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 (NA) = the number of entities in one mole of a substance (NA = 6.022x1023 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 (Mr) = 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 12C atom– Mr = sum of relative atomic masses (Ar) 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 Na3PO4? Mr (Na3PO4) = 163.94– Na3PO4 → 3 Na+ + PO4

3-

– 3 moles 1 mole– Moles of Na3PO4 = 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 CaCl2 solution is 0.1 M. Calculate the volume of solution containing 4 mmol of Cl-– CaCl2 → Ca2+ + 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 mLWe 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 CaCl2, 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 NH3?

• 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 CaCl2 solution contains 3 osmol of solute per

liter of solution. CaCl2 → 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. C6H12O6 → C6H12O6

1 M does not dissociate → 1 osmol/L

Exercises4) ? osmolarity of 0,15mol/L solution of :• a) NaCl [0,30 M]• b) MgCl2 [0,45 M]• c) Na2HPO4 [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 CaCl2 [150]• c) 300 mM KCl [600]• d) 0,15 M NaH2PO4 [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 H2O = 100g of sol. • 10% KCl → 10g of KCl/100g of solutionb) 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 H2O

[30g NaCl + 570g H2O]

7) 250g 8% Na2CO3, ? mass of Na2CO3 (purity 96%)

[20,83g {96%}]

8) Normal saline solution is 150 mM. What is its percent concentration?

[0,9%]

Homework9) 14g KOH / 100ml MW = 56,1g/mol;

C = ? • [ 2,5M ]

10) C(HNO3) = 5,62M; ρ = 1,18g/cm3 (density), MW = 63g/mol, ? % • [ 30% ]

11) 10% HCl; ρ = 1,047g/cm3, MW=36,5g/mol ? C(HCl)

• [ 2,87M ]

Prefixes for units• giga- G 109

• mega- M 106

• kilo- k 103

• 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/L10-12 10-9 10-6 10-3 mol/L

∀ µg ‹ mg ‹ g10-6 10-3 g

∀ µL ‹ mL ‹ dL ‹ L10-6 10-3 10-1 L

1L = 1dm3 1mL = 1 cm3

Exercise12) cholesterol (MW = 386,7g/mol) 200 mg/dl = ? mmol/L

[5,2 mM]

Conversion of units• Pressure = the force acting normally on unit area of a surface

(in pascals, Pa) 1 kPa = 103 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% N2 21% O2 1% water, inert gases, CO2 (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:

pO2 = 71 mmHgpCO2 = 35 mmHg

• Convert the values to kPa

pO2 = 9,5 kPapCO2 = 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 n1 = n2

V1 x C1 = V2 x C2

• 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• c1 = 0,25 M (= concentration before dilution) dilution 1

: 5 ( five times diluted sample ) → c2 = 0,25 x 1/5 = 0,05 M (= final concentr. )

4) mix equation: • (m1 x p1) + (m2 x p2) = p x (m1 + m2)• m = mass of mixed solution, p = % concentration

Exercises15) Final solution: 190g 10% sol.

? mass (g) of 38% HCl + ? mass (g) H2O you need?[50g HCl]

16) Dilute 300g of 40% to 20% sol.[1+1 = 300g of H2O]

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]

Exercises18) ? 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(H3O+)– a = γ x c– a = activity– γ = activity coefficient– c = concentration (mol /L)– in diluted (mM) solutions: γ = 1 ⇒ a = c

• pH = - log c(H3O+)• c(H3O+) = [H3O+] = molar concentration

Dissociation of water• Even the purest water is not all H2O → about 1 molecule in 500 million transfers a

proton H+ to another H2O molecule, giving a hydronium ion H3O+ and a hydroxide ion OH-:

• Dissociation of water: H2O ↔ H+ + OH-

• H2O + H+ + OH- ↔ H3O+ + OH-

• H2O + H2O ↔ H3O+ + OH-

• The concentration of H3O+ 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 H3O+ and OH- ions.

• Equilibrium constant of water: Keq = [H3O+] x [OH-] / [H2O]2

• Keg x [H2O]2 = [H3O+] x [OH-]

• Keq x [H2O]2 = constant, because [H2O] is manifold higher than[H3O+] or [OH-]

Kw = constant = ionic product of water Kw = [H3O+] x [OH-] = 1.10-14

Ionic product of waterKw = [H3O+] x [OH-] = 10-14

pKW = pH + pOH = 14pK = - log K pH = -log [H3O+] pOH = -log [OH-]

10-14 = [H3O+] x [OH-] / loglog 10-14 = log ([H3O+] x [OH-] ) log(a x b) = log a + log b

log 10-14 = log [H3O+] + log [OH-] -14 = log [H3O+] + log [OH-] / x (-1) 14 = - log [H3O+] - log [OH-]

↓ ↓ ↓ pKW = pH + pOH - log KW = pKW

14 = 7 + 7 in pure water

pKW = pH + pOH = 14• Water: [H3O+] = 10–7 (pH = 7)

[OH-] = 10–7 (pOH = 7)

• Simplification: [H3O+] = [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 H3O+ and OH- will fall until the product equals 1.10-14.

