Titration, Normality and Equivalency Exercise
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Titration, Normality and Equivalency Exercise *
Complete this worksheet to the phrase "STOP POINT - request a quiz.". Do not hand in this
work sheet. When you are ready, you will be given an examination over this material. Complete
the examination by yourself and hand it in to receive credit.†
For stoichiometry problems, one uses a balanced reaction equation. The number of moles of the
limiting reactant and the number of moles of product produced are indicated by the coefficients
of this balanced equation. The coefficients of the balanced reaction is the ratio that the number
of moles involved in the reaction.
For example, in the following reaction have CH4 be the limiting reactant.
CH4 + 2O2 ! CO2 + 2 H2O
The question is asked, "How many moles of H2O are produced for a given number of moles of
CH4?" The coefficients of interest are underlined. Notice that the underlined blank implies a
coefficient of 1. Thus one writes the ratio of these number of moles of CH4 to the number of
moles of H2O as:
moles of CH4 : moles H2O = 1 : 2
or algebraically:
In a titration, the reaction is performed or measured in such a way as to make the coefficients of
both reactants indicate the molar relationship. Key words to look for that indicate this situation
are "endpoint", "titration", "equivalence point" and "exactly neutralizes". For example in the
following reaction, an endpoint was determined after a certain amount of HCl was used.
Ca(OH)2 + 2HCl ! CaCl2 + 2H2O
Thus one has:
moles HCl : moles Ca(OH)2 = 2 : 1
*
In most cases the designation of the species that the formula symbol refers to will follow the
2007 IUPAC convention of placing it in parenthesis.
†
The IUPAC convention is followed for variables and units. Italic type is used to indicate an
algebraic parameter or variable and NORMAL TYPE is used for units.
Titration, Normality and Equivalency Exercise
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or algebraically:
This language can apply to both an acid-base reaction as above or to a redox reaction.
For example, the following reaction in acid solution is taken to an endpoint.
2H2O + 2MnO4– + 5C2O42– ! 2Mn2+ + 10CO32– + 4H+
One may add the spectator ions, K+ and SO42–. and also make the observation that in acid solution
the carbonic acid will decompose to give CO2 by the reaction:
H2CO3 ! H2O + CO2(gas)
This yields:
8H2SO4 + 2KMnO4 + 5K2C2O4 ! 2MnSO4 + 8H2O + 10CO2 + 6K2SO4
Therefore, in the titration (given the reaction above) the ratio is:
moles of KMnO4 : moles of K2C2O4 = 2 : 5
or algebraically:
If the above discussion is not clear, review stoichiometry. DO NOT PROCEED without
understanding how this is done. To do otherwise will be a waste of time and you will
accomplish little.
It is extremely important that you be able to balance redox reactions by the
half reaction method at this point and be able to tell how many electrons are
transferred.
Equivalents and normality
For titrations, there is another method of handling the stoichiometry. This method utilizes the
concept of normality. For some normality simplifies the calculation, but this is not the reason it
is needed. To obtain a meaningful answer from a titration with an unknown reactant, normality
(or something similar) is required*. See if this makes sense to you after doing this exercise.
*
For this reason, the IUPAC recommendation to eliminate normality is sadly mistaken.
Titration, Normality and Equivalency Exercise
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Definition of equivalents and normality:
To obtain the number of equivalents (eq) or the normality (N)* from the number of moles (mol)
or molarity (M) respectively, one must know the stoichiometry of the reaction involved. It is
impossible to make these conversions without this information.
Some preliminary definitions are needed to relate normality and number of equivalents to
molarity and moles.
Acid-Base reactions:
Steps for Acid-Base Reaction:
Step 1:
Determine the number of hydronium ions (H3O+ or H+ for short, remember H+ / H3O+) which are
donated by the acid or, in the case of a base, the number of hydroxide ions removed from the
base. The coefficient in front of the chemical formula is not relevant only the number of
hydronium ions or hydroxide ions that are removed from each formula unit.
