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CHEM 110
Chapter 4
Aqueous Reactions and
Solution Stoichiometry
Dr V Paideya
2015

General Properties of Aqueous
Solutions
• Solutions are mixtures of two
or more pure substances
...

• All other substances are in
the solution are solutes
...


Dissociation

• Ionic compounds are formed between
metals and non-metals
...
g
...

• A non-electrolyte
a covalent compound may
dissolve in water, but does not
dissociate into component ions
eg
...


Electrolytes
• A strong electrolyte dissociates completely
when dissolved in water
...

CH3COOH
CH3CO2−(aq) + H+(aq)

Electrolytes
• Substances that dissociate into ions when
dissolved in water, eg
...
C12H22O11

Exercise 1
What are the calcium and nitrate
concentrations in a 0
...
238 M = 0
...
238 M = 0
...
g
...

a)

BaCl2

b)

MgCO3

c)

PbS

d)

Hg2Cl2

e)

Cu3(PO4)2

Metathesis (exchange)
Reaction
• Reactions in which positive and negative ions
appear to exchange partners also called a double
displacement reaction
...

2K+(aq) + 2I-(aq) + Pb2+(aq) + 2NO3-(aq) 
PbI2(s) + 2K+(aq) + 2NO3-(aq)

Net ionic equation – shows only those ions that are taking
part in the reaction:

Pb2+(aq) + I−(aq)  PbI2(s)

Spectator ions – ions that don’t take part in the reaction,
K+(aq) & NO3-(aq)

12

Exercise 3
Predict whether a reaction will occur in each
of the following cases
...


a)

AlCl3(aq) + KOH(aq) 

b)

K2SO4(aq) + FeBr3(aq) 

c)

CaI2(aq) + Pb(NO3)2(aq) 

13

Exercise 3 - Answer
Solution for (a):
AlCl3(aq) + KOH(aq) 
Step 1: Predict the products
...

AlCl3(aq) + 3 KOH(aq)  Al(OH)3 + 3 KCl
Step 3: Looking at the solubility rules, decide whether NaCl
& MgCl2 are soluble or insoluble
...

Al3+(aq) + 3 OH–(aq)  Al(OH)3(s)

14

Acid-Base Reactions
ACIDS
Arrhenius: Substances that contain an ionisable H and able to
ionise in aqueous solution to form H+ or H3O+ ions
...

Strong acids - ionise completely in solution
- usually strong electrolytes
Weak acids – partially ionized
indicates reversible reaction

CH3COOH(aq) + H2O(l)

CH3CO2–(aq) + H3O+(aq)
15

Acid-Base Reactions
BASES
Arrhenius: Compounds that contain an OH group and are able
to ionise or dissociate in aqueous solution to form OHStrong bases - ionise completely in solution
- strong electrolytes
NaOH(aq)  Na+(aq) + OH− (aq)

Weak bases – partially ionized - reversible reaction
- weak electrolytes
NH3(aq) + H2O(l)
NH4OH(aq)
NH4+(aq) + OH–(aq)
16

Neutralisation Reaction
Neutralisation Reaction – chemical reaction between a
solution of an acid and a base forming a salt and water

ACID + BASE  SALT + WATER
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)

From the net ionic equation:
H+(aq) + OH-(aq)  H2O(l)

17

More Acid-Base
Neutralisation Reactions
Insoluble Bases – can be dissolved by acid-base reaction

Mg(OH)2(s) + 2 H+(aq)  Mg2+(aq) + 2 H2O(l)
Gas Formation – products of an acid-base reaction may be a
gas, whose formation assists to drive the reaction to
completion (e
...
reactions with sulfide ion
s and carbonate
ions)
...
e
...

Oxidation reaction – Ca undergoes an increase in O
...
due
to loss of electrons
...
S
...

