titrations :experiment 3 - university of kwazulu-natalcheminnerweb.ukzn.ac.za/files/chem 110...
TRANSCRIPT
Aqueous
Reactions
• HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
• From the balanced equation the reacting
ratio of HCl to NaOH is 1 : 1
• At the endpoint:
• no. of moles of HCl = no. of moles of NaOH
• i.e MHCl x VHCl = MNaOH x VNaOH
• MNaOH = MHCl x VHCl / VNaOH
MHCl= given value VHCl = 0.025 dm3
VNaOH ( average vol. from titration)
MNaOH = ----A--- mol dm-3
Conc. of NaOH in g dm-3 = A mol dm-3 × molar mass of
NaOH( in g mol-1)3
Aqueous
Reactions
Aqueous Reactions and
Solution Stoichiometry
Chemistry, The Central Science, 2nd edition
Brown; LeMay; Bursten; Murphy; Langford
and Sagatys
4
Aqueous
Reactions
Solutions:
• Homogeneous
mixtures of two or
more pure
substances.
• The solvent is
present in greatest
abundance.
• All other substances
are solutes.
5
Aqueous
Reactions
Dissociation• Ionic compounds are formed between
metals and non-metals.
• Ionic compound separates into ions
e.g. NaCl Na+ + Cl-
6
Aqueous
Reactions
Electrolytes
• Substances that
dissociate into ions
when dissolved in
water are called
electrolytes.
• A nonelectrolyte may
dissolve in water, but
it does not dissociate
into ions when it does
so.7
Aqueous
Reactions
Electrolytes
• A strong electrolyte
dissociates completely
when dissolved in
water.
• A weak electrolyte
only dissociates
partially when
dissolved in water.
8
Aqueous
Reactions
• Strong electrolytes completely ionise in
solution
– Strong acids
– Strong bases
– Soluble salts
Non-electrolyte weak electrolyte strong
electrolyte
9
Aqueous
Reactions
Acids:
• Substances that
increase the
concentration of H+
when dissolved in
water (Arrhenius).
• Proton donors
(Brønsted–Lowry).
13
Aqueous
Reactions
Acids
There are only seven
strong acids:
• Hydrochloric (HCl)
• Hydrobromic (HBr)
• Hydroiodic (HI)
• Nitric (HNO3)
• Sulfuric (H2SO4)
• Chloric (HClO3)
• Perchloric (HClO4)
14
Aqueous
Reactions
Bases:
• Substances that
increase the
concentration of
OH− when dissolved
in water (Arrhenius).
• Proton acceptors
(Brønsted–Lowry).
15
Aqueous
Reactions
Bases
The strong bases
are the soluble salts
of hydroxide ion:
• Alkali metals
• Calcium
• Strontium
• Barium
16
Aqueous
Reactions
Rules for Recognizing Acids
(1)Compounds that do not contain C and have
an H written in front of their formulas. e.g.
H2S. (*****Refer to page 43 in textbook)
(2) Compounds that COOH in their formulas
e.g. CH3COOH is acetic acid.
(3) A positive ion that contains a N atom and
has a H bound to the N is frequently an acid.
NH4Cl is actually two ions, NH4+ and Cl–. The
ammonium ion is an acid.L.Pillay, 2009
17
Aqueous
Reactions
(1) An ionic compound that contains a
hydroxide (OH–) ion. e.g. NaOH, Mg(OH)2
(2) A neutral compound that contains a N atom.
An H+ can bond to the lone pair of electrons,
forming , so the lone pair on the N allows
the molecule to accept an H+ (which is the
definition of base). **e.g. NH3
(3) An ionic compound that does not contain
the following: Cl–, Br–, I–, NO3–, SO4
2–,ClO4–,
or H+. **e.g. Na2CO3
Rules for Recognizing a Base
L.Pillay, 2009
18
Aqueous
Reactions
Solubility Rules
1. All common compounds of Group I, Group 2
and ammonium ions are soluble.
2. All nitrates, acetates, and perchlorates are
soluble.
3. Salts of Ag, Hg(I), and Pb are insoluble. (Pb
halides are soluble in hot water.)
4. Chlorides, bromides and iodides are soluble
5. All sulfates are soluble, except those of
barium, strontium.