Example 1:
Ca(OH)2 + 2HCl ! CaCl2 + 2H2O
For the acid HCl:
Number of H+ = 1 eq mol–1
That is, the number of hydronium ions removed from each HCl is 1.
For the base Ca(OH)2:
Number of H+ ( / OH-) = 2 eq mol–1 (NOTE UNITS)
That is, the number of hydroxides removed from each Ca(OH)2 is 2.
Notice that for a monoprotic acid or base the value for this is always 1.
Example 2:
2NaOH + H2SO4 ! Na2SO4 + 2H2O
For the acid H2SO4:
Number of H+ = 2 eq mol–1
That is, the number of hydroniums removed from each H2SO4 is 2.
*
The IUPAC (2007) formula symbol for molarity is C. The IUPAC unit is M for molarity.
There are no IUPAC formula and unit symbols for normality, therefore the traditional N and
N will be used for the algebraic symbol and the unit symbol respectively. Likewise, the
formula and unit symbols for equivalence will be given as [#eq] and eq. respectively.
Titration, Normality and Equivalency Exercise
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For the base NaOH:
Number of H+ (/ OH-) = 1 eq mol–1
That is, the number of hydroxides removed from each NaOH is 1.
Example 3:
NaOH + H2SO4 ! NaHSO4 + H2O
For the acid H2SO4:
Number of H+ = 1 eq mol–1
That is, the number of hydroniums removed from each H2SO4 is 1.
For the base NaOH:
Number of H+ (/ OH-) = 1 eq mol–1
That is, the number of hydroxides removed from each NaOH is 1.
Do you understand the difference between 2a and 2b? Notice how many H+s are being used.
Example 4:
NaOH + H3PO4 ! Na2HPO4 + 2H2O
For the acid H3PO4:
Number of H+ = 2 eq mol–1
That is, the number of hydroniums removed from each H3PO4 is 2.
For the base NaOH:
Number of H+ (/ OH-) = 1 eq mol–1
That is, the number of hydroxides removed from each NaOH is 1.
Answer the Following Questions:
1)
For the reaction:
NaOH + HCl ! NaCl + H2O
*
How many protons* are transferred to NaOH?
eq mol–1
How many protons does HCl supply?
eq mol–1
Just as H+ is used in the place of H3O+, the word “proton” is often substituted for the words
“hydronium ion”. Furthermore, a polyhydroxyl base is often referred to as a polyprotic base.
Titration, Normality and Equivalency Exercise
2)
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For the reaction:
Mg(OH)2 + 2HCl ! MgCl2 + 2H2O
3)
How many protons are transferred to Mg(OH)2?
eq mol–1
How many protons does HCl supply?
eq mol–1
For the reaction:
Mg(OH)2 + H2SO4 ! MgSO4 + 2H2O
4)
How many protons are transferred to Mg(OH)2?
eq mol–1
How many protons does H2SO4 supply?
eq mol–1
For the reaction:
Mg(OH)2 + 2H2SO4 ! Mg(HSO4)2 + 2H2O
5)
How many protons are transferred to Mg(OH)2?
eq mol–1
How many protons does H2SO4 supply?
eq mol–1
For the reaction:
3Ca(OH)2 + 2H3PO4 ! Ca3(PO4)2 + 6H2O
6)
How many protons are transferred to Ca(OH)2?
eq mol–1
How many protons does H3PO4 supply?
eq mol–1
For the reaction:
Ca(OH)2 + H3PO4 ! CaHPO4 + 2H2O
How many protons are transferred to Ca(OH)2?
eq mol–1
How many protons does H3PO4 supply?
eq mol–1
Titration, Normality and Equivalency Exercise
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Step 2A (eq from mol):
To obtain equivalents from moles, multiply by the H+ number.
