Chemical calculations I Vladimíra Kvasnicová

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

Important 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

Important terms density () = the mass of a substance per unit of volume (kg.m-3 or g.cm-3)  = 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) Avogadro constant (NA) = the number of entities in one mole of a substance (NA = 6.022x1023) molar weight (MW) = mass of one mole of a substance in grams

Important 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 grams) = Mr dilution = process of preparing less concentrated solutions from a solution of greater concentration

Expression of concentration Molarity (c) (mol x l-1 = mol x dm-3 = M ) = number of moles per liter of a solution c = n / V  number of moles / 1000 mL of solution DIRRECT PROPORTIONALITY

! DIRRECT PROPORTIONALITY ! 1M NaOH MW = 40g /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 !

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]

Normality (N) = concentration in terms of equivalent weights of substance (reflect the number of combining or replaceable units). It is not in common use! 1M HCl = 1N HCl 1M H2SO4 = 2N H2SO4 1M H3PO4 = 3N H3PO4 1M CaCl2 = 2N CaCl2 1M CaSO4 = 2N CaSO4

= concentration in moles of substance per 1 kg of solvent 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

the passage of a solvent through a semipermeable membrane is called osmosis the semipermeable membrane separates two solutions of different concentrations http://www.phschool.com/webcodes10/index.cfm?error=1&errortype=default_global&errortcode=

osmotic pressure osmolarity = molarity of all particles dissolved in a solution (= osmotic active particles) http://www.biologycorner.com/resources/osmosis.jpg

Exercise Describe dissociation of the salts: KNO3 → K2CO3 → Na3PO4 → Na2HPO4 → NaH2PO4 → NH4HCO3 → What is the osmolarity of the 1M solutions? K+ + NO3- Σ 2 ions 2 K+ + CO32- Σ 3 ions 3 Na+ + PO43- Σ 4 ions 2 Na+ + HPO42- Σ 3 ions Na+ + H2PO4- Σ 2 ions NH4+ + HCO3- Σ 2 ions

http://www.mhhe.com/biosci/esp/2001_gbio/folder_structure/ce/m3/s3/assets/images/cem3s3_1.jpg

http://campus.queens.edu/faculty/jannr/cells/cell%20pics/osmosisMicrographs.jpg

Osmotic pressure (Pa) π = i x c x R x T i = 1 (for nonelectrolytes) i = number of osmotic effective particles (for strong electrolytes) isotonic solutions = solutions with the same value of the osmotic pressure (c.g. blood plasma x saline ) Oncotic pressure = osmotic pressure of coloidal solutions, e.g. proteins

Exercises 4) ? osmolarity of 0,15mol/L solution of : a) NaCl b) MgCl2 c) Na2HPO4 d) glucose 5) Saline is 150 mM solution of NaCl. Which solutions are isotonic with saline? [= 150 mM = 300 mosmol/l ] a) 300 mM glucose b) 50 mM CaCl2 c) 300 mM KCl d) 0,15 M NaH2PO4 [0,30 M] [0,45 M] [0,15 M] [300] [150] [600]

Percent concentrations generally expressed as parts 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 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 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%]

Exercises 9) 14g KOH / 100ml MW = 56,1g; 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 = 36,5 ? C(HCl) [ 2,87M ]

Conversion of units pmol/L ‹ nmol/L ‹ mol/L ‹ mmol/L ‹ mol/L 10-12 10-9 10-6 10-3 mol/L g ‹ mg ‹ g 10-6 10-3 g L ‹ mL ‹ dL ‹ L 10-6 10-3 10-1 L 1L = 1dm3 1mL = 1 cm3 Exercise 12) 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

Conversion of units Air composition: Air pressure: 1 mmHg = 0,1333 kPa 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

Exercise 13) Partial pressures of blood gases were measured in a laboratory: pO2 = 71 mmHg pCO2 = 35 mmHg Convert the values to kPa. pO2 = 9,5 kPa pCO2 = 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)

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

Exercises 15) final solution: 190g 10% sol. ? mass (g) of 38% HCl + ? mass (g) H2O [50g HCl] 16) dilute 300g of 40% to 20% sol. [1+1 = 300g of H2O] 17) 20g 10% solution of NaOH → 20% sol. ? m (g) of NaOH [2,5g of NaOH]

Exercises 18) ? prep. 250ml of 0,1M HCl from stock 1M HCl 19) 10M NaOH was diluted 1: 20, ? final concentration [0,5M] 20) 1000mg/l glucose 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]