Download Matter and Measurement

Acid and Base Anhydrides
These are compounds that themselves are not acids or
base, but when dissolved in water produce an acid or base
by reacting with water.
Example: SO3(g) can dissolve in water and react with water.
SO3(g) + H2O(l) --> H2SO4(aq)
Reaction of compounds with water are called HYDRATION
reactions.
SO2(g), a pollutant in the atmosphere produced as a
byproduct of the burning of coal, reacts with O2(g) to form
SO3(g)
2SO2(g) + O2 --> 2SO3 (g)
The SO3 formed then react with water in the atmosphere to
form H2SO4.
SO3(g) + H2O(l) --> H2SO4(l)
The consequence of the reaction of SO3 and H2O is ACID
RAIN.
Oxides of alkali and alkaline earth metals (Gr IA and IIA) react
with water to form a base and hence are base anhydrides.
Na2O(s) + H2O(l) --> 2 NaOH(aq)
Oxides of non-metals react with water to form acids and
hence are acid anhydrides.
Cl2O(g) + H2O(l) --> 2 HOCl(aq)
ionic oxides --> base anhydrides
covalent oxides --> acid anhydrides
Other reactions of acids and bases
1) Acids react with carbonates and hydrogen carbonates
(bicarbonates) producing CO2(g), salt and water
NaHCO3(s) + HCl(aq) --> NaCl(aq) + H2O(l) + CO2(g)
Net ionic equation
NaHCO3(s) + H+(aq) + Cl-(aq) --> Na+(aq) + H2O(l) + CO2(g)
2) Acids react with oxides of metals to form salt and water
CuO(s) + H2SO4(aq) --> CuSO4(aq) + H2O(l)
Net ionic equation
CuO(s) + 2H+(aq) --> Cu2+(aq) + H2O(l)
3) Acids react with many metallic elements to form H2(g) and
a salt.
Zn(s) + 2 HCl(aq) --> ZnCl2(aq) + H2(g)
Net ionic equation
Zn(s) + 2 H+(aq) --> Zn2+(aq) + H2(g)
One of the constituents of acid rain is nitric acid. Marble has
the chemical composition of calcium carbonate. Write
balanced chemical equations representing how acid rain
containing nitric acid dissolves marble statues.
Net ionic equation
CaCO3(s) + 2H+ --> Ca2+(aq) + CO2(g) + H2O(l)
Naming Acids & Bases
There are three classes of acids
1) Binary acids - e.g. HCl, HF
replace “ide” with “ic”
2) Oxoacids - contain H, O and a 3rd element
Replace “-ate” with “ic”.
e.g. HNO3 - nitric acid, H2SO4 - sulfuric acid
or replace “-ite” with “-ous”. e.g. HNO2 - nitrous acid
3) Organic acids - CH3COOH acetic acid
Bases
NaOH - sodium hydroxide
Mg(OH)2 - magnesium hydroxide
CH3 C OH
O
Oxidation-Reduction Reaction
Oxidation-Reduction reactions, or REDOX reactions involve
transfer of electrons between reactants
Examples of reactions that fall under the class of redox
reactions are the reactions that take place when a battery
operates, reactions responsible for corrosion of metals,
metabolic reactions
Fe(s) + 2H+(aq) --> Fe2+(aq) + H2(g)
Fe(s) is OXIDIZED to Fe2+
H+ is REDUCED to H2
Charges on Atoms and Molecules
A charged ion forms when the neutral atom or molecule gives
up or accepts electrons.
Ionic compounds are made up of oppositely charged ions.
Solid NaCl exists as a lattice of Na+ and Cl- interacting with
each other; a crystal of potassium permanganate, KMnO4, is
made up of K+ ions and MnO4- ions.
The charge on these ions is real; the atom or molecule is
charged.
Formal Charges
Formal charges are hypothetical charges calculated
assuming that the electrons in a covalent bond are equally
shared by both atoms.
So in HCl, the formal charge on both H and Cl is zero
0
0
H Cl
Formal charges do not account for electronegativities
Oxidation Number or Oxidation State
The oxidation number of an atom in a compound is the
charge the atom would have if the compound was completely
ionic.
For an ionic compound, an atom’s oxidation number is the
same as the charge on the atom.
For example, in NaCl, the oxidation number of Na+ is +1 and
that of Cl- is -1.
