FSC 112-Introductory Chemistry
Dr K.O. Abdulwahab
1
Solutions
• A solution is a mixture solute (s) and a
solvent.
• The amount of solute present in a
solution is usually expressed as
concentration.
• The different ways of expressing
concentration include:- Mass percent,
molality, molarity, mole fraction etc
2
Solution Concentration
• Molarity(M):
Moles solute/1L solution
• Molality (m):
Moles solute/1kg solvent
• Mole fraction (XA):
Moles A*
total moles solution
• Mass percent:
Mass solute
total mass of solution
x 100
*In some applications, one needs the mole fraction of solvent, not solute.
Make sure you find the quantity you need!
Concentration Units Continued
Molarity (M)
M =
moles of solute
liters of solution
Molality (m)
m =
moles of solute
mass of solvent (kg)
4
Molarity
• The most commonly used expression of
concentration is the molarity (M)
• This is defined as :-
• The unit is moldm-3
5
Example on Molarity
Calculate the molarity of a solution that contains 11.5 g of NaOH dissolved in 1.50 L
of solution. [Na = 23, O = 16, H = 1]
M= no of moles/volume of solution
No of moles (n) = mass/Mm
Mm of NaOH= 23+16+1= 40
n= 11.5/40 = 0.2875
M= 0.2875/1.50
Ans = 0.192 moldm-3
6
Questions
1. Calculate the molarity of a solution
prepared by dissolving 1.56 g of gaseous
HCl in enough water to make 26.8 ml of
solution. Ans = 1.59 M
2. How many moles of Ag+ ions are present
in 25 ml of 0.2 M Ag2O solution? Hint:
First find the number of moles of Ag2O,
Then find the no of moles of Ag+
Ans = 0.01 moles
7
Molality (molal)
•
Molality is a concentration unit based on the
number of moles of solute per kilogram of
solvent.
moles of solute
m=
kg of solvent
in dilute aqueous solutions molarity and
molality are nearly equal
Molality
Calculate the molality of a sulfuric acid solution containing
24.4 g of sulfuric acid in 198 g of water. The molar mass of
sulfuric acid is 98.09 g.
First, we find the number of moles of sulfuric acid in 24.4 g of
the acid, using its molar mass as the conversion factor.
The mass of water is 198 g, or 0.198 kg. Therefore,
Molality
The density of a 2.45 M aqueous solution of methanol (CH3OH)
is 0.976 g/mL. What is the molality of the solution? The molar
mass of methanol is 32.04 g.
M= n/v, n= M x v, = 2.45 M x 1 L = 2.45 moles
Molality
Next step is to calculate the mass of water in 1 L of the
solution, using density as a conversion factor. The total mass
of 1 L of a 2.45 M solution of methanol is
Because this solution contains 2.45 moles of methanol, the
amount of water (solvent) in the solution is
Molality
The molality of the solution can be calculated by converting
898 g to 0.898 kg:
Percent by Mass
mass of solute
x 100%
% by mass =
mass of solute + mass of solvent
mass of solute x 100%
=
mass of solution
Mole Fraction (X)
moles of A
XA =
sum of moles of all components
13
Percent by mass
A sample of 0.892 g of potassium chloride (KCl) is dissolved in
54.6 g of water.
What is the percent by mass of KCl in the solution?
Mole Fraction
B
A
A
A
B
A
B
A
A
B
A
B
A
A
Mole Fraction ()
moles of A A
 A = sum of moles of all components
moles of B B
 B = sum of moles of all components
Since A + B make up the
entire mixture, their mole
fractions will add up to one.
A
A
+
+
B
B
 A + B = 1.00
Class Work
In our glass of iced tea, we have added 3 tbsp
of sugar (C12H22O11). The volume of the tea
(water) is 325 mL. What is the mole fraction
of the sugar in the tea solution?
(1 tbsp sugar ≈ 25 g)
•
Solution
First, we find the moles of both the solute and
the solvent.