6. Except for rule 1, carbonates, hydroxides,
sulphides, oxides, silicates, and phosphates
are insoluble. 19
Aqueous
Reactions
Solubility - Example
Ag+ Cu2+ Mg2+
S2-
Cl-
OH-
ppt
ppt
ppt
ppt
sol
ppt
sol
sol
ppt
Salts of Ag+ are insoluble! Rule 3
Sulphides
insoluble
Rule 6Chlorides
soluble *
Rule 4
OH insoluble *
Rule 6
Group II
L.Pillay, 2009
20
Aqueous
Reactions
Solubility Rules
Will a precipitate form when aqueous
solutions of Ca(NO3)2 and NaCl are
mixed?
– Ca(NO3)2 and NaCl are both soluble
– Combinations could be CaCl2 and NaNO3
– CaCl2 would be soluble (Rule **4--)
– NaNO3 would be soluble (Rule **2--)
L.Pillay, 2009
21
Aqueous
Reactions
SAMPLE EXERCISE 1 Using Solubility
Rules
• Classify the following ionic compounds
as soluble or insoluble in water:
• (a) sodium carbonate (Na2CO3),
• (b) lead sulfate (PbSO4).
22
Aqueous
Reactions
• (a)Most carbonates are insoluble, but
carbonates of the alkali metal cations
(such as sodium ion) are an exception
to this rule and are soluble. Thus,
Na2CO3 is soluble in water.
• (b) Although most sulfates are water
soluble, the sulfate of Pb2+ is an
exception. Thus, PbSO4 is insoluble in
water.
23
Aqueous
Reactions
SAMPLE EXERCISE 2
• a) Predict the identity of the precipitate
that forms when solutions of BaCl2 and
K2SO4 are mixed.
• (b) Write the balanced chemical
equation for the reaction.
24
Aqueous
Reactions
(a) Most compounds of SO42– are soluble
but those of Ba2+ are not.
Thus, BaSO4 is insoluble and will
precipitate from solution.
KCl, on the other hand, is soluble.
25
Aqueous
Reactions
• b) From part (a) we know the chemical
formulas of the products, BaSO4 and
KCl.
• The balanced equation with phase
labels shown is
26
Aqueous
Reactions
Types of reactions
• Acid – base reactions
(neutralisation reaction)
• Precipitation reactions
• Redox reactions
(Oxidation –reduction reactions)
• Disproportionation reactions
L.Pillay, 2009
27
Aqueous
Reactions
Solution Chemistry
• It is helpful to pay attention to exactly
what species are present in a reaction
mixture (i.e., solid, liquid, gas, aqueous
solution).
• If we are to understand reactivity, we
must be aware of just what is changing
during the course of a reaction.
28
Aqueous
Reactions
Writing Net Ionic Equations
1.Write a balanced molecular equation.
2.Dissociate all strong electrolytes.
3.Cross out anything that remains
unchanged from the left side to the right
side of the equation, i.e. identify and
cancel spectator ions.
4.Write the net ionic equation with the
species that remain.29
Aqueous
Reactions
Molecular Equation
The molecular equation lists the reactants
and products in molecular form.
Unbalanced equation
Pb(NO3)2 (aq) + KI (aq) PbI2 (s) + 2KNO3 (aq)
Balanced equation
Pb(NO3)2 (aq) + 2KI (aq) PbI2 (s) + 2KNO3 (aq)
30
Aqueous
Reactions
Ionic Equation
• In the ionic equation all strong electrolytes
(strong acids, strong bases, and soluble
ionic salts) are dissociated into their ions.
• This more accurately reflects the species
that are found in the reaction mixture.
Balanced Molecular equation:
Pb(NO3)2 (aq) + 2KI (aq) PbI2 (s) + 2KNO3 (aq)
Balanced Ionic equation:
Pb2+ (aq) + 2NO3- (aq) + 2K+ (aq) + 2I
-(aq)
PbI2 (s) + 2K+ (aq) + 2NO3-(aq) 31
Aqueous
Reactions
Net Ionic Equation
• To form the net ionic equation, cross out
anything that does not change from the
left side of the equation to the right: i.e.
cross out the spectator ions
Pb2+ (aq) + 2NO3- (aq) + 2K+ (aq) + 2I
-(aq)
PbI2 (s) + 2K+ (aq) + 2NO3
-(aq)
32
Aqueous
Reactions
Net Ionic Equation
• The only things left in the equation are
those things that change (i.e. react)
during the course of the reaction.
