1) Expression of concentration
molar concentration
percent concentration
conversion of units
Chemical calculations
Vladimíra Kvasnicová
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
Important terms
solute = a substance dissolved in a solvent in
forming a solution
density (ρ) = the mass of a substance per unit of
volume (kg.m-3 or g.cm-3) ρ = m/V
solvent = a liquid that dissolves another
substance or substances to form a solution
mass
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
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 (in g/mol)
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 (grams / mol) = Mr
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
dilution = process of preparing less
concentrated solutions from a solution of
greater concentration
1M NaOH MW = 40g /mol
=> 1M solution of NaOH = 40g of NaOH / 1L of solution
DIRRECT PROPORTIONALITY
Exercises
1) 17,4g NaCl / 300mL, MW = 58g/mol, C = ?
[1M]
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 !
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
1M
1M
1M
1M
HCl
=
H2SO4 =
H3PO4 =
CaCl2 =
CaSO4 =
1N
2N
3N
2N
2N
HCl
H2SO4
H3PO4
CaCl2
CaSO4
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
the semipermeable membrane
separates two solutions of
different concentrations
the passage of
a solvent through
a semipermeable
membrane is called
osmosis
http://www.phschool.com/webcodes10/index.cfm?error=1&errortype=default_global&
errortcode=
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
→
K+ + NO3-
Σ 2 ions
K2CO3
→
2 K+ + CO32-
Σ 3 ions
Na3PO4
→
3 Na+ + PO43-
Σ 4 ions
Na2HPO4
→
2 Na+ + HPO42-
Σ 3 ions
NaH2PO4
→
Na+ + H2PO4-
Σ 2 ions
NH4HCO3
→
NH4+ + HCO3-
Σ 2 ions
What is the osmolarity of the 1M solutions?
http://www.mhhe.com/biosci/esp/2001_gbio/folder_structure/ce/m3/s3/assets/image
s/cem3s3_1.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 )
http://campus.queens.edu/faculty/jannr/cells/cell%20pics/osmosisMicrographs.jpg
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
[0,30
[0,45
[0,45
[0,15
M]
M]
M]
M]
a) weight per unit weight (W/W) g/g of solution
10% NaOH
→ 10g of NaOH + 90g of H2O = 100g of sol.
5) Saline is 150 mM solution of NaCl.
Which solutions are isotonic with
saline? [= 150 mM = 300 mosmol/l ]
a)
b)
c)
d)
300 mM glucose
50 mM CaCl2
300 mM KCl
0,15 M NaH2PO4
Percent concentrations
• generally expressed as parts of solute
per 100 parts of total solution
(percent or „per one hundred“)
• three basic forms:
[300]
[150]
[600]
[300]
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.
10% KCl
→ 10g of KCl/100g of solution
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/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 = 36,5 g/mol
? C(HCl)
[ 2,87M ]