[#eq] = [#H+ or OH–]@n
Example 5:
There were 0.35 mole of Ca(OH)2 in the reaction: Ca(OH)2 + 2HCl ! CaCl2 + 2H2O. How
many equivalents are there?
[#eq] = [#OH–]@n
[#eq(Ca(OH)2)] = 2 eq mol–1 @ 0.35
[#eq(Ca(OH)2)] = 0.70
Step 2B (N from M):
To obtain normality from molarity, multiply by the H3O+ number. The algebraic equation is:
N = [#H+ or OH–] @ c
Example 6:
A solution of Ca(OH)2 in the reaction: Ca(OH)2 + 2HCl 6 CaCl2 + 2H2O is 0.10 M. What is
the solution's normality?
or:
Do the following conversions:
7)
If one completely neutralizes a 0.1500 M solution of Ca(OH)2, what is the initial normality
of the Ca(OH)2 solution? (“Completely” implies all the H+ or OH– ions are used)
ANS
N
Titration, Normality and Equivalency Exercise
8)
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H3PO4 is reacted with NaOH to form Na3PO4. If one were to start with 3.50 equivalents of
H3PO4, how many moles of H3PO4 is this?
ANS
9)
mol
A 0.1002 M solution of H2SO4 is reacted with NaOH to produce NaHSO4 in a titration. If
24.5 mL of the H2SO4 solution is used, how many moles of H2SO4 solution is used?
ANS
mol
10) What is the molarity of a 1.035 N solution of HCl?
ANS
M
11) In an acid-base reaction in water, what is the normality of a 0.547 M solution of Ca(OH)2?
(In water, Ca(OH)2 reacts completely.)
ANS
N
Titration, Normality and Equivalency Exercise
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12) How many grams of NaOH is contained in 0.500 mL of a 1.00 N solution?
ANS
g
13) What is the normality of a solution of Mg(OH)2 for a reaction in which both OH- are
consumed if one has 1.34 mg of Mg(OH)2 dissolved in 1.00 L of solution?
ANS
N
Titration, Normality and Equivalency Exercise
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Redox Reactions:
The concept of equivalents and normality is also applied to oxidation-reduction reactions. The
conversions are handled the same as the acid-base conversions with the number of electrons
taking the role of the number of hydronium ions. The number of electrons are determined from
the half reaction. This number is the total number of electrons used for the half reaction (See the
footnote in "Balancing Redox Reactions on page 192)
Step 1:
Determine the number of electrons which are donated by the reducing agent or are accepted by
the oxidizing agent.
For example in the reaction:
8H2SO4 + 2KMnO4 + 5K2C2O4 ! 2MnSO4 + 8H2O + 10CO2 + 6K2SO4
The number of electrons for KMnO4 is 5 (or 5
and the number of electrons for K2C2O4 is 2 (or 2
/
)
/
).
NOTICE THE IMPORTANCE OF THE ABILITY TO WRITE HALF-REACTIONS HERE!
Write the half reactions below to confirm the numbers given.
For KMnO4:
For K2C2O4:
Titration, Normality and Equivalency Exercise
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For Step 2A (eq from mol):
To obtain equivalents from moles, multiply by the e– number.
Example 7:
If there are 0.35 mole of KMnO4 and 0.20 mole of K2C2O4, how many equivalents are there of
each? One would have:
[#
] = 5 @ 0.35
= 1.75
[#
] = 2 @ 0.20
= 0.40
For Step 2B (N from M):
To obtain normality from molarity, multiply by the e– number or:
N = [#e–] @ C
Example 8:
If there are 0.350 M KMnO4 solution and 0.020 M K2C2O4 solution, what is the normality of each
solution? One would have:
or:
= 5 @ 0.350
= 1.75
= 2 @ 0.020
= 0.040
Calculate the following:
14) For the reaction in acid (unbalanced):
Cr2O72– + SO32– ! Cr3+ + SO42–
A) Give the balanced half reactions:
For the Cr2O72– ion:
For the SO32– ion:
Titration, Normality and Equivalency Exercise
B)
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How many equivalents of K2Cr2O7 is 38.22 g?
