Unit 7
Reactions in Aqueous
Solutions II:
Calculations
1
Calculations Involving Molarity
• Example 1: If 100.0 mL of 1.00 M NaOH and
100.0 mL of 0.500 M H2SO4 solutions are mixed,
what will the concentration of the resulting
solution be?
• Plan:
– Write a balanced equation
– Construct a reaction summary showing amts of mols
(or mmol) of NaOH and H2SO4.
– Determine amount of solute formed
– Find final (total) volume (sum of volumes mixed)
– Calculate molarity.
2
Recall…
• Molarity = mols of solute
L of solution
OR
= mmol of solute
ml solution
1 mol = 1000mmol
1L
= 1000 ml
or
1mmol = 10-3 mol
Therefore, molarity = no. mol
L
= no. mmol
ml
3
Calculations Involving Molarity
2 NaOH +
Reaction
Ratio:
Before
Reaction:
After
Reaction:
H2SO4  Na 2SO4  2 H2O
2 mol
1 mol
1 mol
2 mol
100ml (1.0mol)
1000mL/ L
100ml (0.5mol)
1000mL/L
0.1 mol NaOH
0.05 mol H2SO4
0 mol
0 mol
0 mol
0 mol
0.05 mol
0.1 mol
Note: The NaOH and H2SO4 neutralize each other exactly
4
Calculations Involving Molarity
• What is the total volume of solution?
100.0 mL + 100.0 mL = 200.0 mL
• What is the sodium sulfate amount, in mol?
0.05 mol
• What is the molarity of the solution?
? mol Na2SO4 = 0.05 mol x 1000ml
L
200 ml
1L
= 0.250 M Na2SO4
5
Calculations Involving Molarity
• Example 2: If 130.0 mL of 1.00 M KOH
and 100.0 mL of 0.500 M H2SO4 solutions
are mixed, what will be the concentration
of KOH and K2SO4 in the resulting
solution?
– You try it!
• What is the balanced reaction?
2 KOH + H 2SO 4  K 2SO 4 + 2 H 2 O
6
Calculations Involving Molarity
2 KOH + H 2SO 4  K 2SO 4 + 2 H 2 O
Reaction
Ratio:
2 mol
Before
130ml (1mol)
Reaction:
1000ml /L
0.13 mol
After
Reaction:
0.03 mol
1 mol
1 mol
2 mol
100ml (0.5mol)
1000mL /L
0.05 mol
0 mol
0 mol
0.05 mol
0 mol
0.1mol
Note: The KOH is in excess and the H2SO4 is the limiting reagent and
would therefore determine how much product is formed.
7
Calculations Involving Molarity
• What is the total volume of solution?
130.0 mL + 100.0 mL = 230.0 mL
• What are the potassium hydroxide and potassium
sulfate amounts?
0.03 mol & 0.05 mol
• What is the molarity of the solution?
? M KOH = 0.03mol x
1000 ml
230.0 mL
1L
? M K2SO4= 0.05 mol
x 1000ml
230.0 mL
1L
= 0.130 M KOH
= 0.217 M K2SO4
8
Calculations Involving Molarity
• Example 3: What volume of 0.750 M
NaOH solution would be required to
completely neutralize 100 mL of 0.250 M
H3PO4?
You try it!
9
Calculations Involving Molarity
3 NaOH + H 3PO 4  Na 3PO 4 + 3 H 2 O
0.250 mol H 3PO 4
? L NaOH = 0.100 L H 3PO 4 