•
 1 mol C12 H 22 O11 
75.g C12 H 22 O11 
 = 0.219 mol
 342 g C12 H 22 O11 
 1 mol H 2 O 
325mL H 2 O 
 = 18.1 mol
 18.0 g H 2 O 
Next, we substitute the moles of both into the
mole fraction equation.
0.219 mol sugar
moles solute
=
= 0.012
sugar =
(0.219 mol + 18.1 mol)
total moles solution
χ
Types of Chemical Reactions
❖Precipitation reactions
❖Acid-base reactions
❖Oxidation-reduction reactions
❖Decomposition reaction
❖Displacement reaction
18
Precipitation Reactions
• Sometimes when two solutions are mixed, an
insoluble substance forms i.e. a solid forms
and separates from the solution.
AgNO3 + HCl → HNO3 +AgCl
• Such a reaction is called a precipitation
reaction.
19
Reaction of potassium
Iodide with lead nitrate
led to the formation of a
yellow precipitate of PbI2
(lead iodide).
20
Precipitation Reactions
In a mixture of K2CrO4(aq) + Ba(NO3)2(aq) to
predict the product, we must remember that
• When ions form a solid compound, the
compound formed must have a zero net
charge. Therefore, K+ and Ba2+ cannot combine
to form the solid, nor could CrO42- and NO3• The possible combinations of a given cations
and anions from the above are: K2CrO4, KNO3,
BaCrO4 , Ba(NO3)2.
• Note that it cannot be K2CrO4 or Ba(NO3)2. The
product will therefore be KNO3 and BaCrO4
21
Precipitation Reactions
• Question: which one of these is likely to be the
precipitate?
Rules for the solubility of salts in water
1. All nitrate (NO3-)salts are soluble.
2. Most salts containing the alkali metal
ions (Li+, Na+, K+, Cs+, Rb+) and the
ammonium ion (NH4+) are soluble.
3. Most chloride, bromide and iodide salts
are soluble except salts containing Ag+
Pb2+ and Hg22+.
22
Precipitation Reactions
4. Most sulphate salts are soluble except
BaSO4, PbSO4, HgSO4 and CaSO4 .
5. Most hydroxide salts are only slightly
soluble except NaOH and KOH.
6. Most sulphides (S2-), carbonates (CO32-),
chromate (CrO42-) and phosphate (PO43-)
salts are only slightly soluble.
23
Stoichiometry of Precipitation
Reactions
Calculate the mass of lead(II) sulphate formed
in the reaction of 145 mL of 0.123 M lead(II)
nitrate and excess sodium sulphate.
• Steps to follow:
1. Write the balanced equation for the
reaction
2. Calculate the number of moles of reactant
3. Determine which reactant is limiting
4. Calculate the moles of products as required
5. Convert to grams/kg/mg or other units as
required.
24
Example
• Calculate the mass of lead(II) sulfate formed in the
reaction of 145 mL of 0.123 M lead(II) nitrate and
excess sodium sulfate.
PbNO3 (aq) + Na2SO4 (aq) → PbSO4 (s) + 2NaNO3 (aq)
1 mole of Pb(NO3)2 will
produce 1 mole of PbSO4
From the question no of mole of Pb(NO3)2
= 0.123 x 0.145 (molarity x volume)= = 0.0178 mole
25
Solution (Ctd)
Since the ratio is 1:1, the no of moles of PbSO4
is 0.0178 moles.
To convert to mass,
• 1 mole of PbSO4 = 303.26 g
• 0.0178 mole = 303.26 * 0.0178 = 5.40 g
26
Acid-Base Reactions
According to Bronsted and Lowry:
⮚An acid is a proton donor
⮚A base is a proton acceptor
Prediction of products is just like the
precipitation reaction.
27
Acid-Base Titrations
• A technique for
determining the
amount of a certain
substance by carrying
out a titration
experiment is called
volumetric analysis.
28
Acid-Base Titrations
• A titration experiment involves the titrant
(solution in the burette) and the titrand.
• The point in the titration where enough
titrant has been added to completely react
with the titrand is called the equivalence or
stoichiometric point.
29
Acid-Base Titrations
• This point is often marked by a colour
change in the indicator. The point where
the indicator actually changes is the end
point.