Pb2+(aq) + 2I-(aq) PbI2 (s)
33
Aqueous
Reactions
Net ionic equations
• Total molecular equation:
2AgNO3(aq) + Cu(s) 2Ag(s) + Cu(NO3)2 (aq)
• Total ionic equation:
2Ag+(aq) + 2NO3
-(aq) + Cu(s) 2Ag(s) + Cu
2+(aq) +
2NO3-(aq)
• Net Ionic equation: (remove spectator ions)
2Ag+(aq) + Cu(s) 2Ag(s) + Cu
2+
L.Pillay, 2009
34
Aqueous
Reactions
Net ionic equations -
exampleCaCl2 + Na2CO3 ?
1.Determine products
CaCl2 + Na2CO3 CaCO3 + NaCl
2.Balance equations and predict solubility
CaCl2(aq) + Na2CO3(aq) CaCO3(s) + 2NaCl(aq)
L.Pillay, 2009
35
Aqueous
Reactions
3. Write total ionic equations
Ca2+(aq) + 2Cl-(aq) + 2Na+(aq) + CO32-(aq)
CaCO3(s) + 2Na+(aq) + 2Cl-(aq)
4. Write net ionic equation
Ca2+ (aq) + CO32-(aq) CaCO3(s)
36
Aqueous
Reactions
Acid-Base (neutralisation)
reactions
• acid + base → salt + water
• Driving force is the reaction of H+ and OH- to
produce H2O
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)L.Pillay, 2009
37
Aqueous
Reactions
Neutralization Reactions
Consider the reaction between a strong acid, HCl and a strong
base, NaOH:
Total molecular equation:
HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l)
Total ionic equation:
H+ (aq) + Cl- (aq) + Na+ (aq) + OH-(aq)
Na+ (aq) + Cl- (aq) + H2O (l)
38
Aqueous
Reactions
Neutralization Reactions
Cancel common reactants and products
H+ (aq) + Cl- (aq) + Na+ (aq) + OH- (aq)
Na+ (aq) + Cl- (aq) + H2O (l)
Net Ionic equation:
H+ (aq) + OH-(aq) H2O (l)
39
Aqueous
Reactions
Gas-Forming ReactionsThe reaction below does not give the expected product,
H2CO3:
NaHCO3(aq) + HCl(aq) NaCl(aq) + H2CO3(aq)
Instead the expected product decomposes to give a
gaseous product (CO2):
**H2CO3 (aq) CO2 (g) + H2O (l)
Overall:
NaHCO3(aq) + HCl(aq) NaCl(aq) + CO2(g) + H2O (l)
Write the total ionic equation here!!
And the net ionic equation is:
HCO3- (aq) + H+ (aq) CO2 (g) + H2O (l)
40
Aqueous
Reactions
Gas-Forming Reactions
• This reaction gives the predicted product, but you had better carry it out in the fumehood, or you will be very unpopular!
• Just as in the previous examples, a gas is formed as a product of this reaction:
Molecular equation is:
**Na2S (aq) + 2HCl (aq) 2NaCl (aq) + H2S (g)
Write the total ionic equation here!!
The net ionic equation is:
S2- (aq) + 2H+ (aq) H2S (g)
41
Aqueous
Reactions
Displacement reactions
• Activity Series:
• Any free metal higher on list will displace
another metal lower on series.
e.g. Cu(s) + FeSO4(aq)
Zn(s) + 2HCl(aq)
Li
K
Ca
Na
Mg
Al
Mn
Zn
Cr
Fe
Cd
Co
Ni
Sn
Pb
H
Cu42
Aqueous
Reactions
Precipitation Reactions
When one mixes ions
that form compounds
that are insoluble (as
could be predicted by
the solubility
guidelines), a
precipitate is formed.
43
Aqueous
Reactions
Precipitation reactions
• Step #1: Determine the possible products using
the general double displacement equation.
AB + CD AD + CB
• Step #2: Predict whether either of the possible
products is water insoluble.
• Step #3: Write the complete balanced equation.
• Step #4: Write the net ionic equation.
L.Pillay, 2009
44
Aqueous
Reactions
AgNO3(aq) + Na2S(aq) ?
1. The possible products from the mixture of
AgNO3(aq) and Na2S(aq) are Ag2S and NaNO3.