Oxidation numbers can also be assigned to atoms in a
covalent compound.
In assigning oxidation number to atoms in a covalent bond,
the relative electronegativity of the two bonded atoms is
taken into account
For example, in HCl, since Cl is more electronegative than H,
Cl is assigned an oxidation state of -1 and H a +1 oxidation
number.
+1
-1
H Cl
Assignment of oxidation numbers assumes no sharing of the
electron pair in the covalent bond.
Rules to assign oxidation numbers
1) The oxidation state on any atom in its elemental form is
zero (H2, O2, Na(s))
2) The sum of the oxidation states of all atoms in a neutral
compound is zero (CH4, NH3)
3) The sum of the oxidation states of all atoms in an ion is
equal to the charge on the ion (NO3-, SO42-)
4) The oxidation state of all Group IA elements on
compounds is +1, Group IIA elements is +2
5) The oxidation state of H in compounds is +1, except in
metal hydrides of very electropositive elements where it is -1
6) The oxidation state of F is always -1
7) The oxidation state of O in a compound is always -2,
except where two O atoms are bonded together, as in H2O2,
(each O is -1), or O is bonded to F as in OF2 (O is +2).
Problem:
Determine the oxidation number of S in each of the following
compounds
a) H2S; b) S8; c) SCl2; d) Na2SO3, e) SO42a) H : +1; S : -2
b) S8 - elemental form of S, oxidation number of S is zero
c) SCl2 - Cl has an oxidation number of -1; S is +2
d) Na2SO3 - Na is +1; O is -2
2(1) + y + 3(-2) = 0
2 + y - 6 = 0; hence y = 4
Na2SO3
S is +4
e) y + (4(-2)) = -2 => y = +6.
Hence in SO42- S is +6
Oxidizing and Reducing Agents
An atom is oxidized (loses electrons) if its oxidation number
increases, and is reduced (gains electrons) if its oxidation
number decreases
An oxidizing agent causes the oxidation of another species
by accepting an electron from it; in the process it is
reduced.
A reducing agent causes the reduction of another species
by giving up an electron to it; in the process it is oxidized.
A strong oxidizing agent can remove electrons from a weak
reducing agent and a strong reducing agent can force
electrons onto a poor oxidizing agent.
Animation
ClO-(aq) + 2H+(aq) + Cu(s) --> Cl-(aq) + H2O(l) + Cu2+(aq)
+1 -2
+1
0
-1
+1 -2
+2
ClO-(aq) + 2H+(aq) + Cu(s) --> Cl-(aq) + H2O(l) + Cu2+(aq)
Types of Redox Reactions
1) Combination Reactions
Most metals react with non-metals to give ionic compounds.
The metals are oxidized and the non-metals reduced.
Example: 2 Na(s) + Cl2(g) --> 2 NaCl(s)
When the metal can form ions of different charges, different
products are possible depending on the experimental
conditions.
Example:
2 Fe(s) + O2(g) --> FeO(s)
Oxidation Number of Fe is +2
4 Fe(s) + 3O2 (g) -->2 Fe2O3(s) Oxidation Number of Fe is +3
Non-metallic compounds can combine with other nonmetallic compounds.
For example, if P4(s) reacts with Cl2(g), the product depends
on the amount of Cl2(g) present.
If the amount of Cl2(g) is low:
0
+3
0
-1
P4(s) + 6Cl2(g) --> 4PCl3(l)
At higher levels of Cl2(g)
0
0
+5 -1
P4(s) + 10Cl2(g) --> 4PCl5(l)
2) Decomposition Reactions: reverse of combination reactions
For example, strong heating can result in a decomposition
reaction.
+1
-2
0
0
2Ag2O(s) --> 4Ag(s) + O2(g)
Products of decomposition reactions can sometimes be
determined by looking for chemical formulas of stable
molecules (e.g. H2O, HCl, CO2, SO2, NaCl) embedded in the
compound being heated.
+1 +5 -2
+1 -1
0
2 KClO3-->2 KCl + 3O2
Note: Check that oxidation-reduction occurs in the reaction.
+1 +3 -2
+3 -2
+1 -2
2HNO2 --> N2O3 + H2O
Is this a redox reaction?
3) Oxygenation Reactions
O2 is a powerful oxidizing agent and combines with most
elements forming oxides.