• The goal is to choose an indicator such
that the end point(where the indicator
changes colour) occurs at the
equivalence point.
30
Titration
a)
b)
c)
Titration of an acid with a base. a) the flask contains acid and a
few drops of phenolphthalein, which is colorless in acidic
conditions, b) the endpoint has been reached (notice the faint
pink color of the solution), finally in c), the solution is well
beyond the end point since more base was added.
31
Example
(a) 0.5 M hydrochloric acid was titrated against 25cm3
of an unknown concentration of sodium hydroxide. Use
the data below to calculate the mean titre.
ANS:
•
Find the mean of the two concordant
results
•
27.6 + 27.6 = 27.6 cm3
2
(b) Calculate the concentration of the sodium
hydroxide.
NaOH + HCl → NaCl +H2O
•
•
•
•
•
Using the mean titre from previous question
27.6 cm3 = 0.0276 dm3
Calculate the moles of HCl used = volume x
Molarity.
0.0276 x 0.5 = 0.0138 moles
This is equal to the moles of NaOH (mole ratio
= 1:1) = 0.0138
Calculate the concentration of NaOH = moles ÷
volume = 0.0138 ÷ 0.025 dm3 = 0.55M or 0.6M
Practice Questions
1. What volume of a 0.1000 molar HCl solution
is needed to neutralise 25.0 ml of 0.35 molar
NaOH (Ans = 87.5 mL)
1. A 10.0 ml sample of sulphuric acid from an
automobile battery requires 35.08 ml of 2.12
molar sodium hydroxide solution for
complete neutralization. What is the molarity
of the sulphuric acid?
34
Oxidation and Reduction Reactions
⮚Oxidation
-
Addition of oxygen
Increase in oxidation number
Removal of hydrogen
Removal of electron
35
Reduction Reactions
⮚Reduction
-
Removal of oxygen
Decrease in oxidation number
Addition of hydrogen
Addition of electrons
36
Oxidation and Reduction Reactions
• Reducing agent causes reduction to take place
and itself is oxidised.
• Oxidizing agent causes oxidation to take place
and itself is reduced.
• The oxidation state / number:
- is the charge the atom would have if the
shared electrons were divided equally
between identical atoms bonded to each
other.
37
Oxidation and Reduction Reactions
Rules for Assigning Oxidation States
1. The oxidation number of a neutral atom or
element is 0
For example Na(s) , O2(g) P4 are all 0
2. (a)The sum of the oxidation numbers of all
the atoms in a neutral compound is zero. E.g
NaCl = 0.
(b) For an ion, the sum of the oxidation
number is equal to the charge on the ion e.g
The oxidation number of PO33- is -3
38
Oxidation and Reduction Reactions
Rules for Assigning Oxidation States
3. The oxidation state of a monoatomic ion is the
same as its charge e.g Na+ = +1, Cl - = -1,
Ca
2+ = +2.
4. Oxygen in a compound usually has an
oxidation number of -2, except in peroxides
where oxygen has an oxidation number of -1.
39
Oxidation and Reduction Reactions
Rules for Assigning Oxidation States
5. Hydrogen usually has a oxidation number of +1,
but when it combines with a metal, it has an
oxidation number of -1.
5. All group 1 metals have oxidation number of +1.
7. All group 2 metals have oxidation number of +2.
40
Classwork
• Determine the oxidation number of
(a) Al in Al2O3
(b) P in H3PO4
(c) S in SO32(d) N in NO3• Determine the reducing and oxidising agents
in the reaction below:
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)
41
Balancing of Redox Equation
• Redox reaction occurring in acidic solutions
MnO4- + Fe2+ → Mn2+ + Fe3+
Step 1. Identify oxidising and reducing agents
and write half equations
• MnO4- gains electrons, it acts as the oxidizing
agent as it is reduced
MnO4- → Mn2+ (reduction)
42
Balancing of Redox Equation
Fe2+ loses electron, it acts as the reducing
agent as it is oxidised
Fe2+ → Fe3+ (oxidation)
• Step 2: For each half-reaction
(a)Balance all the elements except hydrogen
and oxygen
(b)Balance all oxygen using water
MnO4- → Mn2+ + 4H2O
(c) Balance hydrogen using H+
8H+ + MnO4- → Mn2+ + 4H2O
43
Balancing of Redox Equation
(d) Balance the charge using electrons
5e- + 8H+ + MnO4- → Mn2+ + 4H2O
Fe2+ → Fe3+ + e-
Step 3: If necessary, multiply one or both
balanced half reactions by an integer to
equalize the number of electrons
transferred in the two half reactions.
