not solution ). • m X kg ( solvent ) = moles. • m X kg ( solvent ) = Weight in grams / molecular weight. • Molality is useful in describing the ratio of moles of solute to solvent. • The value of ( m ) does not change with temperature but that of ( M ) dose. • We would not even have to use a volumetric flask, because we weight the solvent.

  30. Example I Calculate the molality HCl solution ( 28 % w/w ) ? m X kg ( solvent ) = Weight in grams / molecular weight. Weight of solvent = 100 – 28 = 72 g ( from 28 % ) = 0.072 kg m X kg ( solvent ) = Weight in grams / molecular weight. m X 0.072 = 28 / 36.5 m = 28 / ( 36.5 X 0.072 ) m = 10.65 ( m)

  31. Example II Calculate the molal concentration of 1 % ( w/v ) NaCl solution ? Where : molecular weight = 58.5 & specific gravity = 1.0055 1 % mean 1 g NaCl in 100 ml solution Weight of solution = volume X specific gravity = 100 X 1.0055 = 100.55 g Weight of solvent = weight of solution – weight of solute = 100.55 – 1 = 99.55 g m X kg ( solvent ) = Weight in grams / molecular weight. m X 0.09955 = 1 / 58.5 m = 1 / ( 58.5 X 0.09955 ) m = 1.72 ( m)

  32. Example III To prepare 0.15 m solution of NaCl, to prepare this solution with this conc, how many grams of NaCl should be dissolved in 500 gm of water ? m X kg ( solvent ) = Weight in grams / molecular weight. 0.15 X 0.5 = Weight in grams / 58.5 Weight in grams = 0.15 X 0.5 X 58.5 Weight in grams = 4.39 g Thus, 4.39 g of NaCl is dissolved in 500 gm of H2O to give the desired concentration.

  33. Mole Fraction • Ratio of moles of one constituent solute or solvent of solution to the total number of moles of all constituents ( solute & solvent ). y1 = n1/(n1+n2) y2 = n2 /(n1+n2) Where n1,n2 are the number of moles. y1 is the mol fraction of solute. y2 is the mol fraction of solvent.

  34. Example I What is the mole fraction of both constituent in 1 % ( w/v ) NaCl solution ( specific gravity = 1.0053 ) and what is the molality ? Volume = Weight / Specific Gravity. Weight = Volume X Specific Gravity. Weight = 100 X 1.0053 Weight = 100.53 g solution 1 g NaCl in ( 100 X 1.0053 = 100.53 g solution ) Solvent 100.53 – 1 = 99.53 g solvent. m X kg = Weight in grams / molecular weight. m X 99.53/1000 = 1 / 58.5. m = 1 / ( 58.5 X 0.09953) m = 0.172 ( m )

  35. Cont. Example I What is the mole fraction of both constituent in 1 % ( w/v ) NaCl solution ( specific gravity = 1.0053 ) and what is the molality ? No. of moles = Weight in grams / molecular weight No. of moles of solute = 1 / 58.5 = 0.171 No. of moles of solvent = 99.53 / 18 = 5.529 y1 = n1/(n1+n2) ) y2 = n2 /(n1+n2) Mole fraction of solute = 0.171 / ( 0.171 + 5.529 ) = 0.003 Mole fraction of solvent = 5.529 / ( 0.171 + 5.529 ) = 0.996

  36. Mole Percent • Mole Percent = mole fraction X 100 • In the last example : Mole fraction of solute = 0.171 / ( 0.171 + 5.529 ) = 0.003 Mole Percent = mole fraction X 100 Mole Percent = 0.003 X 100 = 0.3 % Mole fraction of solvent = 5.529 / ( 0.171 + 5.529 ) = 0.996 Mole Percent = mole fraction X 100 Mole Percent = 0.996 X 100 = 99.6 %

  37. Normality ( N ) Equivalent weight is the weight in grams of one equivalent or the quantities of substance that combine with 1.008 gm H+. The EQUIVALENT WEIGHT is the amount of solute needed to be the equivalent of one mole of hydrogen ions. Therefore, the equivalent weight is dependent on the valence of the solute. For solutes with a valence of one (i.e. NaCl) the molecular weight and equivalent weight are the same. When the valence of the solute is more than one (i.e. H3PO4, valence = 3), then the equivalent weight is equal to the molecular weight divided by the valence.