1 L H 3PO 4
3 mol NaOH
1 L NaOH

 0.100 L NaOH
1 mol H 3PO 4 0.750 mol NaOH
You can also do these kinds of problems
following the above plan
10
Titrations
Acid-base Titration Terminology
 Titration – A method of determining
the concentration of one solution by
reacting it with a solution of known
concentration.
 Primary standard – A chemical
compound which can be used to
accurately determine the concentration of
another solution. Examples include KHP
(potassium hydrogen phthalate) is and
sodium carbonate (a base).
11
Titrations
Properties of an ideal primary standard
1.
2.
3.
4.
5.
6.
7.
8.
It must not react with or absorb the components of the
atmosphere e.g CO2, water vapor, oxygen
It must react according to one invariable reaction
Must have a high % purity
It should have a high formula weight to minimize the
effect of error in weighing
It must be soluble in solvent of interest
Should be non-toxic
Should be readily available (inexpensive)
Should be environmentally friendly
12
Titrations
Standard solution – A solution whose
concentration has been determined using
a primary standard.
Standardization – The process in which
the concentration of a solution is
determined by accurately measuring the
volume of the solution required to react
with a known amount of a primary
standard.
13
Titrations
Acid-base Titration Terminology
 Indicator – A substance that exists in different forms
with different colors depending on the concentration of
the H+ in solution. Examples are:
 phenolphthalein: colourless in acid but pink in base
 methyl red: red in acid but yellow in base.
 Equivalence point – The point at which
stoichiometrically equivalent amounts of the acid and
base have reacted.
 End point – The point at which the indicator changes
color and the titration is stopped.
14
Titrations
Acid-base Titration Terminology
15
Calculations for Acid-Base Titrations
• Potassium hydrogen phthalate is a very good
primary standard.
– It is often given the acronym, KHP.
– KHP has a molar mass of 204.2 g/mol.
16
Calculations for Acid-Base Titrations
 A very common mistake is for students to see the
acronym KHP and think that this compound is
made of potassium, hydrogen, and phosphorous.
17
Standardization of a Base Solution
• Example 4: Calculate the molarity of a
NaOH solution if 27.3 mL of it reacts
with 0.4084 g of KHP.
NaOH + KHP  NaKP + H 2 O
1 mol KHP
? mol NaOH = 0.4084 g KHP  1 mol KHP 
? mol NaOH = 0.4084 g KHP  204.2 g KHP 
204.2 g KHP
1 mol NaOH
 0.00200 mol NaOH
1
mol
NaOH
1 mol KHP
 0.00200 mol NaOH
1 mol KHP
0.00200 mol NaOH
? M NaOH =
 0.0733 M NaOH
0.0273 L NaOH
18
Standardization of an Acid Solution
• Example 5: Calculate the molarity of a
sulfuric acid solution if 23.2 mL of it reacts
with 0.212 g of Na2CO3.
Na 2 CO 3 + H 2SO 4  Na 2SO 4 + CO 2 + H 2 O
You do it!
1 mol Na 2 CO3
? mol H 2SO 4 = 0.212 g Na 2 CO3 

106 g Na 2 CO3
1 mol H 2SO 4
 0.00200 mol H 2SO 4
1 mol Na 2 CO3
? M H 2SO 4 
0.00200 mol H 2SO 4
 0.0862 M H 2SO 4
0.0232 L H 2SO 4
19
Calculations for Acid-Base Titrations
• Example 6: An impure sample of
potassium hydrogen phthalate, KHP, had
a mass of 0.884 g. It was dissolved in
water and titrated with 31.5 mL of 0.100 M
NaOH solution. Calculate the percent
purity of the KHP sample. Molar mass of
KHP is 204.2 g/mol.
NaOH + KHP  NaKP + H2O
You do it!
20
Calculations for Acid-Base Titrations
Plan:
• Calculate # mols of NaOH
• Use reaction ratio to determine # mols of
KHP
• Find mass of KHP corresponding to this #
of mols
• Calculate % purity = mass KHP
x 100
mass entire sample
Calculations for Acid-Base Titrations
0.100 mol NaOH
? mmol NaOH = 0.0315 L solution 
1 L sol' n
 0.00315 mol NaOH
1 mol KHP
? g KHP = 0.00315 mol NaOH 

