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Solution Stoichiometry (Lecture 2)

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Title: Solution Stoichiometry (Lecture 2)


1
Solution Stoichiometry(Lecture 2)
  • Mass concentration Dilution and volumetric
    analysis

2
What will I learn?
  • What is mass concentration?(and how is it
    related to molar concentration?)
  • What is dilution?
  • What is volumetric analysis? (or titration?)
  • How to calculate an unknown concentration in a
    volumetric analysis problem?

3
Other Concentration Units
  • Sometimes, concentration can be expressed in mass
    per unit volume

4
Other Concentration Units
  • Sometimes, concentration can be expressed in mass
    per unit volume
  • Units gdm-3

5
Other Concentration Units
  • Sometimes, concentration can be expressed in mass
    per unit volume
  • Concentration (in gdm-3) can be converted to
    molar concentration via the equation

6
Example 1
  • 2.00 g of NaOH is dissolved in water to give a
    final volume of 150.0 cm3.
  • Calculate the concentration (in gdm-3) of the
    NaOH solution formed

7
Example 1
  • 2.00 g of NaOH is dissolved in water to give a
    final volume of 150.0 cm3.
  • Calculate the concentration (in moldm-3) of the
    solution above

8
Example 1
  • 2.00 g of NaOH is dissolved in water to give a
    final volume of 150.0 cm3.
  • Calculate the concentration (in moldm-3) of the
    solution above

9
Example 1
  • 2.00 g of NaOH is dissolved in water to give a
    final volume of 150.0 cm3.
  • Calculate the concentration (in moldm-3) of the
    solution above

10
Example 1
  • 2.00 g of NaOH is dissolved in water to give a
    final volume of 150.0 cm3.
  • Calculate the concentration (in moldm-3) of the
    solution above

11
Dilution
  • Dilution is the process of adding more solvent to
    a solution

Remains constant
decreases
increases
12
Example 1
  • 10.0 cm3 of a NaOH solution of concentration
    1.500 moldm-3 was diluted to 250 cm3 using
    distilled water. Calculate the concentration of
    NaOH after dilution.

Rearranging,
13
Example 1
  • 10.0 cm3 of a NaOH solution of concentration
    1.500 moldm-3 was diluted to 250 cm3 using
    distilled water. Calculate the concentration of
    NaOH after dilution.

14
Example 1
  • 10.0 cm3 of a NaOH solution of concentration
    1.500 moldm-3 was diluted to 250 cm3 using
    distilled water. Calculate the concentration of
    NaOH after dilution.

15
Example 1
  • 10.0 cm3 of a NaOH solution of concentration
    1.500 moldm-3 was diluted to 250 cm3 using
    distilled water. Calculate the concentration of
    NaOH after dilution.

16
Example 1
  • 10.0 cm3 of a NaOH solution of concentration
    1.500 moldm-3 was diluted to 250 cm3 using
    distilled water. Calculate the concentration of
    NaOH after dilution.

17
Example 1
  • 10.0 cm3 of a NaOH solution of concentration
    1.500 moldm-3 was diluted to 250 cm3 using
    distilled water. Calculate the concentration of
    NaOH after dilution.

18
Example 1
  • 10.0 cm3 of a NaOH solution of concentration
    1.500 moldm-3 was diluted to 250 cm3 using
    distilled water. Calculate the concentration of
    NaOH after dilution.

19
Example 2
  • A concentrated solution of H2SO4 has a
    concentration of 3.35 moldm-3. What volume of
    this acid is required to prepare 250.0 cm3 of
    0.130 moldm-3 H2SO4 solution?

20
Example 2
  • A concentrated solution of H2SO4 has a
    concentration of 3.35 moldm-3. What volume of
    this acid is required to prepare 250.0 cm3 of
    0.130 moldm-3 H2SO4 solution?

21
Example 2
  • A concentrated solution of H2SO4 has a
    concentration of 3.35 moldm-3. What volume of
    this acid is required to prepare 250.0 cm3 of
    0.130 moldm-3 H2SO4 solution?

22
Example 2
  • A concentrated solution of H2SO4 has a
    concentration of 3.35 moldm-3. What volume of
    this acid is required to prepare 250.0 cm3 of
    0.130 moldm-3 H2SO4 solution?

Rearranging,
23
Example 2
  • A concentrated solution of H2SO4 has a
    concentration of 3.35 moldm-3. What volume of
    this acid is required to prepare 250.0 cm3 of
    0.130 moldm-3 H2SO4 solution?

24
Example 2
  • A concentrated solution of H2SO4 has a
    concentration of 3.35 moldm-3. What volume of
    this acid is required to prepare 250.0 cm3 of
    0.130 moldm-3 H2SO4 solution?

25
Example 2
  • A concentrated solution of H2SO4 has a
    concentration of 3.35 moldm-3. What volume of
    this acid is required to prepare 250.0 cm3 of
    0.130 moldm-3 H2SO4 solution?