Fe2+ → Fe3+ + e- (*5)
44
Balancing of Redox Equation
Step 4: Add the half reactions and cancel
identical species.
5e- + 8H+ + MnO4- + 5Fe2+ → Mn2+ + 4H2O
+5Fe3+ +5e5Fe2+ + 8H+ + MnO4- → Mn2+ + 4H2O + 5Fe3+
45
Example
Balancing redox equations
Balance the equation showing the oxidation of Fe2+ ions to Fe3+ ions by dichromate ions (Cr2O72-) in an acidic
medium
Fe2+ + Cr2O72- → Fe3+ + Cr3+
Step 1: Identify oxidising and reducing agents and write half reactions
Fe2+ : +2
Fe3+ : +3
Cr2O72- : 2Cr + 7(-2) = -2
2Cr = 12
Cr = +6
Cr3+ : +3
Fe2+ → Fe3+
Iron loses one electron, i.e. it is oxidised. It donates one electron to dichromate, i.e. it is the reducing agent.
Half equation:
Oxidation reaction: Fe2+ → Fe3+ + eCr26+ → Cr3+
Chromium gains three electrons, i.e. it is reduced. It gains its electrons from iron, i.e. it is the oxidising agent.
Half equation:
Reduction reaction: Cr26+ + 3e- → Cr3+
Fe2+ + Cr2O72- → Fe3+ + Cr3+
Step 2: Balance each kind of atom other that H and O
Fe2+ → Fe3+ + e-
Cr2O72- + 3e- → 2Cr3+
Step 3: Balance oxygen atoms by using H2O
Fe2+ → Fe3+ + e-
Cr2O72- + 3e- → 2Cr3+ + 7H2O
Step 4: Balance H atoms by using H+ ions
14H+ + Cr2O72- + 3e- → 2Cr3+ + 7H2O
Fe2+ → Fe3+ + e-
Step 5: Use electrons as needed to obtain a charge that is balanced
Fe2+ → Fe3+ + e-
+2
+2
+3
+2
14H+ + Cr2O72- +3e- → 2Cr3+ + 7H2O
-1
+14
-2
-3
+9
Add three electrons to the reactant side to balance the charges
Fe2+ → Fe3+ + e-
14H+ + Cr2O72- + 6e- → 2Cr3+ + 7H2O
+6
0
+6
Step 6: Ensure the number of electrons gained equals the number of electrons
lost and add the 2 half reactions together
Fe2+ → Fe3+ + e14H+ + Cr2O72- + 6e- → 2Cr3+ + 7H2O
×1
6Fe2+ → 6Fe3+ + 6e14H+ + Cr2O72- + 6e- → 2Cr3+ + 7H2O
14H+ + Cr2O72- + 6Fe2+ + 6e- → 2Cr3+ + 7H2O + 6Fe3+ + 6e-
14H+ + Cr2O72- + 6Fe2+ → 2Cr3+ +7H2O + 6Fe3+
Balanced
Check to ensure that all atoms and charges are balanced.
×6
Balancing Redox Equation in acidic medium
Questions
Cr2O72- (aq) + HSO3- (aq) → Cr3+ (aq) + HSO4- (aq)
CuS (s) + NO3 - (aq)
Cu2+(aq) + SO42- (aq) + NO (g)
Fe2+ + Cr2O72- → Fe3+ + Cr3+
MnO4- + SO2 → Mn2+ + SO42-
Balancing of Redox Equation
• Redox reaction occurring in basic
medium
1. Use the half-reaction method as specified
for acidic solutions to obtain the final
balanced equation as if H+ were present.