(Remember to consider charge for the possible
products.)
AgNO3(aq) + Na2S(aq) Ag2S + NaNO3
2. According to solubility rules, most sulfides are
insoluble, therefore, Ag2S would be insoluble.
Compounds containing Na+ and NO3- are soluble,
NaNO3 is soluble.
AgNO3(aq) + Na2S(aq) Ag2S(s) + NaNO3(aq)L.Pillay, 2009
45
Aqueous
Reactions
3. Balance the equation:
2AgNO3(aq) + Na2S(aq) Ag2S(s) + 2NaNO3(aq)
4.Write the complete ionic equation, describing the aqueous ionic compounds, AgNO3(aq), Na2S(aq) and NaNO3(aq), as ions.
Describe the solid with a complete formula.
2Ag+(aq) + 2NO3-(aq) + 2Na+(aq) + S2-(aq)
Ag2S(s) + 2Na+(aq) + 2NO3-(aq)
L.Pillay, 2009
46
Aqueous
Reactions
• The nitrate and sodium ions have the same
form on each side of the equation, so they are
eliminated as spectator ions.
2Ag+(aq) + S2-(aq) Ag2S(s)
Ex. Pb(NO3)2 (aq) + K2CrO4(aq) ?
L.Pillay, 2009
47
Aqueous
Reactions
Metathesis (Exchange)
Reactions
• Metathesis comes from a Greek word
that means “to transpose”
• It appears the ions in the reactant
compounds exchange, or transpose,
ions
AgNO3 (aq) + KCl (aq) AgCl (s) + KNO3 (aq)
48
Aqueous
Reactions
Oxidation-Reduction Reactions
• An oxidation occurs
when an atom or ion
loses electrons.
• A reduction occurs
when an atom or ion
gains electrons.
49
Aqueous
Reactions
Oxidation Numbers
To determine if an oxidation-reduction
reaction has occurred, we assign an
oxidation number to each element in a
neutral compound or charged entity.
50
Aqueous
Reactions
Oxidation Numbers
• Elements in their elemental form have
an oxidation number of 0.
• The oxidation number of a monatomic
ion is the same as its charge.
51
Aqueous
Reactions
Oxidation Numbers
In order to keep
track of what loses
electrons and what
gains them, we
assign oxidation
numbers.
52
Aqueous
Reactions
Oxidation and Reduction
• Oxidation ( loss of electrons)
• A species is oxidized when it loses electrons.
– Here, zinc loses two electrons to go from neutral
zinc metal to the Zn2+ ion.
53
Aqueous
Reactions
Oxidation and Reduction
• Reduction (gain in electrons)
• A species is reduced when it gains electrons.
– Here, each of the H+ gains an electron and they
combine to form H2.
54
Aqueous
Reactions
Oxidation and Reduction
• What is reduced is the oxidizing agent.
– H+ oxidizes Zn by taking electrons from it.
• What is oxidized is the reducing agent.
– Zn reduces H+ by giving it electrons.
55
Aqueous
Reactions
Assigning Oxidation Numbers
1. Elements in their elemental form have
an oxidation number of 0.
2. The oxidation number of a monatomic
ion is the same as its charge.
56
Aqueous
Reactions
Assigning Oxidation Numbers
3. Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions.
– Oxygen has an oxidation number of −2, except in the peroxide ion in which it has an oxidation number of −1.
**(Refer to page 39 in textbook)
– Hydrogen is −1 when bonded to a metal, +1 when bonded to a nonmetal,e.g. NaH*
57
Aqueous
Reactions
Assigning Oxidation Numbers
– Fluorine always has an oxidation number of −1.
– The other halogens have an oxidation number of −1 when they are negative; they can have positive oxidation numbers, however, most notably in oxyanions.****e.g. ClO4
-
58
Aqueous
Reactions
Assigning Oxidation Numbers
4. The sum of the oxidation numbers in a
neutral compound is 0.
5. The sum of the oxidation numbers in a
polyatomic ion is the charge on the
ion.