26
Example 2
  • A concentrated solution of H2SO4 has a
    concentration of 3.35 moldm-3. What volume of
    this acid is required to prepare 250.0 cm3 of
    0.130 moldm-3 H2SO4 solution?

27
Dilution
  • Useful formulae for dilution

28
Dilution
  • Useful formulae for dilution

29
Volumetric analysis
  • A quantitative analysis (vs qualitative analysis)
  • Analysis gt calculation / manipulation of the
    results to obtain meaningful data
  • Volumetric analysis gt accurate measurement of
    volume is required

30
Volumetric analysis
31
Volumetric analysis
32
Volumetric analysis
33
Volumetric analysis
  • End point is the point in the titration when the
    indicator undergoes a sharp colour change.
  • Equivalence point is the point when
    stoichiometric amounts / volume of reactants have
    been mixed
  • The two points are not the same
  • However, an ideal indicator is one where the
    equivalence point and end point is very close to
    each other
  • (and a sharp colour change occurs when a small
    extra amount of reactant has been added)

34
Calculations in VA
  • Similar to that encountered before (in normal
    stoichiometry calculations)
  • Difference concentration and reacting volumes
    are now involved

35
Number of moles
Titration volume
mass (m)
number of moles (n)
number of particles (N)
volume (V)
36
Calculations in VA
  • Given

since
37
  • Example 120.0 cm3 of a solution of barium
    hydroxide, Ba(OH)2, of unknown concentration was
    placed in a conical flask and titrated with a
    solution of hydrochloric acid which had a
    concentration of 0.0600 moldm-3. The volume of
    the acid required was 25.0 cm3. Calculate the
    molar concentration of the barium hydroxide.

Balanced equation
Given c and V
c??
Given V
38
  • Example 120.0 cm3 of a solution of barium
    hydroxide, Ba(OH)2, of unknown concentration was
    placed in a conical flask and titrated with a
    solution of hydrochloric acid which had a
    concentration of 0.0600 moldm-3. The volume of
    the acid required was 25.0 cm3. Calculate the
    molar concentration of the barium hydroxide.

Balanced equation
39
  • Example 120.0 cm3 of a solution of barium
    hydroxide, Ba(OH)2, of unknown concentration was
    placed in a conical flask and titrated with a
    solution of hydrochloric acid which had a
    concentration of 0.0600 moldm-3. The volume of
    the acid required was 25.0 cm3. Calculate the
    molar concentration of the barium hydroxide.

Balanced equation
40
  • Example 120.0 cm3 of a solution of barium
    hydroxide, Ba(OH)2, of unknown concentration was
    placed in a conical flask and titrated with a
    solution of hydrochloric acid which had a
    concentration of 0.0600 moldm-3. The volume of
    the acid required was 25.0 cm3. Calculate the
    molar concentration of the barium hydroxide.

Balanced equation
From equation
41
  • Example 120.0 cm3 of a solution of barium
    hydroxide, Ba(OH)2, of unknown concentration was
    placed in a conical flask and titrated with a
    solution of hydrochloric acid which had a
    concentration of 0.0600 moldm-3. The volume of
    the acid required was 25.0 cm3. Calculate the
    molar concentration of the barium hydroxide.

Balanced equation
42
  • Example 120.0 cm3 of a solution of barium
    hydroxide, Ba(OH)2, of unknown concentration was
    placed in a conical flask and titrated with a
    solution of hydrochloric acid which had a
    concentration of 0.0600 moldm-3. The volume of
    the acid required was 25.0 cm3. Calculate the
    molar concentration of the barium hydroxide.

Balanced equation
43
  • Example 120.0 cm3 of a solution of barium
    hydroxide, Ba(OH)2, of unknown concentration was
    placed in a conical flask and titrated with a
    solution of hydrochloric acid which had a
    concentration of 0.0600 moldm-3. The volume of
    the acid required was 25.0 cm3. Calculate the
    molar concentration of the barium hydroxide.

Balanced equation
44
  • Example 120.0 cm3 of a solution of barium
    hydroxide, Ba(OH)2, of unknown concentration was
    placed in a conical flask and titrated with a
    solution of hydrochloric acid which had a
    concentration of 0.0600 moldm-3. The volume of
    the acid required was 25.0 cm3. Calculate the
    molar concentration of the barium hydroxide.

Balanced equation
45
  • Example 120.0 cm3 of a solution of barium
    hydroxide, Ba(OH)2, of unknown concentration was
    placed in a conical flask and titrated with a
    solution of hydrochloric acid which had a
    concentration of 0.0600 moldm-3. The volume of
    the acid required was 25.0 cm3. Calculate the
    molar concentration of the barium hydroxide.