2. To both sides of the equation obtained,
add a number of OH- ions that is equal to
H + ions. (You want to eliminate H+ by
turning is into H2O)
50
Balancing of Redox Equation
3. Eliminate the number of H2O
molecules that appear in both sides of
the equation.
4. Check that the elements and charges
are balanced.
51
Balancing of Redox Equation in Basic solution
Write a balanced equation to represent the oxidation of iodide ion (I-) by permanganate ion (MnO4-) in basic
solution to yield molecular iodine (I2) and manganese (IV) oxide (MnO2)
Skeletal equation:
I- + MnO4-
I2 + MnO2
Step 1: Identify oxidising and reducing agents and write half equations
I- loses an electron: it acts as the reducing agent as it is oxidised
I- → I2 + e-
Oxidation reaction
MnO4- gains 3 electrons: it acts as the oxidising agent as it is reduced
MnO4- + 3e- → MnO2
Reduction reaction
Step 2: Balance each kind of atom other than H and O
2I- → I2 + eMnO4- + 3e- → MnO2
Step 3: Balance the O atoms by using H2O
2I- → I2 + e-
MnO4- + 3e- → MnO2 + 2H2O
Step 4: Balance H atoms by using H+ ions
2I- → I2 + e-
4H+ + MnO4- + 3e- → MnO2 + 2H2O
Step 5: Since the reaction occurs in a basic medium, for each H+ ion we add an
equal number of OH- ions to both sides of the equation. Where H+ and OHions appear on the same side of the equation, they may be combined to form
H2O.
4H+ + 4OH- + MnO4- + 3e- → MnO2 + 2H2O + 4OH4H2O + MnO4- + 3e- → MnO2 + 2H2O + 4OH2H2O + MnO4- + 3e- → MnO2 + 4OHStep 6: Use electrons as needed to obtain a charge that is balanced
2I- → I2 + e-
-2
0
-2
-1
-1
Add one electron to the product side to balance charges
2I- → I2 + 2e-
2H2O + MnO4- + 3e- → MnO2 + 4OH-
0
Already balanced!
-1
-4
-3
0
-4
-4
Step 7: Ensure the number of electrons gained equals the number of electrons lost
and add the two half reactions together
2I- → I2 + 2e-
×3
2H2O + MnO4- + 3e- → MnO2 + 4OH-
×2
6I- → 3I2 + 6e4H2O + 2MnO4- + 6e- → 2MnO2 + 8OH2MnO4- + 6I- + 4H2O → 2MnO2 + 3I2 + 8OHBalanced
Check to ensure that all atoms and charges are balanced.
Question
Balance the following redox equation which occurs in basic solution
Mn2+ + H2O2 → MnO2 + H2O
Step 1: Identify the oxidising and reducing agents and write half reactions
Mn2+ → MnO2
Mn2+ : +2
MnO2: Mn +2(-2) = 0
Mn = +4
Mn2+ loses two electrons: it acts as the reducing agent as it is oxidised
Mn2+ → MnO2 + 2e-
H2O2 → H2O
Oxidation reaction
H2O2: 2(+1) + 2O = 0
O = -1
H2O: 2(+1) + O = 0
O = -2
H2O2 gains one electron: it acts as the oxidising agent as it is reduced
H2O2 + e- → H2O
Reduction reaction
Step 2: Balance each kind of atom other than H and O.
