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CHEMISTRY
Chemistry studies properties and transformation of the
matter.
Matter is formed by atoms. More atoms can form
molecules.
Elements are formed of the same atoms. They can be
metals or non metals.
Molecules can be formed of same or different atoms.
Compounds are formed by molecules constituted by
different atoms.
Periodical Table collects all the elements.
ATOM
It is formed by a central nucleus (positive charge) and
some surrounding electrons (charge -1).
Nucleus has positive charge and is formed essentially of
proton (charge 1+) and neutrons (no charge).
Protons give the atomic number of the element, while
Neutrons contribute to the atomic mass.
Atomic mass is the sum of protons (mass 1) + neutrons
(mass 1). Electrons have no mass.
Periodic Table reports elements and their atomic (Z =
atomic number) and mass number (W).
Protons and Electrons are responsible of a lot of
properties of elements, ions and molecules.
Elements of VIII group – Noble gases
He = Helium (Z =2), Ne = Neon (Z = 10), Ar = Argon (Z = 18),
Kr = Kripton (Z = 36), Xe = Xenon (Z = 54), Rn = Radon (Z = 86).
The name noble gas indicates their high stability. Reactions
involving noble gas are really very unusual.
Each electron has a specific energy and occupies a particular shell
around the nucleus. External shells containing 8 electrons are
particularly stable, like noble gases.
Compounds are formed by elements tending to reach the noble gas
corresponding.
For example Chlorine having 7 electrons in the last shell tends to
catch 1 electron more to assume the Ar configuration.
Sodium (Z=11) having 1 alone electron in the last shell, has difficulty
to hold it and tends to release it to assume to Ne configuration.
Ions Cl- and Na+ are respectively formed. They attract each other
forming the compound NaCl = sodium chloride.
HOW DIFFERENT ATOMS FORM
MOLECULES:
THE CHEMICAL BOND
• Elements belonging to the same group of
periodical table have similar properties.
• Ionic or Electrostatic bond: Na+ + Cl- =NaCl
(sodium gives an electron to chlorine, ions of
opposite charge attract each other).
• Homopolar or Covalent bond: Cl + Cl =Cl2 (Two
chlorine atoms put together a couple of electrons
• Coordination bond: Cu(NH3)42+. Each :NH3 put in
common two electrons bonding Cu2+
• Others less important (Hydrogen, Van der Walls...)
NOMENCLATURE 1
•Basic Oxides = Metals + Oxygen; i.e. CaO = calcium oxide
•Acid Oxides = Non Metals + Oxygen; i.e. CO2 = carbon dioxide
•Bases or Hydroxides = Oxides + H2O; i.e. CaO+ H2O =
Ca(OH)2 = calcium hydroxide
•Acids = acids oxides + H2O; CO2 + H2O= H2CO3 = carbonic
acid (acids containing oxygen)
•Acids = Non metal + hydrogen = i.e. HCl = Hydrochloric acid
NOMENCLATURE 2
•Acids containing oxygen = the name changes depending on the
kind of oxides:
SO2 (sulphur dioxide) forms solphorous acid, while
SO3 (sulphur trioxide) forms solphoric acid.
Cl2O + H2O = 2 HClO (ipochlorous acid)
Cl2O3 + H2O = 2 HClO2 (chlorous acid)
Cl2O5 + H2O = 2 HClO3 (chloric acid)
Cl2O7 + H2O = 2 HClO4 (perchloric acid)
NOMENCLATURE 3
Acid + Bases = Salt
HCl + NaOH = NaCl (sodium chloride) + H2O
HClO+ NaOH = NaClO (sodium ipochlorite) + H2O
HClO2+ NaOH = NaClO2 (sodium chlorite) + H2O
HClO3+ NaOH = NaClO3 (sodium chlorate) + H2O
HClO4+ NaOH = NaClO4 (sodium perchlorate) + H2O
N2O3+H2O=2HNO2 (nitrous acid)+2 NaOH=2 NaNO2 (sodium
nitrite)
N2O5+H2O=2HNO3 (nitric acid)+2 NaOH= 2 NaNO3 (sodium
nitrate)
NOMENCLATURE 4
Acid + Bases = Salt
H2S (sulphydric acid) + 2 NaOH = Na2S (sodium sulphide)
+ 2H2O
H2SO3 (Sulphurous acid)+ 2 NaOH = Na2SO3 (sodium solphite)
+ 2H2O
H2SO4 (Sulphuric acid)+ 2NaOH = Na2SO4 (sodium solphate)
+ 2 H2O
What is a formula?
