Basic Chemistry IV
Vladimíra Kvasnicová
Problems – add formulas
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sodium sulfite
potassium phosphate
ammonium hydrogen phosphate
lithium dihydrogen phosphate
calcium hydrogen carbonate
silver sulfide
zinc sulfate
potassium permanganate
sodium hypobromite
barium nitrate
hydrargyric chloride
Homework solution:
• sodium sulfite
• Na2SO3
• potassium phosphate
• K3PO4
• ammonium hydrogen phosphate
• (NH4)2HPO4
• lithium dihydrogen phosphate
• LiH2PO4
• calcium hydrogen carbonate
• Ca(HCO3)2
• silver sulfide
• Ag2S
• zinc sulfate
• ZnSO4
• potassium permanganate
• KMnO4
• sodium hypobromite
• NaBrO
• barium nitrate
• Ba(NO3)2
• hydrargyric chloride
• HgCl2
Problems – add formulas
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sodium tetraborate decahydrate
potassium aluminium sulfate
sodium aluminium sulfate dodecahydrate
ammonium carbonate
calcium sulfate hemihydrate (hemi = ½)
zinc sulfate heptahydrate
potassium dichromate
potassium magnesium fluoride
ammonium magnesium phosphate
led(II) chloride fluoride
cupric biscarbonate difluoride (bis = twice)
Homework solution:
• sodium tetraborate decahydrate • Na2B4O7 . 10 H2O
• potassium aluminium sulfate
• KAl(SO4)2
• sodium aluminium sulfate
dodecahydrate
• NaAl(SO4)2 . 12 H2O
• ammonium carbonate
• (NH4)2CO3
• calcium sulfate hemihydrate
• CaSO4 . ½ H2O
• zinc sulfate heptahydrate
• ZnSO4 . 7 H2O
• potassium dichromate
• K2Cr2O7
• potassium magnesium fluoride
• KMgF3
• ammonium magnesium phosphate • NH4MgPO4
• led(II) chloride fluoride
• PbClF
• cupric biscarbonate difluoride
• Cu3(CO3)2F2
Chemical properties of
inorganic compounds
organic compounds
high melting points
low melting points
most inorganic compounds are
soluble in water
most organic compounds are
insoluble in water
(containing alkali metal, NH4+, NO2-, NO3-, halides,...)
(hydrocarbons, deriv. with long hydrophobic tail)
not soluble in organic liquids
soluble in organic liquids
most inorganic compounds conduct
an electric current
don´t conduct electricity
Examples of WATER INSOLUBLE inorganic compounds:
• carbonates, phosphates (except those containing alkali metals or NH4+)
• SrII, BaII, PbII, Hg2I sulfates
Solubility of compounds
• solubility = the amount of a species that dissolves in a
given quantity of water at a given temperature to give
a saturated solution
soluble = compound dissolves to the extent of
1 g or more per 100 ml
slightly soluble = less than 1 g but more than 0.1 g per 100 ml
insoluble = compound dissolves less than 0.1 g per 100 ml
Chemical reactions
= chemical changes
stoichiometry = the reactants combine in simple wholenumber ratios (see stoichiometric coefficients: a, b, c, d)
aA+bB→cC+dD
• the single arrow (→) is used for an irreversible reaction
• double arrows (↔) are used for reversible reactions
chemical equilibrium = a state of a reversible chemical
reaction in which the concentrations of reactants and
products are not changing with time
ie
the rates of both the forward and back reactions are equal
Chemical reactions
equilibrium constant (K) describes the ratio of
concentrations of products and reactants in the
equilibrium
aA+bB↔cC+dD
K = [C]c [D]d / [A]a [B]b
• K is constant for given reaction and fixed temperature
• the definition of K is called Guldberg-Waage´s law
(= law of chemical equilibrium)
Chemical reactions
• chemical reaction is described by a chemical equation
• the equation is balanced if substance amount of each
element is the same on both sides of a chemical equation
⇒ elements are balanced in the order:
metal – nonmetal – hydrogen - oxygen
Conservation law (Law of conservation of mass / energy)
= the total magnitude of mass (or energy or charge)
remains unchanged even though there may be exchanges
of that property between components of the system
(the sum of masses of reactants equals
to the sum of masses of products)
Chemical reactions
• NEUTRALIZATION = acid-base reactions
H2SO4 + 2 NaOH → Na2SO4 + 2 H2O
acid +
base →
salt +
water
• PRECIPITATION
NaCl + AgNO3 → AgCl (s) + NaNO3
→ insoluble product = precipitate is formed
• REDOX reactions = oxidative-reduction reactions
→ oxidation states of elements are changed !!!