0
0
+1 -2
4Li(s) + O2(g) -->2 Li2O(s)
In the presence of excess O2, electropositive elements can
also form peroxides (the peroxide ion is O2-)
0
0
+1 -1
K(s) + O2(g) --> KO2(s)
Many binary compounds containing hydrogen can be
oxidized by O2 to form water and an oxide.
-3 +1
0
+5 -2
+1 -2
4PH3(g) + 8O2(g) --> P4O10(s) + 6H2O(g)
A class of oxygenation reactions is combustion reactions
where C and H containing compounds are burnt in O2 to form
CO2 and H2O
-4 +1
0
+4 -2
+1 -2
CH4(g) + 2O2(g) --> CO2(g) +2 H2O(g)
Metabolism of glucose, C6H12O6(s), to form CO2(g) and
H2O(g)
4) Hydrogenation reactions
H2 is a good reducing agent.
i) Nonmetallic elements are reduced by H2
0
0
-3 +1
P4(s) + H2(g) --> PH3(g)
ii) However, when reacting with very electropositive elements
H2 acts as an oxidizing agent, being reduced itself
0
0
+1 -1
Na(l) + H2(g) --> NaH(s)
iii) H2 reacts with metal oxides to yield the metal and water
+1
-2
0
0
+1 -2
Ag2O(s) + H2(g) --> 2Ag(s) + H2O (g)
iv) H2 reacts with non-metal oxides to form water and the
non-metal bonded to H
+2 -2
0
-4 +1
+1 -2
CO(s) + 3H2(g) --> CH4(s) + H2O (g)
5) Displacement Reactions
Reactions when one element displaces another from a
compound.
+1 +5 -2
0
+2 +5 -2
0
2AgNO3(aq) + Cu(s) --> Cu(NO3)2(aq) + Ag(s)
Net ionic reaction
+1
0
+2
0
2Ag+(aq) + Cu(s) --> Cu+2(aq) + Ag(s)
Can we predict which element will displace another?
Cu is a more electropositive element that Ag
In general, the more electropositive element displaces the
less electropositive element.
The ACTIVITY series is a list of metals (and hydrogen)
arranged in decreasing ease of oxidation
Elements higher in the activity series can displace elements
lower in the series.
Predict what happens when
i) Fe(s) is added to a solution of Cu(NO3)2
ii) Cu(s) is added to a solution of Fe(NO3)3
Why should you not cook tomatoes in an aluminum pan?
Why does iron rust in an acidic solution?
Displacement reactions with electronegative elements
The more electronegative element displaces the less
electronegative element, acting as an oxidizing agent.
0
+1 -1
0
+1 -1
Cl2 (g) + 2 KI(aq) --> I2(s) + KCl(aq)
0
-1
0
-1
Cl2 (g) + 2 I- (aq) --> I2(s) + Cl-(aq)
Cl is reduced from 0 in Cl2 to -1 in KCl and I is oxidized from
-1 in KI to 0 in I2
I2 (s) + 2 KCl(aq) --> ????
6) Disproportionation Reactions
A reaction in which the same compound undergoes both
oxidation and reduction.
+1 -1
+1 -2
0
2 H2O2(l) --> 2 H2O(l) + O2(g)
A final note on redox reaction:
To determine if a reaction is a redox reaction calculate the
oxidation numbers of elements in the reactants and
products.
If the oxidation numbers of elements change during the
reaction indicating that both oxidation and reduction occurs
then the reaction is a redox reaction.
Determining Concentrations of Solutions
We can use our knowledge of the type of chemical reactions
and stoichiometry to determine concentrations of solutions.
Determine the volume of the solution of known concentration
that is required to completely react with a given volume of the
solution of unknown concentration.
Method of determining the concentration of solutions is
called TITRATION.
Can be an acid-base reaction, precipitation reaction, redox
reaction
Prepare a STANDARD solution, of known concentration.
Measure a known volume of the solution whose
concentration is to be determined
Add sufficient amount of the standard solution to
completely react with the volume measured of the solution
of unknown concentration.
The most commonly used unit for concentration of solutions
is MOLARITY which is defined as the number of moles of
solute in a liter of solution.
Molarity =
moles of solute
= mol L-1 (M)
Volume of solution in liters
Molarity of a solution made by dissolving 23.4 g of sodium
sulfate, Na2SO4, in enough water to form 125 mL of solution.