59
Aqueous
Reactions
Group1
Group18
H+1-1
Group 2Group
13Group
14Group
15Group
16Group
17
He
Li+1
Be+2
B+3
C+4, +2
-1-4
NAll from +5 to -3
O-1-2
F-1
Ne
Na+1
Mg+2
Group3
Group4
Group5
Group6
Group7
Group8
Group9
Group10
Group11
Group12
Al+3
Si+4-4
P+5+3-3
S+6+4
+2, -2
Cl+7, +5
+3+1, -1
Ar
K+1
Ca+2
Sc+3
Ti+4+3+2
V+5+4
+3, +2
Cr+6+3+2
Mn+7, +6
+4+3, +2
Fe+3+2
Co+3+2
Ni+2
Cu+2+1
Zn+2
Ga+3
Ge+4+2-4
As+5+3-3
Se+6+4-2
Br+7, +5
+3+1, -1
Kr
Rb+1
Sr+2
Y+3
Zr+4+3
Nb+5+4+2
Mo+6+5
+4, +3
Tc+7+5+4
Ru+8, +5
+4+3
Rh+4+3
Pd+4+2
Ag+1
Cd+2
In+3
Sn+4+2-4
Sb+5+3-3
Te+6+4-2
I+7, +5
+3+1, -1
Xe
Cs+1
Ba+2
Lu+3
Hf+4+3
Ta+5+4+3
W+6+5+4
Re+7+5+4
Os+8, +6
+4+3, +2
Ir+4+2
Pt+4+2
Au+3+1
Hg+2+1
Tl+3+1
Pb+4+2
Bi+5+3
+6+4-2
At+7, +5
+3+1, -1
Rn
Fr+1
Ra+2
Lr+3
The Common Oxidation Numbers of the Elements
60
Aqueous
Reactions
SAMPLE EXERCISE 4.8 Determining
Oxidation Numbers
• Determine the oxidation number of
sulfur in each of the following:
• (a) H2S,
• (b) S8 ,
• (c) SCl2,
• (d) Na2SO3,
• (e) SO42–.
61
Aqueous
Reactions
Oxidation Numbers
• Nonmetals tend to have negative
oxidation numbers, although some are
positive in certain compounds or ions.
Oxygen has an oxidation number of −2,
except in the peroxide ion in which it has
an oxidation number of −1.
Hydrogen is −1 when bonded to a metal,
+1 when bonded to a nonmetal.
62
Aqueous
Reactions
• a) When bonded to a nonmetal,
hydrogen has an oxidation number of
+1 . Because the H2S molecule is
neutral, the sum of the oxidation
numbers must equal zero.
• Letting x equal the oxidation number of
S, we have 2(+1) + x = 0.
• Thus, S has an oxidation number of –2.
63
Aqueous
Reactions
Oxidation-reduction (redox)
reactionsBreathalyser Test
3CH3CH2OH + 2K2Cr2O7 + 8H2SO4 →
3CH3COOH + 2Cr2(SO4)3 + K2SO4 + 11 H2O
• Ethanol → acetic acid (oxidised)
• Dichromate → chromic ion
Orange → Green
66
Aqueous
Reactions
Balancing Oxidation-Reduction
Equations
Perhaps the easiest way to balance the
equation of an oxidation-reduction
reaction is via the half-reaction method.
67
Aqueous
Reactions
Balancing Oxidation-Reduction
Equations
This involves treating (on paper only) the
oxidation and reduction as two separate
processes, balancing these half reactions,
and then combining them to attain the
balanced equation for the overall reaction.
68
Aqueous
Reactions
Half-Reaction Method- acid
medium
1. Assign oxidation numbers to
determine what is oxidized and what is
reduced.
2. Write the oxidation and reduction half-
reactions.
69
Aqueous
Reactions
Half-Reaction Method
3. Balance each half-reaction.
a. Balance elements other than H and O.
b. Balance O by adding H2O.
c. Balance H by adding H+.
d. Balance charge by adding electrons.
4. Multiply the half-reactions by integers so that the electrons gained and lost are the same.
70
Aqueous
Reactions
Half-Reaction Method
5. Add the half-reactions. **Cancel what is common to both sides.
6. Make sure the equation is balanced according to mass.
7. Make sure the equation is balanced according to charge.
71
Aqueous
Reactions
Half-Reaction Method
Consider the reaction between MnO4− and C2O4
2− :
MnO4−(aq) + C2O4
2−(aq) Mn2+(aq) + CO2(aq)
72
Aqueous
Reactions
Half-Reaction Method
First, we assign oxidation numbers.