Chemical reactions
oxidation-reduction reactions = REDOX reactions
→ oxidation states of elements are changed !!!
• two changes:
one element is oxidized (its ox. state raises)
other element is reduced (its ox. state lowers)
2 HCl + Zn → ZnCl2 + H2
Zn0 → Zn+II
zinc was oxidized (0 → +II)
2 e-
H+I → H0
hydrogen was reduced (+I → 0) 1 e-
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 (concentration = solute / solution)
Important terms
density (ρ) = the mass of a substance per unit of
volume (kg.m-3 or g.cm-3) ρ = m/V
(cm-3 = ml)
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 (in g/mol)
Important terms
• relative atomic mass (Ar) = mass of an atom
expressed in atomic mass unit
(atomic mass unit (u) = 1/12 of the mass of 12C atom;
the atomic mass unit is also called dalton: Da)
• relative molecular mass (Mr) = sum of
relative atomic masses (Ar) of all atoms that
comprise a molecule
MW (g/mol) = Mr (no units)
= their numerical values are the same
(the mass expressed in Da is the same as in g/mol)
prefix
symbol
value
substance amount (n)
in moles:
1 mol
= 1 000 mmol
= 1 000 000 µmol
= 1 000 000 000 nmol
= 1 000 000 000 000 pmol
1 mol
= 103 mmol
= 106 µmol
= 109 nmol
= 1012 pmol
1 mmol = 10-3 mol = 0.001 mol
1 µmol = 10-6 mol = 0.000 001 mol
Chemical reactions + concentration
problems
Calculate the mass of KOH and the volume of 96% H2SO4
needed for preparation of 25 g of K2SO4
Molar masses (MW):
KOH
56.11 g/mol
H2SO4
98.08 g/mol
K2SO4
174.26 g/mol
1. write related chemical equation
2. balance the equation
3. find the stoichiometric ratio of
reacting substance amounts
4. convert substance amounts to
mass (KOH) or volume (H2SO4)
Density of 96% H2SO4: 1.8355 g/cm3
16.1 g KOH and 7.99 mL of 96% H2SO4
Chemical reactions
problems
5 mL of 0,5 M KMnO4 were mixed with 25 mL of Na2SO3, the
reaction processed quantitatively. Calculate concentration
of Na2SO3 (both: molarity and %).
KMnO4 + Na2SO3 + H2O → MnO2 + Na2SO4 + KOH
n=
2
3
1
2
3
2
1. balance the chemical equation
Molar masses (MW):
Na2SO3
126 g/mol
KMnO4
158 g/mol
0.15M = 1.89% Na2SO3
2. find the stoichiometric ratio of
reacting substance amounts
3. convert the substance amount of
Na2SO3 to molar concentration
4. convert the molar concentration of
Na2SO3 to % concentration (W/V)
Chemical reactions
homework
100 g of FeS was used for the chemical reaction:
FeS + HCl → FeCl2 + H2S
1.
balance the equation
2.
calculate the volume of HCl (36%, density = 1,18 g/ml)
needed for the reaction
3.
