Chemical calculations in medicine Josef Fontana
Chemical calculations Expression of concentration molar concentration percent concentration conversion of units Osmotic pressure, osmolarity Dilution of solutions Calculation of ph strong and weak acids and bases buffers Calculation in a spectrophotometry Calculation in a volumetric analysis
Basic terms Solute = a substance dissolved in a solvent in forming a solution Solvent = a liquid that dissolves another substance or substances to form a solution Solution = a homogeneous mixture of a liquid (the solvent) with a gas or solid (the solute) Concentration = the quantity of dissolved substance per unit quantity of solution or solvent
Basic terms Density (ρ) = the mass of a substance per unit of volume (kg.m -3 or g.cm -3 ) ρ = m / V Mass: m = n x MW (in grams) Amount of substance (n) = a measure of the number of entities present in a substance (in moles). n = m/mw Also used mmol, µmol, nmol, Avogadro constant (N A ) = the number of entities in one mole of a substance (N A = 6.022x10 23 particles in 1 mol) Molar weight (MW) = mass of 1 mole of a substance in grams or relative molecular weight Mr (g/mol)
Basic terms Relative molecular mass (M r ) = the ratio of the average mass per molecule of the naturally occurring form of an element or compound to 1/12 of the mass of 12 C atom M r = sum of relative atomic masses (A r ) of all atoms that comprise a molecule MW (in g/mol) = Mr (no units) Dilution: process of preparing less concentrated solutions from a solution of greater concentration
Expression of concentration Molar concentration or Molarity (c) (mol x l -1 = mol x dm -3 = M ) express the number of moles of a substance per liter of a solution c = n / V number of moles / 1000 ml of solution DIRRECT PROPORTIONALITY
Expression of concentration 1M NaOH MW = 40 g/mole => 1M solution of NaOH = 40g of NaOH / 1L of solution 0,1M solution of NaOH = 4g of NaOH / 1L of solution Preparation of 500 ml of 0,1M NaOH: 0,1M solution of NaOH = 4g of NaOH / 1 L of solution 2g of NaOH / 0.5 L of solution! DIRRECT PROPORTIONALITY!
Number of ions in a certain volume Problem 1: How many moles of Na + ions are in 3.95 g of Na 3 PO 4? Mr (Na 3 PO 4 ) = 163.94 Na 3 PO 4 3 Na + + PO 4 3-3 moles 1 mole Moles of Na 3 PO 4 = 3.95 / 163.94 = 2.4 x 10-2 moles Moles of Na + = 2.4 x 10-2 x 3 = 7.2 x 10-2 moles Problem 2: Molarity of CaCl 2 solution is 0.1 M. Calculate the volume of solution containing 4 mmol of Cl - CaCl 2 Ca 2+ + 2 Cl - 0.1 M = 0.1 mol in 1 L 0.004 mol in X L X = 0.004/0.1 = 0.04 L But in one mole of the solution, there are two moles of Cl - 0.04/2 = 0.02 L = 20 ml We need smaller volume of the solution
Exercises 1) 17,4g NaCl / 300mL, MW = 58g/mol, C =? [1M] 2) Solution of glycine, C = 3mM, V = 100ml.? mg of glycine are found in the solution? [22,5mg] 3) Solution of CaCl 2, C = 0,1M. Calculate volume of the sol. containing 4 mmol of Cl -. [20ml]
Homework What is the molarity of a solution made by dissolving 3 mol of NaCl in enough water to make 6 L of solution? How many grams of NaOH are required for the preparation of 500 ml of 6.0 M solution? MW (NaOH) = 40 g/mol How many liters of 15 M aqueous ammonia solution do you need to get 450 mmol of NH 3? 4.5 g glucose in 2.5 L of solution, MW (glucose) = 180 g/mol c =?