MnO4− + C2O4
2- Mn2+ + CO2
+7 +3 +4+2
Since the manganese goes from +7 to +2, it is reduced
( decrease in oxidation number).Since the carbon goes from +3 to +4, it is oxidized
(increase in oxidation number).73
Aqueous
Reactions
Oxidation Half-Reaction
C2O42− CO2
To balance the carbon, we add a
coefficient of 2:
C2O42− 2 CO2
74
Aqueous
Reactions
Oxidation Half-Reaction
C2O42− 2 CO2
The oxygen is now balanced as well.
To balance the charge, we must add 2
electrons to the right side.
C2O42− 2 CO2 + 2 e−
75
Aqueous
Reactions
Reduction Half-Reaction
MnO4− Mn2+
The manganese is balanced; to balance
the oxygen, we must add 4 waters to
the right side.
MnO4− Mn2+ + 4 H2O
76
Aqueous
Reactions
Reduction Half-Reaction
MnO4− Mn2+ + 4 H2O
To balance the hydrogen, we add 8 H+
to the left side.
8 H+ + MnO4− Mn2+ + 4 H2O
77
Aqueous
Reactions
Reduction Half-Reaction
8 H+ + MnO4− Mn2+ + 4 H2O
To balance the charge, we add 5 e− to
the left side.
5 e− + 8 H+ + MnO4− Mn2+ + 4 H2O
78
Aqueous
Reactions
Combining the Half-Reactions
Now we evaluate the two half-reactions
together:
C2O42− 2 CO2 + 2 e−
5 e− + 8 H+ + MnO4− Mn2+ + 4 H2O
To attain the same number of electrons
on each side, we will multiply the first
reaction by 5 and the second by 2.79
Aqueous
Reactions
Combining the Half-Reactions
5 C2O42− 10 CO2 + 10 e−
10 e− + 16 H+ + 2 MnO4− 2 Mn2+ + 8 H2O
When we add these together, we get:
10 e− + 16 H+ + 2 MnO4− + 5 C2O4
2−
2 Mn2+ + 8 H2O + 10 CO2 +10 e−
80
Aqueous
Reactions
Combining the Half-Reactions
10 e− + 16 H+ + 2 MnO4− + 5 C2O4
2−
2 Mn2+ + 8 H2O + 10 CO2 +10 e−
The only thing that appears on both sides are the electrons. Subtracting them, we are left with:
16 H+ + 2 MnO4− + 5 C2O4
2−
2 Mn2+ + 8 H2O + 10 CO2
81
Aqueous
Reactions
Balancing in Basic Solution
• If a reaction occurs in basic solution, one
can balance it as if it occurred in acid.
• Once the equation is balanced, add OH−
to each side to “neutralize” the H+ in the
equation and create water in its place.
• If this produces water on both sides, you
might have to subtract water from each
side.82
Aqueous
Reactions
eg. MnO4- + H2C2O4 +... → Mn2+ + CO2 +...
1.Oxidation Numbers
2.Half-reactions
MnO4- → Mn2+
H2C2O4 → CO2
3. Balance Equations
Balance all elements except hydrogen and
oxygen by proper multiplication of
compounds or ions.
MnO4- → Mn2+
H2C2O4 → 2CO2
L.Pillay, 2009
83
Aqueous
Reactions
Balance oxygen by adding water, H2O, as
necessary.
MnO4- → Mn2+ + 4H2O
H2C2O4 → 2CO2
Balance hydrogen by adding hydrogen ions,
H+, as necessary.
8H+ + MnO4- → Mn2+ + 4H2O
H2C2O4 → 2CO2 + 2H+
L.Pillay, 2009
84
Aqueous
Reactions
4 & 5. Balance charge by adding electrons &
combine
(8H+ + MnO4- + 5 e-) x 2 → Mn2+ + 4H2O
(H2C2O4 → 2CO2 + 2H+ + 2e-) x 5
Multiplying and combining the two equations
16H+ + 2MnO4- + 5H2C2O4 + 10e- → 2Mn2+ +
8H2O + 10CO2 + 10H+ + 10e-
After cancellation
6H+ + 2MnO4- + 5H2C2O4 → 2Mn2+ + 8H2O +
10CO285
Aqueous
Reactions
6H+ + 2MnO4- + 5H2C2O4 → 2Mn2+ + 8H2O +
10CO2
6. Add number of OH- ion equal to number of H+
ions on both sides of overall reaction
6H+ + 2MnO4- + 5H2C2O4 + 6OH- →
2Mn2+ + 8H2O + 10CO2 + 6OH-
7 & 8. Combine hydrogen ions (protons) and
hydroxide ions to form water when they appear on
the same side of the equation
2MnO4- + 5H2C2O4 --> 2Mn2+ + 2H2O + 10CO2 + 6OH-
L.Pillay, 2009
86
Aqueous
Reactions
Concentration of
Solutions
The concentration of a solution is the amount
of solute present in a given quantity of
solvent or solution.