•The formula gives the identity to a
Compound.
•It tell us which elements are present in the compound
•It gives the quantitative composition of the compound
•It allows to calculate the molecular mass
•From its formula it is possible to know the properties
of the compound
Example:
Na2SO4 (acid, bases, salt, oxidant, reducing,…)
It is a salt, formed by sodium, sulphur and oxygen with
a ratio 2:1:4, its mm is 23x2+32+4x16=142.
It is not an oxidant reagent, but it can be reduced to Na2SO3 or to
S or to Na2S.
What is a reaction?
The transformation of a few
compounds to others. Reactions
evolve to a lower energy status
In any reaction, the material balance must be verified
Frequent types of reactions are:
1) Acid – base reactions. 2) Precipitation reactions.
3) Complex formation reactions. 4) Redox reactions.
Property of compounds
Molecular compounds
ionic compounds
Molecular compounds remain no dissociated
Ionic compounds dissociate in ions
Ions are pieces of compounds which have an electrical charge
They can be positive (cations) and negative (anions), both
are able to carry electrical current.
Acids and bases dissociate in Protons (H+) or hydroxyl ions (OH-)
and anions and cations respectively.
i.e.: HCl (hydrochloric acid) dissociates in H+ and Cl-, while
NaOH (sodium hydroxide) dissociates in OH- and Na+.
Salts dissociate in cations and anions (i.e. NaCl = Na+ + Cl-)
Some easy rules
Each element can have one or
more oxidation numbers.
Oxidation numbers have a sign too, it can be + or -.
The highest oxidation number corresponds to the
periodical table group.
The lowest one can be obtained by subtracting 8 from
the highest (– 8).
i.e. The highest oxidation number of chlorine is +7,
the lowest is –1. Metals have generally only zero or the
oxidation number corresponding to the periodical table
group. Compounds have no charge. Ions have a charge.
Some examples of easy reactions
Reaction between aluminium sulphate and barium nitrate.
Formulae: aluminium belongs to the third group and has +3;
Sulphate is the ion of sulphuric acid then is SO4= (charge –2),
The formula of aluminium sulphate is Al2(SO4)3.
Barium belongs to the second group then can be Ba2+, while
nitrate being the ion of nitric acid is NO3-. Formula is Ba(NO3)2
Reaction is:
Al2(SO4)3 + 3 Ba(NO3)2 = 3 BaSO4 + 2 Al (NO3)3
This means that (2x27+3x96) g of aluminium sulphate reacts with
3(137,3+ 2x62) g of Ba(NO)2 to give what????
By the way, compare the number with the atomic mass.
Some easy
calculations 1
In any reaction, the material
balance must be verified
By putting in a becker 10 g of zinc and by adding HCl, hydrogen
Develops. Write the reaction and calculate how much zinc chloride
Will be formed.
Write the reaction: zinc belongs to the second group, its oxidation
numbers can be 0 or +2.
Chlorine is in the VII group, in the lowest oxidation state is –1.
Zn + 2 HCl = ZnCl2 + H2 (gas)
Calculations: 1atomic mass of zinc (65,37) reacts with 2 molecular
Mass of HCl (35,45=70,90) to give 1 molecular mass of ZnCl2
(134,27) and 1 molecular mass of H2 (2).
Some easy calculations 2 (follows from 1)
But zinc put in the reaction was 10 g (not 65,3) then to know how
much ZnCl2 is formed, it is necessary to divide 10 / mw of Zinc
(65,37) to have the number of moles (= 0,153). The same number of
moles of ZnCl2 (see reaction coefficients). The mass of ZnCl2 will
be = number of moles (0,153) x mm of the salt (134,27) = 20,5 g
In the above reaction H2 (gas) was formed. Some rules for gas exist.
They connect volume, pressure and temperature of a gas through
a general equation : PV = nRT, where P is pressure, V volume and T
The absolute temperature (°C + 273,14°K). n = the number of moles
Of the gas. R is a constant. When Volume and temperature are
expressed in litres and atmospheres, respectively R = 0,0821.
