Chemical calculations IVladimíra Kvasnicová

1) Expression of concentration molar concentration

percent concentration

conversion of units

2) Osmotic pressure, osmolarity

3) Dilution of solutions

4) Calculation of pH strong and weak acids and bases

buffers

5) Calculation in a spectrophotometry

6) 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

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 HCl1M H2SO4 = 2N H2SO4

1M H3PO4 = 3N H3PO4

1M CaCl2 = 2N CaCl2

1M CaSO4 = 2N CaSO4

Molality (mol.kg –1)= concentration in moles of substance per1 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

http://www.phschool.com/webcodes10/index.cfm?error=1&errortype=default_global&errortcode

=

the passage ofa solvent througha semipermeable

membrane is called osmosis

the semipermeable membrane separates two

solutions of different concentrations

http://www.biologycorner.com/resources/osmosis.jpg

osmotic pressur

e

osmolarity = molarity of all particles dissolved in a solution (= osmotic active particles)

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 glucoseb) 50 mM CaCl2

c) 300 mM KCl d) 0,15 M NaH2PO4

[0,30 M][0,45 M][0,45 M][0,15 M]

[300][150][600][300]

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 expressionin 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%]

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 ]

Exercises

Conversion of units

• pmol/L ‹ nmol/L ‹ mol/L ‹ mmol/L ‹ mol/L10-12 10-9 10-6 10-3 mol/L

g ‹ mg ‹ g10-6 10-3 g

L ‹ mL ‹ dL ‹ L10-6 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:

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]

18) ? prep. 250ml of 0,1M HCl from stock 1M HCl[25ml of 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]

Exercises