Concentration can be expressed in various
ways; most commonly used is molarity.
Other ways of expressing units of
concentration are: parts per million and mass
(weight) percent. 88
Aqueous
Reactions
Molarity• Two solutions can contain the same compounds but
be quite different because the proportions of those
compounds are different eg. 3M and 6M HCl.
• Molarity is one way to measure the concentration of a
solution.
• Molarity (M) or molar concentration is the
number of moles of solute in 1 litre (L) of
solution.moles of solute (n)
volume of solution in liters (V)Molarity (M) =
89
Aqueous
Reactions
• Percent Concentration
• Can be expressed in several ways. Three
common methods are:
1. mass percent (%m/m) =mass solute(g) x 100%
mass solution(g)
2. volume percent (%v/v) =volume solute x 100%
volume solution
3. mass/volume percent (%m/v)= mass solute, g x 100
volume solution, mL
91
Aqueous
Reactions
Mixing a Solution
• To create a solution of a
known molarity, one
weighs out a known mass
(and, therefore, number of
moles) of the solute.
• The solute is added to a
volumetric flask, and
solvent is added to the line
on the neck of the flask.
92
Aqueous
Reactions
Example
How many grams of potassium dichromate(K2Cr2O7) are required to prepare a 250 mLsolution whose concentration is 2.16 M?
n = M x v= 2.16 mol /1L x 0.25 L
= 0.540 mol
mass = n x molar mass= 0.540 mol x 294.2 g /1mol
= 159 g
93
Aqueous
Reactions
Dilution
• One can also dilute a more concentrated solution by– Using a pipet to deliver a volume of the solution to
a new volumetric flask, and
– Adding solvent to the line on the neck of the new flask.
94
Aqueous
Reactions
Dilution
The molarity of the new solution can be determined from the equation
Mconc Vconc = Mdil Vdil, (for dilution only!)
where Mconc and Mdil are the molarity of the concentrated and dilute solutions, respectively, and Vconc and Vdil are the volumes of the two solutions.
96
Aqueous
Reactions
What is the new concentration of the
solution?
MconcVconc = MdilVdil
(0.020 M)(0.500 dm3) = (Md)(0.1000 dm3)
Md = 0.010 M
99
Aqueous
Reactions
Dilution of Solutions
Dilution is a procedure for preparing a less concentrated
solution from a more concentrated one.
moles of solute before dilution = moles of solute after dilution
Example
Describe how you would prepare 5.00 x 102 mL ofa 1.75 M H2SO4 solution starting with an 8.61 Mstock solution of H2SO4.
**Mconc = 8.61 M , Vconc = ?Mdil = 1.75 M , Vdil = 5.00 x 102 mL
100
Aqueous
Reactions
Vconc =Mdil Vdil
Mconc=
1.75 M x 500 mL8.61 M
= 102 mL
We must dilute 102 mL of 8.61 M H2SO4 withsufficient water to give a final volume of 5.00 x102 mL in a 500 mL volumetric flask to obtainthe desired concentration.
101
Aqueous
Reactions
Titration
Titration is an
analytical
technique in
which one can
calculate the
concentration
of a solute in
a solution.
102
Aqueous
Reactions
THE STOICHIOMETRY OF
REACTIONS IN AQUEOUS
SOLUTION: TITRATIONS
• The reaction of MnO4- with Sn2+ in
acidic solution follows the equation:2MnO4
- + 5Sn2+ + 16H+ 2Mn2+ + 5Sn4+ + 8H2O
•
• (a) How many mL of 0.02000 mol dm-3
KMnO4 solution would be needed to
react with all the tin in 25.00 mL of
0.05000 mol dm-3 SnCl2 solution?