Avogadro law: equal volumes of different gases at the same pressure
And temperature contain the same number of molecules. A mole =
6 x 1023 molecules at 1 atm and 273°C occupies 22,4 litres.
Some calculations 3 (follows from 2)
1) The compound ferric alum has the formula:
K2SO4Fe2(SO4)324H2O. Calculate the percentage of iron present.
By knowing the formula, the mm (mole mass) can be obtained.
Atomic mass are yield from the tables. K = 39,1; S = 32; O = 16;
Fe = 55,85; H = 1.
MM of the compound is = 1005,7. In it 2 Fe atoms (2 x 55,85=
111.7) are present. By dividing 111,7/1005,7 and multiplying by
100, the percentage of Fe in the alum is obtained.
2) Calculate how many moles of CO2 can be obtained by heating
150 g of CaCO3 (calcium carbonate).
By heating CaCO3, the following reaction takes place:
CaCO3 = CaO + CO2. This means that 1 mole of CO2 (i.e. 44 g) is
produced by 1 mole of CaCO3 (100 g). By dividing 150 g by 100
(1,5), the number of moles of CaCO3 and then of CO2 is obtained.
To obtain the weight of CO2, 1,5 is multiplied by 44 (mm of CO2).
Some calculations 4 (follows from 3)
Reaction with formation of a slight soluble compound. Some
compounds are not too much soluble in water and their solubility
is regulated as discussed in next slides.
Slight soluble substances are BaSO4 (it is used for x-ray inspection),
Fe2O3 (rust), Pb3O4 (minium), CaCO3 (marmor), CaC2O4 (calcium
oxalate, is an organic substance, responsible to the formation of
renal stones), CaSO4 (gypsum or chalk), and others.
Problem: An operator has to prepare 50 g of Barium sulphate.
Calculate what amount of BaCl2 and Na2SO4 must mixer to obtain
the required quantity of BaSO4.
The reaction is: BaCl2 + Na2SO4 = BaSO4 + 2 NaCl.
One mole of BaSO4(329,34) is obtained from 1 of BaCl2 and Na2SO4
50 /329,34 = 0,152. This number by 208,24 and by 142, respectively,
Gives the weight of BaCl2 and Na2SO4 to use.
Some definitions
System is a part of matter subject of study. It can be homogeneous
(if in all part has the same properties) or heterogeneous (when its
parts have different properties.
A phase is a homogeneous part of a system. When the composition
of a phases vary, but homogeneity remains, the phase is called
Solution.
A solution is a homogeneous mixture of two or more substances.
The main component is the solvent, the minor components are
called solute.
Concentration can be defined as the quantity of a solute present in
a fixed quantity of solvent. There are different possibility :
% weight; % volume; Molarity; Normality; Molality.
Total or free (at equilibrium) concentrations will be distinguished.
Total Concentration 1
% weight = g of solute per 100 g of solvent.
% volume = g of solute per 100 cc of solvent.
Molarity (M or mol dm-3)= number mol of solute per litre of solution.
Normality (N) = number equivalents (to define) of solute per litre
(dm3) of solution.
Molality (m) = number mol of solute per kg of solvent.
Easy calculations
1) Calculate the molarity of a solution of NaCl 5% in volume.
The problem indicates that 5 g of NaCl are solved in 100 cm3 of a
solvent (Water). It is the same to say 50 g per dm3. It is sufficient to
divide 50 per mm of NaCl (mm =23+35,45=58,45) to have 0.855 M
2) How many grams of Na2SO4 (mm=142) are in 10 cc of its solution
0.25 M? Answer: 0.01x0.25x142 = 0.355.
Total Concentration 2
Molarity corresponds to the presence of 1 mm(molecular mass)
in 1.00 dm3 of solution. The mm is exactly defined.
Normality corresponds to the presence of 1 equivalent in 1.00 dm3
of solution. The equivalent weight is not a constant, but it depends
on the reaction taking place, then it can assume different values.
However, some general rules exist: (weq = equivalent weight)
• For an atom, weq = ma/oxidation number. I.e. weqAl= Al/3=
26.98 /3, corresponding to the reaction 2Al+6HCl=2AlCl3+3H2.
Weq is the weight corresponding in the reaction to 1 H.
• For an acid, weq = mm of acid / number of H
• For a basis, weq = mm of basis / number of OH
• For H3PO4, weq can be mm/1; mm/2 or mm/3 depending on the
protons involved in the reaction.