Mn2+ → MnO2 + 2e-
Already balanced in this case
H2O2 + e- → H2O
Step 3: Balance the O atoms by using H2O
2H2O + Mn2+ → MnO2 + 2eH2O2 + e- → 2H2O
Step 4: Balance the H atoms by using H+
2H2O + Mn2+ → MnO2 + 2e- + 4H+
2H+ + H2O2 + e- → 2H2O
Step 5: For each H+ ion, add equal no. of OH- to both sides of equation
2H2O + Mn2+ + 4OH- → MnO2 + 2e- + 4H+ + 4OH2H2O + Mn2+ + 4OH- → MnO2 + 2e- + 4H2O
Mn2+ + 4OH- → MnO2 + 2e- + 2H2O
2H+ + 2OH- + H2O2 + e- → 2H2O + 2OH2H2O + H2O2 + e- → 2H2O + 2OHH2O2 + e- → 2OHStep 6: Use electrons as needed to obtain a charge that is balanced
Mn2+ + 4OH- → MnO2 + 2e- + 2H2O
+2
-4
0
-2
0
-2
-2
H2O2 + e- → 2OH-
-1
0
-2
-1
-2
Add one electron to the reactant side to balance charges
H2O2 + 2e- → 2OHStep 7: Ensure no. of electrons gained equals no. of electrons lost and add the half reactions together
Mn2+ + 4OH- → MnO2 + 2e- + 2H2O
H2O2 + 2e- → 2OHMn2+ + H2O2 + 2OH- → MnO2 + 2H2O
Questions
Write balanced equations to represent the following reactions in a basic
solution
(a) Fe(OH)2 + MnO4- → MnO2 + Fe(OH)3
(b) Bi(OH)3 + SnO22- → SnO32- + Bi
Stoichiometry of Redox Reactions
Example
1. A piece of iron wire weighing 0.1568 g is
converted into Fe2+ (aq) and requires 26.24 ml
of a KMnO4 (aq) solution for its titration. What
is the molarity of the KMnO4?
• The balanced equation is:
5Fe2+ + MnO4- + 8H+
5Fe3+ + Mn2+ + 4H2O
5 moles of Fe2+ = 1 mole of MnO460
Solution contd
No of moles of Fe = mass/ Mm = 0.1568/ 56 =
0.002808 moles
0.002808 Fe = 1/5* 0.002808 MnO4- =
0.0005616 moles
Molarity = mole/volume = 0.0005616/0.02624
= 0.02140 M
61
Practice Questions
2). 24.5 ml of KMnO4 (aq) solution is required to
titrate 0.350 sodium oxalate in the redox
reaction. What is the molarity of the KMnO4?
[Hint balance the equation first]
C2O42- + MnO4- + H+
Mn2+ + H2O + CO2
3. A 25.00 ml sample of Fe2+ requires 34.77 ml of
0.05876 molar K2Cr2O7 for its titration. What is
the molarity of the Fe in the sample?
Fe2+ + Cr2O72Fe3+ + Cr3+
62
2. Decomposition
Example: NaCl
Cl Na
→
General:
Cl
+
Na
AB → A + B
Compound = Element + Element
Ex. Decomposition Reaction
Other examples of Decomposition Reactions
•
•
•
2H₂O(I ) → 2H₂(g) + O₂(g)
CaCO₃(s) → CaO(s) + CO₂(g)
2KClO₃(s) → 2KCl(s) + 3O₂(g)
Single Replacement Reaction
Single Replacement Reactions
•
•
Write and balance the following single
replacement reaction equations:
Zn(s) +2 HCl(aq) → ZnCl2 + H2(g)
• 2 NaCl(s) + F2(g) → 2 NaF(s) + Cl2(g)
• 2 Al(s)+ 3 Cu(NO3)2(aq)→ 3 Cu(s)+ 2 Al(NO3)3(aq)
Double Displacement
Example: MgO + CaS
Mg
+
Ca
O
General:
S
→
Mg
S
+
Ca
O
AB + CD → AD + CB
Double Replacement Reactions
•
•
Think about it like “foil”ing in algebra, first
and outer ions go together + inside ions go
together
Example:
AgNO3(aq) + NaCl(s) → AgCl(s) + NaNO3(aq)
•
Another example:
K2SO4(aq) + Ba(NO3)2(aq) → KNO3(aq) + BaSO4(s)
2
Practice
•
1.
2.
3.
4.
5.
6.
Predict the products:
HCl(aq) + AgNO3(aq) →
CaCl2(aq) + Na3PO4(aq) →
Pb(NO3)2(aq) + BaCl2(aq) →
FeCl3(aq) + NaOH(aq) →
H2SO4(aq) + NaOH(aq) →
KOH(aq) + CuSO4(aq) →