• Addition of a base to pure water → increasing OH- and H3O+ will fall until the product

equals 1.10-14.• If [H+] decreases, [OH-] increases KW is still 10-14

• If [OH-] decreases, [H+] increases = constant !• Example:

– Lemon juice has a [H3O+] of 0.01 M. What is the [OH-] ?– [H3O+ ] 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 H3O+ 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/[H3O+] =-log [H3O+]• e. g. The pH of a solution whose

[H3O+ ] is 1.10-4 is 4.• pH + pOH = 14

9.0 – 10.0hand soap

7.35 – 7.45blood

7.0pure water

6.0urine

5.0coffee

2.9vinegar

1.5 – 2.0gastric juice

0.5lead-acid battery

pHSubstance

pH of strong acids and bases• Strong acids are those that react

completely with water to form H3O+ and anion (HCl, H2SO4, HClO4, HNO3,…)

• Generally: HA → H+ + A-

• [HA] = [H+]• E. g. HCl + H2O ↔ H3O+ + Cl-

• pH = - log c(H+) • = - log cHA

• 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: c1 = 0,1 M → c2 = 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 cBOH

• 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 H3O+

ions. Most of the molecules of the weak acids remain in the molecular form (uncharged)

• Kdis ≤ 10–2, HA ↔ H+ + A-

Kdis = [H+] [A-] [H+] = [A-] [HA] = cHA Kdis = Ka

[HA] Ka = [H+]2

cHA Ka x cHA = [H+]2 / log log (Ka x cHA ) = 2 x log [H+] log Ka + log cHA = 2 x log [H+] / ½ ½ log Ka + ½ log cHA = log [H+] / x (-1) -½ log Ka - ½ log cHA = - log [H+] - log Ka = pKa

½ pKa - ½ log cHA = pH

=> pH = ½ pKa - ½ log cHA

• Weak acids (HA) [HA] ≠ [H+] Kdis ≤ 10–2

– HA ↔ H+ + A-

– pH = ½ pKa - ½ log cHA

• Weak bases (BOH) [BOH] ≠ [OH-] – Kdis = [B+] [OH-] [BOH]

– BOH ↔ B+ + OH-

– pOH = ½ pKb - ½ log cBOH

• 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 cHA

• pH = ½ pKa - ½ log cHA

• BASES: pOH = - log cBOH

• pOH = ½ pKb - ½ log cBOH

• pH = 14 – pOH

Exercises1) 0,1M HCl, pH = ?, [H+] = ?

[10-1 M, pH =1]

2) 0,01M KOH, pH = ?, [H+] = ? [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 NH4OH; pK = 4,74; pH = ?[pOH = 2,72; pH = 11,3]

5) 0,1M lactic acid; pH = 2,4; Ka = ?[pK=3,8; Ka = 1,58 x 10-4]

Homework6) strong acid: pH = 3 c = ?

7) strong base: pH = 11 c = ?

8) dilution of a weak acid: c1 = 0,1M → c2 = 0,01M Calculate the change of pH caused by the dilution.

9) dilution of a strong acid:c1 = 0,1M → c2 = 0,01MCalculate 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. CH3COOH + CH3COONa)• Weak base + its salt (i.e. NH4OH + NH4Cl)• Two salts of an oxoacid (i.e. HPO4

2- + H2PO41-)

• Henderson – Hasselbalch equation• pH = pKA + log (cS x VS / cA x VA) → for acidic

buffer• pOH = pKB + log (cS x VS / cB x VB) pH = 14 –

pOH → for basic buffer

Exercises• 200 mL 0.5 M acetic acid + 100 mL 0.5 M sodium

acetate → buffer, pKA = 4.76, pH = ?• [ pH = 4,46 ]2) 20 mL 0.05 M NH4Cl + 27 mL 0.2 M NH4OH →

buffer, K = 1.85 x 10-5, pH = ?• [pH = 10]3) The principal buffer system of blood is a

bicarbonate buffer (HCO3- / H2CO3). Calculate a

ratio of HCO3- / H2CO3 components if the pH is 7.38

and pK(H2CO3) = 6.1. Homework