Weq calculation is different for an oxidant or a reducing (see bellow)
Total Concentration 3 -REDOX
Oxidant is a compound which takes electrons and passes in its
reduced form: Ox + n1 e = red.
Reducing is a compound which releases electrons and turns in its
oxidated form: Red = Ox + n2 e.
As free electrons cannot be present in a condensed state of matter,
an oxidation cannot take place if, contemporary, a reduction does not
occur, by involving the same number of electrons:
The reaction can be: Ox1 +Red2 = Red1 + Ox2, if n1 = n2
Weq of an oxidant or of a reducing is calculated as follows: Weq =
Mm oxidant (or reducing) / electrons number involved in the reaction.
For example: IO3- can be reduced to I2 or to I-, according to the
half -reactions: 2 IO3- +10 e= I2; weq = 2 mm/10 electrons or
: IO3- + 5 e = I-; weq = mm/5 electrons
Redox reaction balance rules
1) All elements not combined have oxidation number zero
2) Oxygen has oxidation number –2, (except peroxydes, i.e. H2O2)
3) Hydrogen has oxidation number +1, (except hydrures, i.e. LiH)
4) A compound has always charge zero.
5) An ion has always the charge corresponding to its properties
6) It is possible to have fractional oxidation numbers
7) In a redox reaction the released electron number must be equal to
those taken by the oxidant.
To balance a redox reaction
1.Find the oxidant and the reducing in the proposed reaction
2.Write separately half reaction for both
3.Balance the ion where is the oxidant.
4.Calculate the redox number variation and add the
corresponding electrons number on the part of the ox species
5.Charge balance: if the reaction occurs in acid field balance
with H+, in alkaline field balance with OH6.Mass balance: write the corresponding H2O on the other side
of H+, or OH-. Now the half reaction of oxidant is balanced.
7.Treat in the same way the half reaction of the reducing.
8.Find the minimum common multiple between the acquired
and released electrons and balance definitively both half
reaction.
EXAMPLES 1
Reaction between Cu2+ + Zn = Cu + Zn2+
Copper is the oxidant because reduces its oxidation number from
2+ to zero, while zinc is the reducing by passing from Zn to Zn2+
Half reactions are:
Cu2+ +2e = Cu; Zn = Zn2+ + 2e. No balance of charge and mass is
necessary because they are already balanced. The number of
exchanged electrons is 2 in both cases then the minimum common
multiple is 2 and the complete reaction is: Cu2+ + Zn = Cu + Zn2+
Reaction between Ag++ Cu = Ag + Cu2+(not balanced)
Half reaction of the oxidant: Ag++ e = Ag
Half reaction of the reducing: Cu = Cu2+ + 2e
The half reactions are balanced for charge and mass, but the electron
number is not the same, then half reaction of Ag must be multiplied
by two, to have: 2Ag++ Cu = 2Ag + Cu2+
EXAMPLES 2 – acid field
Reaction between: MnO4- + Fe2+ + H+ = Mn2+ + Fe3+
Oxidant is permanganate ion and reducing is iron (II).
H. reaction ox: MnO4- = Mn2+, Manganese passes from +7 to +2, by
Acquiring 5e, and then: MnO4- + 5 e = Mn2+.
Right there is charge –6, left is +2 and H+ is used to balance charges,
so that it is necessary to add 8 H+ right and balance with H2O left:
MnO4- + 5 e + 8 H+ = Mn2+ + 4 H2O (half reaction balanced)
H. reaction red: Fe3+ + e = Fe2+ (half reaction balanced).
Minimum common multiple between 1 and 5 is 5, then at end:
MnO4- +5 Fe2+ + 8 H+ = Mn2+ + 5 Fe3+ + 4 H2O
EXAMPLES 3 – acid field
1) Reaction between: dichromate ion and iron (II) in acid field:
Cr2O7= + Fe2+ + H+ = Cr2+ + Fe3+
S. reaction ox:Cr2O7= = Cr3+, to balance chrome, it is Cr2O7= = 2Cr3+,
Chrome assumes 3 e x 2Cr = 6 e to have:
Cr2O7= + 6 e= 2 Cr3+,
there are 8- right and 6+ left and H+ is used to balance charges,
it is necessary to add 14 H+ right (to balance charges) and 7 H2O left
To balance mass, i.e. Cr2O7= + 6 e + 14 H+ = 2 Cr3+ + 7 H2O.