  38. The equivalent weight ( one equivalent ) of an acid or base is that contains 1 g atom. • Equivalent weight = molecular weight / n. n = number of replaceable H or OH for acid or base. n = number of electrons lost or gained in oxidation, reduction reaction. • Equivalent weight = atomic weight / number of equivalent per atomic weight. e.g. F & O2 1 equivalent of F = molecular weight = 19 1 equivalent of O2 = molecular weight / Valency = 16 / 2 = 8

  39. Normality ( N ) = grams of equivalent weight of solute / Liter of solution ( not solvent ) L X N = equivalent N = weight in gram / (equivalent weight x volume)

  40. Concentration Expression :

  41. Buffer

  42. pH Example I What is the pH of solution with [ H+] = 32 X 10-5M/L ? pH = - log [ H+] pH = - log 32 X 10-5 pH = 3.495 pH = - log [ H+] or pH = - log [ H3O+ ]

  43. Example II The hydronium ion concentration of 0.1 M solution was found to the 1.32 X 10-3 M, What is the pH of the solution? pH = - log [ H3O+ ] pH = - log [ 1.32 X 10-3 ] pH = 2.88 Example III If the pH of a solution is 4.72, what is the hydrogen ion concentration? pH = - log [ H+ ] - log [ H+ ] = 4.72 log [ H+ ] = - 4.72 take anti-log for both side [ H+ ] = 1.91 X 10-5

  44. pH = 0 H+ = 1 M/l = 1 X 100 = 1 pH = 1 H+ = 1 X 10-1 = 0.1 M/L pH = 2 H+ = 1 X 10-2 = 0.01 M/L pH = 3 H+ = 1 X 10-3 = 0.001 M/L Note : the change in one pH unit means 10 fold change in [H+]. pH + pOH = 14

  45. Introduction • When a minute trace of hydrochloric acid is added to pure water, a significant increase in hydrogen-ion concentration occur immediately. • In a similar manner, when a minute trace of sodium hydroxide is added to pure water, it cause correspondingly large increase in the hydroxyl-ion concentration. • These change take place because water alone cannot neutralize even trace of acid or base, i.e. it has no ability to resist change in hydrogen-ion concentration or pH. • Therefore, it is said to be unbuffered. • E.g. CO2 + H2O H2CO3 decrease pH from 7 to 5.8 These change of pH are of great concern in pharmaceutical preparation also NaCl solution ability to resist change of pH. To ensure stability and solubility, we used to control pH by using a buffer

  46. Buffer : Compound or a mixture of compound which by presence in solution to resist change in pH up of addition of small quantities of acid or base or a solvent.

  47. Adjustment of pH by the buffer • pH important for stability & solubility, therefore should be adjusted pH by buffer . • Also pH play important role in : 1-Parenteral dosage form. 2-Eye drops. 3-Nasal drops.

  48. Adjustment of pH by the buffer Degree of acidity and alkalinity depends on the relative concentration of H+ ion and OH- ion. if H+ > OH- = acidic H+ = OH- = neutral H+ < OH- = alkaline Acidity and alkali may be strong or weak: 1- weak acid, pH 3.5- 7 2- strong acid, pH 0-3.5 3- weak base, pH 7-10 4- strong base, pH 10.5-14 0 3.5 7 10.5 14

  49. The product of H+ ion and OH- ion in the any aqueous liquid is constant i.e. Kw = [ H+] [ OH- ] Increase of one tend to decrease of another.

  50. Some notes about buffer: 1-Buffers solution should be prepared using freshly boiled and cooled water. 2- Buffers solution should be stored in containers of Alkali-free glass. 3-Buffers solution should be discarded no later than three months from the date of manufacture.

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