1. Reactions in Aqueous Solutions II: Calculations
Unit 7
Reactions in Aqueous Solutions II: Calculations 1 1 1 1 1 1 1 1

2. Calculations Involving Molarity
Example 1: If mL of 1.00 M NaOH and mL of 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 2 2 2 3 3 3 3

3. Recall… Molarity = mols of solute L of solution OR = mmol of solute
ml solution
1 mol = 1000mmol or 1mmol = 10-3 mol
1L = ml
Therefore, molarity = no. mol = no. mmol
L ml

4. Calculations Involving Molarity
Reaction
Ratio:
2 mol
1 mol
Before
Reaction:
100ml (1.0mol)
1000mL/ L
100ml (0.5mol)
1000mL/L
0.1 mol NaOH
0.05 mol H2SO4
0 mol
After
0.05 mol
0.1 mol
Note: The NaOH and H2SO4 neutralize each other exactly 3 3 3 3 4 4 4 4

5. Calculations Involving Molarity
What is the total volume of solution?
100.0 mL mL = 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 ml
L ml L
= M Na2SO4 4 4 4 4 5 5 5 5

6. Calculations Involving Molarity
Example 2: If mL of 1.00 M KOH and mL of 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? 5 5 5 6 6 6 6

7. Calculations Involving Molarity
Reaction
Ratio:
2 mol
1 mol
Before
Reaction:
130ml (1mol)
1000ml /L
100ml (0.5mol)
1000mL /L
0.13 mol
0.05 mol
0 mol
After
0.03 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. 6 6 6 7 7 7 7

8. Calculations Involving Molarity
What is the total volume of solution?
130.0 mL mL = 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 ml
230.0 mL L = M KOH
? M K2SO4= 0.05 mol x 1000ml
230.0 mL L = M K2SO4 7 7 7 8 8 8 8

9. Calculations Involving Molarity
Example 3: What volume of M NaOH solution would be required to completely neutralize 100 mL of M H3PO4?
You try it! 8 8 8 9 9 9 9

10. Calculations Involving Molarity
You can also do these kinds of problems following the above plan 9 9 9 10 10 10 10

11. 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). 10 10 10 11 11 11 11

12. Titrations Properties of an ideal primary standard
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

13. 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.

14. 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. 10 10 10 11 11 11 11

15. Titrations
Acid-base Titration Terminology 10 10 10 11 11 11 11

16. 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 g/mol. 11 11 11 12 12 12 12

17. 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.

18. Standardization of a Base Solution
Example 4: Calculate the molarity of a NaOH solution if 27.3 mL of it reacts with g of KHP. 12 12 13 13 13 13

19. Standardization of an Acid Solution
Example 5: Calculate the molarity of a sulfuric acid solution if 23.2 mL of it reacts with g of Na2CO3.
You do it! 14 15 14 15 15 15

20. Calculations for Acid-Base Titrations
Example 6: An impure sample of potassium hydrogen phthalate, KHP, had a mass of g. It was dissolved in water and titrated with 31.5 mL of M NaOH solution. Calculate the percent purity of the KHP sample. Molar mass of KHP is g/mol.
NaOH + KHP  NaKP + H2O
You do it! 16 16 17 17 17 17

21. 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

22. Calculations for Acid-Base Titrations18 18 19 20 20 20

23. 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.

24. 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.

25. 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.

26. 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 NaOH  CH3COO.Na + HOC6H4COO.Na + H2O
aspirin Sodium Ethanoate Sodium-2-hydroxybenzoate

27. 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 HCl  NaCl H2O
Sodium hydroxide + 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.

28. Balancing REDOX Reactions
Na + Cl Na+ + Cl-
Na  Na+ + e-
Cl + e-  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

29. Rules for balancing redox reactions:
Consider the following reaction:
Cr2O72-(aq) + I-(aq)  Cr3+(aq) I2 (s) (in acidic solution)
Write as much of the unbalanced equation as possible, omitting spectator ions
Divide reaction into 2 half-reactions, each of which contains the oxidized & reduced forms of one of the species:
Cr2O  Cr3+
I  I2
Reduction
Oxidation

30. Rules for balancing redox reactions:
Balance atoms & charges in each half reaction:
Balance atoms other than O and H:
Cr2O  2Cr3+
Balance O atoms by adding H2O molecules:
Cr2O  2Cr H2O
Balance H atoms by adding H+ ions:
14H Cr2O  2Cr H2O
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.
Only add H+ in acidic solution
6e- +

31. Rules for balancing redox reactions:
Adding H+, OH-, or H2O to balance oxygen or hydrogen
In acidic or neutral solution:
In basic solution:
To Balance O
____________________________
For each O needed, add one H2O
To Balance O
__________________________
For each O needed, add one H2O
And then
And then
To Balance H
___________________________
For each needed, add one H+
To Balance H
______________________________
For each H needed, add one H2O to side needing H
and
Add one OH- to other side

32. Rules for balancing redox reactions:
Balance atoms & charges in each half reaction:
For the other half reaction:
I  I2
Balance atoms other than O and H:
2I  I2
Balance O atoms by adding H2O molecules:  not needed
Balance H atoms by adding H+ ions:  not needed
Balance charge by adding e-.
2I  I2 + 2e-

33. Rules for balancing redox reactions:
If necessary, multiply one or both half reactions by an integer to make # e-s gained = # e-s lost
6e- + Cr2O  2Cr3+
3 (2I-  I2 + 2e-) 
6I  3I2 + 6e-
Add balanced half-reactions & include states of matter, cancel stuff that appears on both sides:
6e- + 14H Cr2O  2Cr H2O
6I- (aq) H+ (aq) + Cr2O72- (aq)  3I2 (s) + 2Cr3+ (aq) + 7 H2O (l)
NOTE: electrons must always cancel out
Check that atoms & charges are balanced

34. 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

35. 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

36. 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 M KMnO4 is required to oxidize 35.0 mL of M HCl? The balanced reaction is: 52 51 54 54 53

37. 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 53 52 55 55 54

38. 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 M potassium dichromate solution. What is the concentration of the iron (II) sulfate solution?
The balanced equation is:
You do it! 53 54 56 56 55

39. Calculations for Redox Titrations54 55 57 57 56

40. 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) mL of Na2CO3(aq) was required.
Calculate the concentration of the ammonia in the cloudy ammonia solution.