Expression of concentration Molality (mol.kg 1 ) concentration in moles of substance per 1 kg of solvent Osmolality (mol.kg 1 or osmol.kg -1 ) concentration of osmotic effective particles (i.e. particles which share in osmotic pressure of solution) it is the same (for nonelectrolytes) or higher (for electrolytes: they dissociate to ions) as molality of the same solution Osmolarity (osmoles / L) osmolality expressed in moles or osmoles per liter
Osmotic pressure (Pa) OP is a hydrostatic pressure produced by solution in a space divided by a semipermeable membrane due to a differential concentrations of solute Osmosis is the movement of solvent from an area of low concentration of solute to an area of high concentration! Free diffusion is the movement of solute from the site of higher concentration to the site of lower concentration! π = i x c x R x T i = number of osmotic effective particles (for strong electrolytes) i = 1 (for nonelectrolytes)
Osmotic pressure (Pa) Osmolarity is a number of moles of a substance that contribute to osmotic press. of solution (osmol/l) The concentration of body fluids is typically reported in mosmol/l Osmolarity of blood is 290 300 mosmol/l Isotonic solutions Solutions with the same value of the osmotic press. (blood plasma x saline) Oncotic pressure osmotic pressure of coloidal solutions, e.g. proteins The figure is found at http://en.wikipedia.org/wiki/osmotic_pressure
Osmolarity - examples Example 1: A 1 M NaCl solution contains 2 osmol of solute per liter of solution. NaCl Na + + Cl - 1 M does dissociate 1 osmol/l 1 osmol/l 2 osmol/l in total Example 2: A 1 M CaCl 2 solution contains 3 osmol of solute per liter of solution. CaCl 2 Ca 2+ + 2 Cl - 1 M does dissociate 1 osmol/l 2 osmol/l 3 osmol/l in total Example 3: The concentration of a 1 M glucose solution is 1 osmol/l. C 6 H 12 O 6 C 6 H 12 O 6 1 M does not dissociate 1 osmol/l
Exercises 4)? osmolarity of 0,15mol/L solution of : a) NaCl [0,30 M] b) MgCl 2 [0,45 M] c) Na 2 HPO 4 [0,45 M] d) glucose [0,15 M] 5) Saline is 150 mm solution of NaCl. Which solutions are isotonic with saline? [= 150 mm = 300 mosmol/l ] a) 300 mm glucose [300] b) 50 mm CaCl 2 [150] c) 300 mm KCl [600] d) 0,15 M NaH 2 PO 4 [300]
Percent concentrations Generally expressed as part of solute per 100 parts of total solution (percent or per one hundred ). Three basic forms: a) weight per unit weight (W/W) g/g of solution 10% NaOH 10g of NaOH+90g of H 2 O = 100g of sol. 10% KCl 10g of KCl/100g of solution b) volume per unit volume (V/V) ml/100ml of sol. 5% HCl = 5ml of HCl / 100ml of sol. c) weight per unit volume (W/V) g/100 ml (g/dl; mg/dl; μg/dl; g % ) The most frequently used expression in medicine 20% KOH = 20g of KOH / 100 ml of sol.
Exercises 6) 600g 5% NaCl,? mass of NaCl, mass of H 2 O [30g NaCl + 570g H 2 O] 7) 250g 8% Na 2 CO 3,? mass of Na 2 CO 3 (purity 96%) [20,83g {96%}] 8) Normal saline solution is 150 mm. What is its percent concentration? [0,9%]
Homework 9) 14g KOH / 100ml MW = 56,1g/mol; C =? [ 2,5M ] 10) C(HNO 3 ) = 5,62M; ρ = 1,18g/cm 3 (density), MW = 63g/mol,? % [ 30% ] 11) 10% HCl; ρ = 1,047g/cm 3, MW=36,5g/mol? C(HCl) [ 2,87M ]
Prefixes for units giga- G 10 9 mega- M 10 6 kilo- k 10 3 deci- d 10-1 centi- c 10-2 milli- m 10-3 micro- μ 10-6 nano- n 10-9 pico- p 10-12 femto- f 10-15 atto- a 10-18
Conversion of units pmol/l nmol/l µmol/l mmol/l mol/l 10-12 10-9 10-6 10-3 mol/l µg mg g 10-6 10-3 g Exercise 12) cholesterol (MW = 386,7g/mol) 200 mg/dl =? mmol/l µl ml dl L 10-6 10-3 10-1 L [5,2 mm] 1L = 1dm 3 1mL = 1 cm 3
Conversion of units Pressure = the force acting normally on unit area of a surface (in pascals, Pa) 1 kpa = 10 3 Pa Dalton s law = the total pressure of a mixture of gasses or vapours is equal to the sum of the partial pressures of its components Partial pressure = pressure of one gas present in a mixture of gases Air composition: 78% N 2 21% O 2 1% water, inert gases, CO 2 (0,04%) Air pressure: 1 atm = 101 325 Pa (~ 101 kpa) = 760 Torr (= mmhg) 1 mmhg = 0,1333 kpa 1 kpa = 7,5 mmhg
Exercises 13) Partial pressures of blood gases were measured in a laboratory: po 2 = 71 mmhg pco 2 = 35 mmhg Convert the values to kpa po 2 = 9,5 kpa pco 2 = 4,7 kpa
Conversion of units Energy content of food: 1 kcal = 4,2 kj 1 kj = 0,24 kcal 14) A snack - müesli bar (30g) was labelled: 100g = 389 kcal. Calculate an energy intake (in kj) per the snack. 490kJ / 30g