calculate the mass of FeCl2 prepared by the reaction
Ar(Fe) = 55,8
Ar(S) = 32,1
Ar(Cl) = 35,5
Ar(H) = 1,0
Practice 1
1)
write the chemical equation of this reaction:
potassium hydroxide reacts completely with
phosphoric acid
2)
balance the chemical equation
3)
calculate molar weights (MW) of all reactants and
products
relative atomic masses (Ar):
potassium (39), oxygen (16), hydrogen (1), phosphorus (31)
4)
calculate the total mass of reactants and compare
it with the total mass of products
Practice 1 - SOLUTION
1)
KOH + H3PO4 → K3PO4 + H2O
2)
3 KOH + H3PO4 → K3PO4 + 3 H2O
3)
Mr (KOH) = 39 + 16 + 1 = 56 ⇒ MW (KOH) = 56 g/ mol
Mr (H3PO4) = 3x1 + 31 + 4x16 = 98 ⇒ MW (H3PO4) = 98 g/mol
Mr (K3PO4) = 3x39 + 31 + 4x16 = 212 ⇒ MW (K3PO4) = 212 g/mol
Mr (H2O) = 2x1 + 16 = 18 ⇒ MW (H2O) = 18 g/mol
4)
reactants:
products:
3x56 + 98 = 266 g/mol
212 + 3x18 = 266 g/mol
MW (reactants) = MW (products)
(the Law of conservation of masses confirmed)
Practice 2
Task: 20 g of NaOH was mixed with 0.25 mol of H2SO4
1)
2)
3)
4)
write the chemical equation of this reaction
what is the type of this reaction?
balance the equation
write the simplest ratio of substance amounts of reacting
compounds
5) calculate the substance amount of NaOH used in this reaction
Ar (Na) = 23, Ar(O) = 16, Ar(H) = 1
6) give the ratio of substance amounts of reactants used in this
reaction (express the ratio in the whole numbers)
7) was the reaction completed?
8) calculate the mass of H2SO4 used (Ar of sulfur = 32)
9) calculate the volume of 96% H2SO4 used (density: 1.8355 g/cm3)
10) calculate the mass of salt produced by this reaction
11) describe the dissociation of the salt
Practice 2 - SOLUTION
NaOH + H2SO4 → Na2SO4 + H2O
neutralization
2 NaOH + H2SO4 → Na2SO4 + 2 H2O
n(NaOH) / n(H2SO4) = 2/1
Mr(NaOH) = 40 ⇒ MW (NaOH) = 40 g/mol
m = n x MW ⇒ n = m/MW = 20/40 = 0.5 mol
6) n(NaOH) / n(H2SO4) = 0.5 / 0.25 = 1 / 0.5 = 2 /1
7) yes (see point 4)
8) Mr(H2SO4) = 98 ⇒ MW (H2SO4) = 98 g/mol; n (used) = 0.25 mol
m = n x MW = 0.25 x 98 = 24.5 g
9) 96% H2SO4 (density: 1.8355 g/cm3) ⇒ 1 ml = 1.8355 g of 96%
⇒ 1.8355 x 0.96 = 1.762 ⇒ 1 ml = 1.762 g of H2SO4
⇒ 1.762 g = 1 ml ⇒ 24.5 g = 13.9 ml
10) n(H2SO4) /n(Na2SO4) = 1/1 ⇒ 0.25 mol of Na2SO4 produced (= n)
MW (Na2SO4) = 142 g/mol ⇒ m = n x MW = 0.25 x 142 = 35.5 g
11) Na2SO4 → 2 Na+ + 1 SO42- (3 ions)
1)
2)
3)
4)
5)
Chemical calculations
- important terms • reactants / products
• stoichiometric coefficients
• substance amount
• relative atomic / molecular mass
• molar mass
• density
• concentration (molar, percent)
read more in the textbook
General and Inorganic Chemistry
for Medical Students
- Radim Černý pages: 7-9, 34-40, 63-104
see http://vyuka.lf3.cuni.cz
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