mol NaOH
204.2g KHP
 0.643 g KHP
1 mol KHP
g KHP
0.643 g
% KHP =
100% 
100%  72.7 %
g sample
0.884 g
22
Back Titration
• Allows the user to find the concentration of
a reactant by reacting it with an excess
volume of another reactant of known
concentration. The resulting mixture is
then titrated back, taking into account the
molarity of the excess which was added.
– E.g. Adding excess acid to a base and
determining how much base was present by
titrating the unreacted acid.
23
Back Titration
• Back titrations can be used for many
reasons, including:
– when the sample is not soluble in water,
– when the sample contains impurities that
interfere with forward titration,
– when the end-point is more easily identified
than in forward titration.
24
Back Titration
• Consider using titration to measure the amount
of aspirin in a solution.
– Using titration it would be difficult to identify the end
point because aspirin is a weak acid and reactions
may proceed slowly.
• Using back titration the end-point is more easily
recognised in this reaction, as it is a reaction
between a strong base and a strong acid.
– This type of reaction occurs at a high rate and thus
produces an end-point which is abrupt and easily
seen.
25
Back Titration
• Step 1: Involves reacting the aspirin solution
with a measured amount of sodium hydroxide;
– an amount that will exceed the amount of aspirin
present.
– Because the hydrolysis reaction occurs at a very
low rate at room temperature it will be heated to
increase the reaction rate.
CH3COOC6H4COOH + 2NaOH  CH3COO.Na + HOC6H4COO.Na + H2O
aspirin
Sodium Ethanoate
Sodium-2-hydroxybenzoate
26
Back Titration
• Step 2: back titration of the un-hydrolyzed sodium
hydroxide solution with hydrochloric acid. This process
reacts the excess sodium hydroxide with hydrochloric
acid.
NaOH
+
Sodium hydroxide +
HCl

NaCl
+
H 2O
Hydrochloric acid  Sodium Chloride + Water
• By the method of back titration the amount of
hydrochloric acid needed to neutralize the
unreacted sodium hydroxide in the solution can
be determined.
– Knowing this and the amount of sodium hydroxide
that was added the amount of aspirin that reacted
with the sodium hydroxide can be determined.
27
Balancing REDOX Reactions
Na
 Na+ + eCl + e-  ClNa + Cl Na+ + Cl-
Oxidation half reaction
Reduction half reaction
Overall reaction
• One way to balance REDOX reactions is
by the half reaction method:
– Divides the overall reaction into oxidation and
reductions reactions
Rules for balancing redox reactions:
Consider the following reaction:
Cr2O72-(aq) + I-(aq)  Cr3+(aq) + I2 (s) (in acidic solution)
1.
Write as much of the unbalanced
equation as possible, omitting
spectator ions
2.
Divide reaction into 2 halfreactions, each of which
contains the oxidized & reduced
forms of one of the species:
Cr2O72I-
 Cr3+ Reduction
 I2
Oxidation
Rules for balancing redox reactions:
3. Balance atoms & charges in each half reaction:
a)
Balance atoms other than O and H:
Cr2O72-  2Cr3+
b)
Balance O atoms by adding H2O molecules:
Cr2O72-  2Cr3+ + 7 H2O
c)
Balance H atoms by adding H+ ions:
Only add H+ in
14H+ + Cr2O72-  2Cr3+ + 7 H2O acidic solution
c)
Balance charge by adding e-. They are added to left in
reduction half reaction because the reactants gains them;
they are added to the right in the oxidation half-reaction
because the reactant loses them.
6e- + 14H+ + Cr2O72-  2Cr3+ + 7 H2O
Rules for balancing redox reactions:
Adding H+, OH-, or H2O to balance oxygen or hydrogen
In acidic or neutral solution:
To Balance O
____________________________
For each O needed, add one H2O
And
then
To Balance H
___________________________
For each needed, add one H+
In basic solution:
To Balance O
__________________________
For each O needed, add one H2O
And
then
To Balance H
______________________________
For each H needed, add one H2O
to side needing H
and
Add one OH- to other side
Rules for balancing redox reactions:
3.
Balance atoms & charges in each half reaction:
For the other half reaction:
I-  I2
a) Balance atoms other than O and H:
2I-  I2
b) Balance O atoms by adding H2O molecules:  not
needed
c) Balance H atoms by adding H+ ions:  not needed
b) Balance charge by adding e-.
2I-  I2 + 2e-
Rules for balancing redox reactions:
4. If necessary, multiply one or both half reactions by
an integer to make # e-s gained = # e-s lost
6e- + Cr2O72-  2Cr3+
3 (2I-  I2 + 2e-)