S. reaction red: For iron is easy: Fe2+ = Fe3+ +e. We need to multiply
by six the iron half reaction, to have definitely:
Cr2O7= + 6 Fe2+ + 14 H+ = 2 Cr2+ + 6 Fe3+ + 7 H2O
2) Reaction between thiosulphate and iodine: S2O3= + I2 = S4O6= + IOx. Number of S right is +2, but left is + 2.5, but there are 4 S, so
that, 2 S2O3= = S4O6= + 2 e and I2 + 2 e = 2 I-.
Complete reaction is: 2 S2O3= + I2 = S4O6= + 2 I-
EXAMPLES 4 – acid field
Dissolution of copper (II) sulphur by means of nitric acid.
S in CuS has oxidation number –2 and can be oxydated by HNO3,
Nitrogen has ox number +5 and it can be reduced to NO (+2).
Oxidation half reaction:
CuS = Cu2+ + S + 2 e (balanced for charge and mass)
Reduction half reaction: NO3- + 3 e + 4 H+ = NO + 2 H2O (balan.)
Minimum common multiple between 3 and 2 is 6, so that complete
reaction is: 3 CuS + 2 NO3- + 8 H+ = 2 NO + 4 H2O + 3 Cu2+ + 3 S
The reaction can be written also in molecular form, as follows:
3 CuS + 8 HNO3 = 2 NO + 4 H2O + 3 Cu(NO3)2 + 3 S.
EXAMPLES 4 – alkaline field
The mechanism is the same that in acid. Only the charge is balanced
By OH-, instead than H+
Example 1
Arsenic (As) oxidation to arseniate (AsO43-) by means of sodium
hypochloride (ClO-) in alkaline solution.
As + 8 OH- = AsO43- + 5 e + 4 H2O
ClO- + 2 e + H2O = Cl- + 2 OH- (Between 5 and 2 is 10)
Complete ionic reaction:
2 As + 5 ClO- + 6 OH- = 2 AsO43- + 5 Cl- + 3 H2O
Complete molecular reaction:
2 As + 5 NaClO + 6 NaOH = 2 Na3AsO4 + 5 NaCl + 3 H2O
EXAMPLES 5 – alkaline field
Example 2
Reduction of bismuth hydroxide [Bi(OH)3] by means of sodium
stannite Na2SnO2 in alkaline solution.
Bismuth can have oxidation numbers: +5, +3, 0;
Tin can have oxidation number +4, +2, 0. The reagents are Bi(+3)
and Sn(+2). If bismuth is reduced it will beBi0, while tin will be
oxidated, becoming Sn(+4), that means stannate SnO3=.
The half – reactions:
Bi(OH)3 + 3 e = Bi0 + 3 OHSnO2= + 2 OH- = SnO3= +2 e + H2O
Reaction: 2Bi(OH)3 + 3 SnO2= = 2 Bi0 + 3 SnO3= + 3H2O
Molecular reaction:
2Bi(OH)3 + 3 Na2SnO2 = 2 Bi0 + 3 Na2SnO3 + 3H2O
Chemical equilibrium – Reactions 1
In the previously studied slides, the mass of the reaction products
were calculated, by assuming the complete transformation of the
reagents in the corresponding products.
This is not always true. Most of the reactions are of equilibrium, i.e.
The transformation of the reagents into the products is not complete,
But part of the reagents remain together with part of products.
Put in a bottle CO and H2O in appreciable concentration. At start,
they react quickly. The speed decreases by increasing the reaction
products concentration. The following reaction takes place:
CO + H2O  CO2 + H2. The symbol  indicates a double arrow,
i.e. the reaction can give from right to left or vice versa.
The equilibrium reactions occur frequently, overall if the reaction
takes place in a single phases (homogeneous), for example for gases
Chemical equilibrium – Reactions 2
Chemical reactions are equilibrium reaction, except those involving
Formation of a gas or a precipitate (a slight soluble compound).
The separation of phases makes complete a reaction.