23.4 g x 1 mol Na2SO4
142 g Na2SO4
= 0.165 mol Na2SO4
Molarity of solution = 0.165 mol Na2SO4 = 1.32 M
0.125 L solution
In solution Na2SO4 completely dissociates to Na+ and SO42Na2SO4(s) --> 2 Na+(aq) + SO42-(aq)
Molarity of Na+(aq) (1.32 M x 2) = 2.64 M
Molarity of SO42-(aq) = 1.32 M
Preparing solutions of known concentrations
How many grams of Na2SO4 are required to make 350 mL of
0.500 M Na2SO4?
First determine how many moles of Na2SO4 are contained in
350 mL of a 0.500 M solution.
Moles of Na2SO4 = 0.500 mole Na2SO4 x 0.350 L solution
1 L solution
= 0.175 moles Na2SO4
Grams of Na2SO4 = 0.175 moles Na2SO4 x 142 g Na2SO4
1 mole Na2SO4
= 24.9 g Na2SO4
Dilutions
A more dilute solution of a specific concentration can be
prepared from solution of higher and known concentration.
Number of moles before dilution = Number of moles after
dilution
Since, moles = molarity (mol/L) x volume (L)
For the dilution
Initial molarity x initial volume = final molarity x final volume
Mi x Vi = Mf x Vf
How many milliliters of 3.0M H2SO4 are required to make 450
mL of 0.10 M H2SO4?
Mi x Vi = Mf x Vf
Vi = Mf x Vf
Mi
= 0.10 M x 450 mL
= 15 mL
3.0 M
To prepare the 0.10 M solution, measure out 15 mL of the
3.0M H2SO4 (also known as the stock solution) and add water
till the total volume is 450 mL.
Determining the concentation of an HCl solution
via an acid-base titration
HCl(aq) + NaOH(aq) --> NaCl(aq) + H2O(l)
H+(aq) + OH-(aq) --> H2O(l)
Net ionic equation
As more NaOH is added, more of the HCl is consumed.
The point at which just enough NaOH has been added so that
all the HCl has been consumed is called the EQUIVALENCE
point.
Knowing the equivalence point, the stoichiometry of the
reaction, the concentration of the NaOH solution, we can
determine the concentration of the HCl solution.
To indicate when the equivalence point has been reached, a
very small amount of a chemical compound called an
INDICATOR is added to the HCl solution, before the NaOH is
added.
At the equivalence point, all the HCl has reacted.
Addition of a very small amount of NaOH results in a
dramatic color change of the indicator. This is called the END
POINT of the titration.
Based on the volume of NaOH that must be added to reach
the end point we can determine the concentration of the HCl.
In an acid-base titration 45.7 mL of 0.500M H2SO4 is required
to neutralize 20.0mL of a NaOH solution. Determine the
concentration of the NaOH solution
We know that at the equivalence point of the titration enough
H2SO4 has been added so that all the NaOH has reacted.
This means that the number of moles in 45.7 mL of 0.500M
H2SO4 is enough to react with all the NaOH.
We do need to know the stoichiometry of the reaction
between H2SO4 and NaOH.
H2SO4 + 2 NaOH --> Na2SO4 + 2 H2O
This means that 1 mole of H2SO4 reacts with 2 moles of
NaOH.
Moles of H2SO4 added = 0.0457 L x 0.500M = 0.00228 mol
Hence the number of moles of NaOH = 2 x 0.00228
= 0.00456 mol
Hence, molarity of NaOH solution
0.00456 mol NaOH
0.0200 L NaOH
= 2.28 M
The quantity of Cl- in a water supply is determined by
titrating the sample with Ag+.
a) How many grams of chloride ion are in a sample of the
water, if 20.2 mL of 0.100M Ag+ is required to react with all
the chloride ions in the sample?
b) If the sample has a mass of 10.0g,what percent chloride
does it contain?
Ag+(aq) + Cl-(aq) --> AgCl(s)
Moles of Ag+ added = 0.100 M x 0.00202 L = 2.02 x 10-3 mol
Hence, moles of Cl- in sample = 2.02 x 10-3
grams of Cl- in sample = 2.02 x 10-3 mol x 35.5g/mol Cl= 7.17 x 10-2 g Clb) % Cl- in sample = 7.17 x 10-2 g Cl10.0 g solution
x 100% = 0.717 % Cl-