2Ca(s) + O2(g) → 2CaO(s)
20

Oxidation State Changes
Assign oxidation states to all the elements in the compounds
that are involved in the reaction below:
0

0

Ca2+ and O2-

2Ca(s) + O2(g)  2 CaO(s)
Ca loses 2 electrons

O gains 2 electrons

21

Rules for Assigning Oxidation Numbers
1
...

e
...
Na = 0 and Cl2 = 0
2
...

e
...
Na+ = +1 and Cl- = -1
3
...
of oxygen is usually -2 except in the peroxide
ion in which it has an oxidation no
...


b) Oxidation no
...

c) Oxidation no
...
Other
halogens have an oxidation no
...
22

Rules for Assigning Oxidation Numbers
4
...

e
...
Pb(NO3)2, KCl, H2SO4

5
...

e
...
NH+, NO3-, Cr2O72-

23

Oxidising & Reducing Agents
Oxidising Agent or Oxidant:

• Causes the oxidation of another substance;
• Gains electrons, i
...
they are reduced;
• Examples: MnO4-, and Cr2O72-

Reducing Agent or Reductant:

• Causes the reduction of another substance;
• Loses electrons, i
...
it is oxidised;
• Example: NaH and CaH2

24

Exercise 5
• Identify the reducing and oxidising agents in the
following reaction:
5SO32- + 2MnO4- + 6H+  5SO42- + 2Mn2+ + 3H2O

25

Oxidation & Reduction Half Reactions
Separate the overall REDOX reaction oxidation and reduction
half reaction:
0

+2

+2

0

Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Half Reactions – REDOX reactions are linked by gain/loss of eOxidation Reaction:

Zn(s)  Zn2+(aq) + 2 e-

Reduction Reaction:

Cu2+(aq) + 2 e-  Cu(s)

Overall Reaction:

26
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)

Exercise 6
Show the oxidation and reduction half reaction that
occur, and write the overall ionic equation for the
reaction of Mn with lead(II) nitrate solution to
produce Pb(s) and Mn(NO3)2(aq):
0

+2

-1

0

+2

-1

Mn(s) + Pb(NO3)2(aq)  Pb(s) + Mn(NO3)2(aq)
Oxidation Half-Reaction:
Reduction Half-Reaction:
Overall Ionic Equation:
27

Balancing REDOX Reactions
HALF REACTION METHOD – ACID MEDIUM

Follow the steps below:
1
...
Balance
a) all elements atoms except H & O;
b) O by adding H2O; &
c) H by adding H+
3
...
Multiply half reactions by appropriate coefficients so that
electrons cancel & add reactions together
...
Assign oxidation numbers to determine what is
oxidised and what is being reduced
...

Since C goes from +3 to +4, it is oxidised
...
Divide the equation into two half reactions: one
for oxidation and one for reduction
...
Balance each half reaction
Oxidation
a
...

C2O42-  2CO2
b
...
H+ not added as there is no H in the equation
...
To balance the charge, we must add 2 electrons to the
right-hand side of equation
...
The manganese is balanced
MnO4-  Mn2+
b
...

MnO4-  Mn2+ + 4H2O
c
...
To balance the charge, we add 5e- to the left-hand side
...
Multiply the half reactions by integers if
necessary so that the numbers of electrons are
equal on each side
...
Add the two half-reactions and simplify by
cancelling species that appear on bot sides of the
combined equation
...
Balance reaction exactly as you would for ACIDIC
MEDIUM, i
...
steps 1-4
2
...

3
...


35

Balancing REDOX Reactions
HALF REACTION METHOD – BASIC MEDIUM
Method 2

Follow the steps below:
1
...
Balance
a) All elements other than H & O;
b) O by adding H2O; &
c) H by adding H+
d) H+ by adding OH- to both sides of the equation to
form H2O
3
...
Multiply half reactions by appropriate coefficients so that
electrons cancel & add reactions together
...