6I-  3I2 + 6e-
5. Add balanced half-reactions & include states of
matter, cancel stuff that appears on both sides:
+ Cr2O72-  2Cr3+ + 7 H2O
6I-  3I2 + 6e6I- (aq) + 14H+ (aq) + Cr2O72- (aq)  3I2 (s) + 2Cr3+ (aq) + 7 H2O (l)
6e- + 14H+
NOTE: electrons must always cancel out
6. Check that atoms & charges are balanced
Calculations for Redox Titrations
• One method of analyzing samples quantitatively for the
presence of oxidizable or reducible substances is by
redox reactions.
– Concentration of a solution determines by carefully reacting it
with a standard solution of oxidizing or reducing agent
– E.g. potassium permanganate (KMnO4)
• Has an intense purple colour, therefore acts as its own indicator.
(a)
(b)
(a) Nearly colouless FeSO4
solution is titrated with deep
purple KMnO4 (b) the end
point is the point at which
the solution becomes pink,
owing to a very small
excess of KMnO4
Calculations for Redox Titrations
• Potassium dichromate, K2CrO7 is another
common oxidizing agent used in redox
titrations
• However, an indicator must be used
• K2CrO7 is orange and its reduced product,
Cr3+ is green
Calculations for Redox Titrations
• Just as we have done stoichiometry with
acid-base reactions, it can also be done
with redox reactions.
• Example 7: What volume of 0.200 M
KMnO4 is required to oxidize 35.0 mL of
0.150 M HCl? The balanced reaction is:
2 KMnO 4  16 HCl  2 KCl + 2 MnCl 2  5 Cl2  8 H2O
36
Calculations for Redox Titrations
Plan:
• Find # mols of HCl
• Use reaction ratio to determine # mols KMnO4
• Use given molarity of KMnO4 and # mols to find volume
2 KMnO 4  16 HCl  2 KCl + 2 MnCl 2  5 Cl2  8 H2O
0.035 L HCl 0.150 M HCl   0.00525 mol HCl
0.00524 mol HCl  2 mol KMnO 4   0.000656 mmol KMnO 4
 16 mol HCl 

1L
0.000656 mol KMnO 4 
 0.200 mol KMnO 4

  0.00328 L  3.28mL

37
Calculations for Redox Titrations
• Example 8: A volume of 40.0 mL of iron (II)
sulfate is oxidized to iron (III) by 20.0 mL of
0.100 M potassium dichromate solution.
What is the concentration of the iron (II)
sulfate solution?
• The balanced equation is:
2
6 Fe  Cr2O7  14 H +  6 Fe3+  2 Cr3+  7 H 2O
2+
You do it!
38
Calculations for Redox Titrations
20.0 mL 0.100 M Cr2O7

2
  2.00 mmol Cr O
2
2
7

2+


6
mmol
Fe
2
2+

2.00 mmol Cr2 O 7 

12
.
0
mmol
Fe
2 
 1 mmol Cr2 O 7 
12.0 mmol Fe2+
 0.300 M Fe2+
40.0 mL
39
Back (Indirect) Titration to Determine the
Concentration of a Volatile Substance
A student was asked to determine the concentration of ammonia, a volatile
substance, in a commercially available cloudy ammonia solution used for
cleaning.
First the student pipetted 25.00mL of the cloudy ammonia solution into a
250.0mL conical flask.
50.00mL of 0.100M HCl(aq) was immediately added to the conical flask which
reacted with the ammonia in solution.
The excess (unreacted) HCl was then titrated with 0.050M Na2CO3(aq).
21.50mL of Na2CO3(aq) was required.
Calculate the concentration of the ammonia in the cloudy ammonia
solution.
40