Balanced equation
46
  • Example 120.0 cm3 of a solution of barium
    hydroxide, Ba(OH)2, of unknown concentration was
    placed in a conical flask and titrated with a
    solution of hydrochloric acid which had a
    concentration of 0.0600 moldm-3. The volume of
    the acid required was 25.0 cm3. Calculate the
    molar concentration of the barium hydroxide.

Balanced equation
ALTERNATIVELY
47
  • Example 120.0 cm3 of a solution of barium
    hydroxide, Ba(OH)2, of unknown concentration was
    placed in a conical flask and titrated with a
    solution of hydrochloric acid which had a
    concentration of 0.0600 moldm-3. The volume of
    the acid required was 25.0 cm3. Calculate the
    molar concentration of the barium hydroxide.

Balanced equation
?
48
  • Example 120.0 cm3 of a solution of barium
    hydroxide, Ba(OH)2, of unknown concentration was
    placed in a conical flask and titrated with a
    solution of hydrochloric acid which had a
    concentration of 0.0600 moldm-3. The volume of
    the acid required was 25.0 cm3. Calculate the
    molar concentration of the barium hydroxide.

Balanced equation
?
49
  • Example 120.0 cm3 of a solution of barium
    hydroxide, Ba(OH)2, of unknown concentration was
    placed in a conical flask and titrated with a
    solution of hydrochloric acid which had a
    concentration of 0.0600 moldm-3. The volume of
    the acid required was 25.0 cm3. Calculate the
    molar concentration of the barium hydroxide.

Balanced equation
?
50
  • Example 120.0 cm3 of a solution of barium
    hydroxide, Ba(OH)2, of unknown concentration was
    placed in a conical flask and titrated with a
    solution of hydrochloric acid which had a
    concentration of 0.0600 moldm-3. The volume of
    the acid required was 25.0 cm3. Calculate the
    molar concentration of the barium hydroxide.

Balanced equation
?
51
  • Example 2What volume of NaOH of concentration
    0.500 moldm-3 is required to neutralise 22.50
    cm3 of 0.262 moldm-3 sulphuric acid?

Balanced equation
?
52
  • Example 2What volume of NaOH of concentration
    0.500 moldm-3 is required to neutralise 22.50
    cm3 of 0.262 moldm-3 sulphuric acid?

Balanced equation
?
53
  • Example 2What volume of NaOH of concentration
    0.500 moldm-3 is required to neutralise 22.50
    cm3 of 0.262 moldm-3 sulphuric acid?

Balanced equation
?
54
  • Example 2What volume of NaOH of concentration
    0.500 moldm-3 is required to neutralise 22.50
    cm3 of 0.262 moldm-3 sulphuric acid?

Balanced equation
?
55
  • Example 2What volume of NaOH of concentration
    0.500 moldm-3 is required to neutralise 22.50
    cm3 of 0.262 moldm-3 sulphuric acid?

Balanced equation
?
56
  • Example 3Calculate the volume of nitric acid of
    concentration 0.200 moldm-3 required to react
    completely with 4.0 g of copper(II) oxide.

Balanced equation
Given mass
v??
Given c
57
  • Example 3Calculate the volume of nitric acid of
    concentration 0.200 moldm-3 required to react
    completely with 4.0 g of copper(II) oxide.

Balanced equation
58
  • Example 3Calculate the volume of nitric acid of
    concentration 0.200 moldm-3 required to react
    completely with 4.0 g of copper(II) oxide.

Balanced equation
59
  • Example 3Calculate the volume of nitric acid of
    concentration 0.200 moldm-3 required to react
    completely with 4.0 g of copper(II) oxide.

Balanced equation
?
60
  • Example 3Calculate the volume of nitric acid of
    concentration 0.200 moldm-3 required to react
    completely with 4.0 g of copper(II) oxide.

Balanced equation
?
61
  • Example 3Calculate the volume of nitric acid of
    concentration 0.200 moldm-3 required to react
    completely with 4.0 g of copper(II) oxide.

Balanced equation
?
Rearranging,
62
  • Example 3Calculate the volume of nitric acid of
    concentration 0.200 moldm-3 required to react
    completely with 4.0 g of copper(II) oxide.

Balanced equation
?
63
  • Example 3Calculate the volume of nitric acid of
    concentration 0.200 moldm-3 required to react
    completely with 4.0 g of copper(II) oxide.

Balanced equation
?
64
  • Example 3Calculate the volume of nitric acid of
    concentration 0.200 moldm-3 required to react
    completely with 4.0 g of copper(II) oxide.

Balanced equation
?
65
What will I learn?
  • What is mass concentration?(and how is it
    related to molar concentration?)
  • What is dilution?
  • What is volumetric analysis? (or titration?)
  • How to calculate an unknown concentration in a
    volumetric analysis problem?

66
End of Lecture 2
  • Often greater risk is involved in postponement
    than in making a wrong decision Harry A Hopf
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