Example: decomposition of Thiosulfate
+2 -2

+1

0

+4 -2

+1 -2

S2O32- + 2 H+  S + SO2 + H2O
+2 reduced to 0

+2 oxidised to +4

38

Displacement Reaction
A
REDOX
reaction
where
one
element
in
molecular/atomic form displaces another element in
ionic form
...


Nonmetals: Magnesium displaces hydrogen

Mg(s) + 2 HCl(aq)  MgCl2(aq) + H2(g)
2
...

The most well-known peroxide is hydrogen peroxide, H2O2
...

40
O22-(aq) + 2H2O(l)  2OH-(aq) + H2O2(aq)

Molarity
• Two solutions can contain the same compounds but be
quite different because the proportions of those
compounds are different eg
...

• Molarity is one way to measure the concentration of a
solution
...
3 g of sodium sulfate in 500 ml of
water
...
3 g
142
...
171 mol

Molarity of Na2SO4 = 0
...
5 L = 0
...
342 M
42

Mixing a Solution

43

Exercise 10
How many grams of potassium dichromate
(K2Cr2O7) is required to prepare a 250 mL solution
whose concentration is 2
...
16 mol L-1 x 0
...
540 mol
mass = n x molar mass
= 0
...
2 g mol-1
= 159 g
44

Solution Dilution

• The preparation of a lower concentration solution from a
higher concentration solution by adding a solvent
...

n1 = n2
• The molarity of the new solution can be determined from
the equation
Mconc x Vconc= Mdil x 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
...
020 M K2Cr2O7 solution after
diluting solution to a final volume of 1L
...
020 M)(0
...
1000 dm3)
Mdil = 0
...
00 x 102 mL of a 1
...
61 M stock solution of H2SO4
...
Three common
methods are:
1
...
volume percent (%v/v) = volume solute x 100%
volume solution
3
...

• The completeness of reaction is denoted by a change in
colour of the indicator
...
02 mL

Indicator – An intensely-coloured, organic dye that exhibits
different colours in acidic & basic medium
Standard solution – a solution of known concentration

Equivalence point – the stoichiometric point, i
...
Equal
amounts of reactants
End point – volume at which the indicator changes colour
For an accurate titration:
Equivalence point ≈ End point
51

Exercise 12
A 25,00 cm3 of 0,0846 mol dm-3 HCl are neutralised by
30,00 cm3 of approx
...

Calculate:

i)the exact molarity of the NaOH solution, and
ii)the concentration of NaOH in g dm-3 of solution
...
6201 g
...
The whole solution
was titrated against 0
...
50 ml was required for the complete reaction
...

 Indirect titration method
 Direct reaction is slow, analyte is insoluble and contains
impurities
 Involves:
 The addition of excess reagent
...
When 25,0 cm3 of this diluted
solution was back titrated with 15,75 cm3 of 0,200 mol dm-3
NaOH were required for neutralisation
...
e
...


57

Solution
Mconc

X

Vconc = Mdil

Molarity of HCl soln
...
00625 mol
Moles of NaOH
= (0,200 mol NaOH x

15,75 cm3 𝑠𝑜𝑙𝑛
1000 cm3 soln

)(

1 mol HCl
)
1 mol NaOH

= 0
...

Moles of HCl reacted with CaCO3 =0
...
00315
= 0,0031 mol HCl
Mass of CaCO3 = n x MM
= 0,0031 mol HCl x

1 mol CaCO3
2 mol HCl

x 100 g mol-1 CaCO3

= 0,155 g CaCO3 (in 25,0 cm3 soln)
Therefore in 100 cm3 soln:
0,155 g CaCO3 x
% Purity =

0,62 g
0,72 g

100 cm3 soln
25 cm3 soln

= 0,62 g CaCO3

x 100 % = 86 %

59

Summary of Concepts









Aqueous Solutions
Ionic equations
Acids and Bases
Precipitation reactions
Redox reactions
Disproportionation
Peroxide Chemistry
The stoichiometry of reactions in
aqueous solution

60



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