• 104
Aqueous
Reactions
• From the balanced equation the reacting
ratio of MnO4- to Sn2+ is 2 : 5
• i.e moles of MnO4- / moles of Sn2+ = 2/5
• i.e MMnO4- x VMnO4
- = 2/5 MSn2+ x VSn2+
• i.e. VMnO4- = 2/5 MSn2+ x V Sn2+ / MMnO4
- +
= 2/5 x 0.05000 mol dm-3 x 0.02500 dm3 / 0.02000
mol dm-3
= 0.02500 dm3
= 25.00 mL
105
Aqueous
Reactions
• (b) A 0.5500 g sample of solder, is an alloy
containing lead and tin, was dissolved in
acid and all the tin was converted to Sn2+.
The solution was then titrated with 0.02000
mol dm-3 KMnO4 solution. The titration
required 27.73 mL of the KMnO4 solution.
Calculate the percent by mass of tin in the
solder.
106
Aqueous
Reactions
• From the balanced equation2MnO4
- + 5Sn2+ + 16H+ 2Mn2+ + 5Sn4+ + 8H2O
• the reacting ratio: mol of Sn2+/ mol
of MnO4- = 5 / 2
• mol of Sn2+= 5/2 x mol of MnO4-
•
= 5/2 x MMnO4- x VMnO4-
• = 5/2 x 0.02000 mol dm-3 x 0.02773 dm3
•
• = 0.001387 mol of Sn
•
107
Aqueous
Reactions
• Mass of Sn = mol of Sn x molar mass
• = 0.001387 mol x 118.7 g mol-1
• = 0.1646 g
•% Sn in Solder = mass of Sn / mass of solder x 100
• = 0.1646 g / 0.5500 g x 100
• = 29.93 %
108
Aqueous
Reactions
• Example
• What mass of Ag2CO3 (275.7g/mol) is formed
when 25.00mL of 0.200M AgNO3 are mixed
with 50.00mL of 0.0800M Na2CO3?
• Solution:
• Na2CO3(aq) + 2AgNO3(aq) Ag2CO3(s) + 2NaNO3(aq)
•
• Mixing these 2 solutions will result in one of three
possible outcomes:
• (1) an excess of AgNO3 will remain after the reaction
is complete.
• (2) an excess of Na2CO3 will remain after reaction is
complete.
• (3) neither reagent will be in excess.
•
109
Aqueous
Reactions
• Determine amount of Ag2CO3 produced from each
reactant:
• From the stoichiometric relationship
• n Na2CO3 = n Ag2CO3
• 1 1
• n Ag2CO3 = 1 x nNa2CO3
• i.e. n Ag2CO3 = 1 x 0.0800 mol dm-3 x 0.050 dm3
•
• = 0.004 mol
•110
Aqueous
Reactions
• n AgNO3 = n Ag2CO3
2 1
• n Ag2CO3 = 0.200 mol dm-3 x 0.025dm3
2
= 0.005 mol AgNO3 x 1
• 2
= 0.0025 mol 111
Aqueous
Reactions
• Compare the amounts of Ag2CO3 produced form
each reactant:
• From: Na2CO3 0.004 mol
• AgNO3 0.0025 mol
• AgNO3 - the limiting reagent
• The amount of Ag2CO3 produced will be limited
by the amount of AgNO3 available.
• mass Ag2CO3 = 0.0025 mol x 275.7 g/mol
• = 0.689g Ag2CO3112
Aqueous
Reactions
• Initial amounts are
• Amount AgNO3, nAgNO3 = 0.200 M x 0.025dm3
• = 0.005mol AgNO3
• amount Na2CO3, nNa2CO3 = 0.0800 M x 0.050 dm3
• = 0.004 mol Na2CO3
•
• From the stoichiometric relationship
• n Na2CO3 = nAgNO3
• 1 2
• n AgNO3 = 2 x nNa2CO3
• i.e. n AgNO3 = 2 x 0.004 mol
• = 0.008 mol
= the amount of AgNO3 that would
be required to react with Na2CO3.
•
113
Aqueous
Reactions
• From the stoichiometric relationship
• n Na2CO3 = nAgNO3
• 1 2
• n AgNO3 = 2 x nNa2CO3
• i.e.n AgNO3 = 2 x 0.004 mol
• = 0.008 mol
= the amount of AgNO3
that would be required to react with
Na2CO3.114