Dilution of solutions Concentration of a substance lowers, substance amount remains the same 1) useful equation n 1 = n 2 V 1 x C 1 = V 2 x C 2 2) mix rule % of sol.(1) parts of sol.(1) % of final sol. % of sol.(2) parts of sol.(2)
Dilution of solutions Concentration of a substance lowers, substance amount remains the same 3) expression of dilution 1 : 5 or 1 / 5 1 part (= sample) + 4 parts (= solvent) = 5 parts = total volume c 1 = 0,25 M (= concentration before dilution) dilution 1 : 5 ( five times diluted sample ) c 2 = 0,25 x 1/5 = 0,05 M (= final concentr. ) 4) mix equation: (m 1 x p 1 ) + (m 2 x p 2 ) = p x (m 1 + m 2 ) m = mass of mixed solution, p = % concentration
Exercises 15) Final solution: 190g 10% sol.? mass (g) of 38% HCl +? mass (g) H 2 O you need? [50g HCl] 16) Dilute 300g of 40% to 20% sol. [1+1 = 300g of H 2 O] 17) You have 20g of 10% solution of NaOH and you want to produce 20% sol. How many grams of NaOH you add? [2,5g of NaOH]
Exercises 18)? prep. 250ml of 0,1M HCl from stock 1M HCl [25ml of 1M HCl] 19) 10M NaOH was diluted 1: 20,? final concentr. [0,5M] 20) 1000mg/l Glc was diluted 1: 10 and then 1 : 2? final concentration [50mg/l] 21) what is the dilution of serum in a test tube containing 200 μl of serum 500 μl of saline 300 μl of reagent [1 : 5]
Calculation of ph ph = - log a(h 3 O + ) a = γ x c a = activity γ = activity coefficient c = concentration (mol /L) in diluted (mm) solutions: γ = 1 a = c ph = - log c(h 3 O + ) c(h 3 O + ) = [H 3 O + ] = molar concentration
Dissociation of water Even the purest water is not all H 2 O about 1 molecule in 500 million transfers a proton H + to another H 2 O molecule, giving a hydronium ion H 3 O + and a hydroxide ion OH - : Dissociation of water: H 2 O H + + OH - H 2 O + H + + OH - H 3 O + + OH - H 2 O + H 2 O H 3 O + + OH - The concentration of H 3 O + in pure water is 0.000 0001 or 1.10-7 M The concentration of OH - is also 1.10-7 M. Pure water is a neutral solution, without an excess of either H 3 O + and OH - ions. Equilibrium constant of water: K eq = [H 3 O + ] x [OH - ] / [H 2 O] 2 K eg x [H 2 O] 2 = [H 3 O + ] x [OH - ] K eq x [H 2 O] 2 = constant, because [H 2 O] is manifold higher than [H 3 O + ] or [OH - ] K w = constant = ionic product of water K w = [H 3 O + ] x [OH - ] = 1.10-14
Ionic product of water K w = [H 3 O + ] x [OH - ] = 10-14 pk W = ph + poh = 14 pk = - log K ph = -log [H 3 O + ] poh = -log [OH - ] 10-14 = [H 3 O + ] x [OH - ] / log log 10-14 = log ([H 3 O + ] x [OH - ] ) log(a x b) = log a + log b log 10-14 = log [H 3 O + ] + log [OH - ] -14 = log [H 3 O + ] + log [OH - ] / x (-1) 14 = - log [H 3 O + ] - log [OH - ] pk W = ph + poh - log K W = pk W 14 = 7 + 7 in pure water
pk W = ph + poh = 14 Water: [H 3 O + ] = 10 7 (ph = 7) [OH - ] = 10 7 (poh = 7) Simplification: [H 3 O + ] = [H + ] = c(h + ) => ph = log c(h + ) ph = 0 14 ph 0 -------------- 7 --------------14 acidic neutral basic Addition of an acid to pure water increasing H 3 O + and OH - will fall until the product equals 1.10-14. Addition of a base to pure water increasing OH - and H 3 O + will fall until the product equals 1.10-14. If [H + ] decreases, [OH - ] increases K W is still 10-14 If [OH - ] decreases, [H + ] increases = constant! Example: Lemon juice has a [H 3 O + ] of 0.01 M. What is the [OH - ]? [H 3 O + ] x [OH - ] = 1.10-14 1.10-2 x [OH - ] = 1.10-14 / : 1.10-2 [OH - ] = 1.10-12
ph scale Exponential numbers express the actual concentrations of H 3 O + and OH - ions. In 1909, S. P. L. Sørensen proposed that only the number in the exponent shall be used to express acidity. Sørensen s scale is known as the ph scale (power of hydrogen) ph = log 1/[H 3 O + ] =-log [H 3 O + ] e. g. The ph of a solution whose [H 3 O + ] is 1.10-4 is 4. ph + poh = 14 Substance lead-acid battery gastric juice vinegar coffee urine pure water blood hand soap ph 0.5 1.5 2.0 2.9 5.0 6.0 7.0 7.35 7.45 9.0 10.0
ph of strong acids and bases Strong acids are those that react completely with water to form H 3 O + and anion (HCl, H 2 SO 4, HClO 4, HNO 3, ) Generally: HA H + + A - [HA] = [H + ] E. g. HCl + H 2 O H 3 O + + Cl - ph = - log c(h + ) = - log c HA Calculations: 1) 0.1 M HCl, ph =? 2) Strong acids ph: a) 1.6 c =? (mol/l) b) 3.0 c =? 3) Dilution of a strong acid: c 1 = 0,1 M c 2 = 0,01 M, ph =? Strong bases are those that are ionized completely (NaOH, KOH, LiOH): Generally: BOH B + + OH - [BOH] = [OH - ] E. g. NaOH Na + + OH - poh = - log c BOH ph = 14 poh Calculations: 1) 0.01 M KOH, ph =? 2) Strong bases ph: a) 11 c =? b) 10.3 c =? 3) 50 ml of a solution contains 4mg of NaOH. MrNaOH = 40, ph =?