Equilibrium reaction means that after the start, the reaction product
React among them to have again the reagents. In the above example
At start CO and H2O produce CO2 + H2, but when a appreciable
Concentration of CO2 + H2 is formed, they react to form again
CO + H2O. A point exists when the speed of the direct reaction will
Be equal to the reverse one. This is the equilibrium point. This means
that no static equilibrium, but a dynamic equilibrium exists.
Really, chemical reactions occur on the basis of energetic content.
If reagents have more energy than the products the reaction takes
Place, till both energy contents will be equal. This means that a
mathematic relation regulates the trend of the reaction.
The mass action law
At equilibrium and at constant temperature, the ratio among the
concentrations of the reaction products and the reagents, each of
them at exponent equal to the stoichiometric coefficient, is a
constant, generally indicated by k
The equilibrium reaction:
aA+b B c C +d D
is regulated by the constant: k = (Cc Dd) / (Aa Bb)
The value of k is constant and it is independent of the reagents or
products concentration. It is only depending on the temperature.
The mass action law – calculation 1
Example 1:
2.94 moles of I2 and 8.1 of H2 form at equilibrium at a constant
temperature 5.64 moles of HI. Calculate the equilibrium constant.
The reaction is: I2 + H2  2 HI and the constant k=cHI2/(cH2 cI2)
It is possible to draw up the following table:
Table
H2
I2
Initial moles
8.1
2.94
Formed moles
-----Used moles
2.82(5.64/2) 2.82
At equilibrium
5.28
0.12
K = (5.64)2 / (5.28 x 0.12) = 50.2
HI
--5.64
---5.64
The mass action law – calculation 2
Example 2: At a constant temperature, the constant of the reaction:
I2 + H2  2 HI is k = 0.02. Calculate the equilibrium concentration
For all the present species, starting by 0.48 moles of both I2 + H2.
Prepare a table similar to that above formulated.
Table
H2
I2
Initial moles
0.48
0.48
Formed moles
-----Used moles
x
x
At equilibrium
0.48-x
0.48-x
HI
--2x
---2x
Write the constant k = 0.02= (2x)2/[(0.48-x) (0.48-x)].
It can be calculated x = 0.032 moles
Concentrations: cHI=0.064; cH2= 0.448 = cI2
Total and free concentration 1
It is unusual that a chemical species alone is present in a solution.
As previously described, different types of bonds can take place
Among atoms or ions, according to the properties of them.
Elements put right in the periodical table tend to catch an electron
To reach a stable configuration. They become anions, i.e. Cl-.
The presence of the negative charge or the presence of two electrons
(like :NH3) give them the property to hook a cation to form a new
Species, called a complex ion. It can be charged or neutral.
The formation of a complex follows the equilibrium rules. For
instance, a complex can be formed between Fe3+ and Cl-, as follows:
Fe3+ + Cl-  FeCl2+ with a k = c FeCl2+ / (cFe3+ cCl-).
The same occurs between Cu2+ + NH3  Cu (NH3)2+.
Total and free concentration 2
In the example of the above slide, it is possible to write:
CFe = cFe + cFeCl and in the similar way: CCu = cCu + cCu(NH3)
Charges are omitted. C = total concentration and c = the free one or
Equilibrium concentration. Sometimes [Cu2+] is used to indicate the
Free concentration of the ion copper (II).
Pay attention in all the equilibrium constants and / or in all the
thermodynamic expression, the free concentration must be used.
Example: Calculate the free concentration of silver and cyanide in
a solution 0.1 M of KAg(CN)2, knowing k = 1 x 1021.
As KAg(CN)2 completely dissociates in K+ and Ag(CN)2-,
the equilibrium : Ag+ + 2 CN-  Ag(CN)2-, is to study.
It has k = [Ag(CN)2-] [Ag+]-1 [CN)-]-2 = 1 x 1021
(follows)
Total and free concentration 3
(follows)
The table relative to the species can be built:
Table
Initial moles
Formed moles
Used moles
At equilibrium
Ag(CN)20.1
--x
0.1 -x
Ag+
--x
--x
CN--2x
---2x
k = [Ag(CN)2-] [Ag+]-1 [CN-]-2 = (0.1-x)/(x) (2x)2= 1 x 1021,
By resolving the above expression, it is obtained [Ag+]= 2.9x10-8M
[CN-]= 5.85x10-8 and [Ag(CN)2-] remain 0.1M, because 2.9x10-8 is
negligible with respect to 0.1M.