Weak acids (HA) - [HA] [H + ] Weak acids react only to a slight extent with water to form relatively few H 3 O + ions. Most of the molecules of the weak acids remain in the molecular form (uncharged) K dis 10 2, HA H + + A - K dis = [H + ] [A - ] [H + ] = [A - ] [HA] = c HA K dis = K a [HA] K a = [H + ] 2 c HA K a x c HA = [H + ] 2 / log log (K a x c HA ) = 2 x log [H + ] log K a + log c HA = 2 x log [H + ] / ½ ½ log K a + ½ log c HA = log [H + ] / x (-1) -½ log K a - ½ log c HA = - log [H + ] - log K a = pk a ½ pk a - ½ log c HA = ph => ph = ½ pk a - ½ log c HA
Weak acids (HA) [HA] [H + ] K dis 10 2 HA H + + A - ph = ½ pk a - ½ log c HA Weak bases (BOH) [BOH] [OH - ] K dis = [B+] [OH-] [BOH] BOH B + + OH - poh = ½ pk b - ½ log c BOH ph of basic solutions: ph + poh = 14 ph = 14 - poh
Important equations ph = - log c(h + ) pk = - log K ph + poh = 14 ACIDS: ph = - log c HA ph = ½ pk a - ½ log c HA BASES: poh = - log c BOH poh = ½ pk b - ½ log c BOH ph = 14 poh
Exercises 1) 0,1M HCl, ph =?, [H + ] =? 2) 0,01M KOH, ph =?, [H + ] =? [10-1 M, ph =1] [10-12 M, ph = 12] 3) 0,01M acetic acid, K = 1,8 x 10 5, ph =? [pk = 4,74; ph = 3,4] 4) 0,2M NH 4 OH; pk = 4,74; ph =? [poh = 2,72; ph = 11,3] 5) 0,1M lactic acid; ph = 2,4; K a =? [pk=3,8; K a = 1,58 x 10-4 ]
Homework 6) strong acid: ph = 3 c =? 7) strong base: ph = 11 c =? 8) dilution of a weak acid: c 1 = 0,1M c 2 = 0,01M Calculate the change of ph caused by the dilution. 9) dilution of a strong acid: c 1 = 0,1M c 2 = 0,01M Calculate the change of ph caused by the dilution.
ph of buffers Buffers are solutions that keep the ph value nearly constant when acid or base is added. Buffer solution contains: Weak acid + its salt (i.e. CH 3 COOH + CH 3 COONa) Weak base + its salt (i.e. NH 4 OH + NH 4 Cl) Two salts of an oxoacid (i.e. HPO 4 2- + H 2 PO 4 1- ) Henderson Hasselbalch equation ph = pk A + log (c S x V S / c A x V A ) for acidic buffer poh = pk B + log (c S x V S / c B x V B ) ph = 14 poh for basic buffer
Exercises 200 ml 0.5 M acetic acid + 100 ml 0.5 M sodium acetate buffer, pk A = 4.76, ph =? [ ph = 4,46 ] 2) 20 ml 0.05 M NH 4 Cl + 27 ml 0.2 M NH 4 OH buffer, K = 1.85 x 10-5, ph =? [ph = 10] 3) The principal buffer system of blood is a bicarbonate buffer (HCO 3 - / H 2 CO 3 ). Calculate a ratio of HCO 3 - / H 2 CO 3 components if the ph is 7.38 and pk(h 2 CO 3 ) = 6.1. Homework