ANS
C)
A 0.12 M solution of H2SO3 is what normality?
ANS
N
D) What is the molarity of a 0.1900 N K2Cr2O7 solution?
ANS
E)
M
If one has 0.200 moles of H2SO3, how many equivalents does one have?
ANS
eq
Titration, Normality and Equivalency Exercise
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15) For the reaction in acid (not balanced):
Fe2+ + MnO4– ! Fe3+ + Mn2+
A) Give the balanced half reactions:
For the Fe2+ ion:
For the MnO4– ion:
B)
How many grams are there in 1.50 eq. of FeSO4?
ANS
C)
g
What is the normality of a solution which contains 79.0 g of KMnO4 in 2.50 L of solution?
ANS
N
D) What is the molarity of a solution which has 0.150 eq of FeSO4 dissolved in 750.0 mL of
solution?
ANS
M
Titration, Normality and Equivalency Exercise
E)
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What is the normality of a solution which contains 24.99 g of FeSO4 in 1.333 L of solution?
ANS
N
Titration using equivalents and normality:
Notice that if one uses equivalencies and normality for a titration reaction problem, then one
need not know the reaction involved to get an answer, since at an end point:
X eq reactant #1 = Y eq reactant #2
and also since X N @ Y L = Z eq (the value for normality times the volume in L is equal to the
number of equivalents) then:
N1V1 = N2V2
provided the volumes are in the same units.
Example 9:
An unknown acid was titrated to a sharp endpoint with 0.115 N base. 25.00 mL of acid was used
and 34.55 mL of base was used. What is the normality of the acid?
Nacid × 25.00 mL = 0.115 N × 34.55 mL
Nacid = 0.159 N
Calculate the following:
16) An unknown acid is titrated with a standardized base solution. The concentration of the
base is 0.1022 N. 45.15 mL of the base solution was used to titrate
25.00 mL of the acid. What is the normality of the acid solution?
ANS
N
Titration, Normality and Equivalency Exercise
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17) In a titration to determine the concentration of Ca(OH)2 in a solution, both hydroxide ions
(OH–) were neutralized. 10.00 mL of the Ca(OH)2 solution were titrated with a standard
acid. This standard acid was 0.1100 N and 23.98 mL of it was used for the titration. What
is the molarity of the Ca(OH)2 solution?
ANS
M
18) In the reaction between FeSO4 (molar mass = 151.93) and KMnO4 in acidic solution,
Fe2(SO4)3 and MnSO4 are produced. A laboratory worker reacted 6.08 g of solid FeSO4
with a solution of KMnO4 until an endpoint is reached. 37.98 mL of the solution was used
in the titration. What is the normality and molarity of the solution?
ANS
N
ANS
M
19) An oxalic acid (H2C2O4) solution was used to standardize a potassium permanganate
(KMnO4) solution in a redox titration. Before using the oxalic acid as a standard, however,
the oxalic acid was standardized in a acid-base titration using NaOH solution. In the first
(acid-base) titration, 25.00 mL of acidic oxalic solution was used and 45.34 mL of NaOH
solution of 0.1011 N was used. Both protons are titrated in the acid-base reaction. In the
second (redox) titration, 25.00 mL of acidic oxalic solution was used and 35.03 mL of
potassium permanganate solution was used.
A) Write the balanced equation for the acid-base reaction.
B)
Calculate the normality and molarity of the oxalic acid in the acid-base case.
ANS
N
ANS
M
Titration, Normality and Equivalency Exercise
C)
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Write half reactions and the balanced equation for the redox reaction.
For H2C2O4:
For KMnO4:
D) What is the normality of the oxalic acid in the redox case?
ANS
E)
N
Calculate the normality and molarity of the potassium permanganate in the redox case.
ANS
N
ANS
M
20) A sample of sodium oxalate of mass 0.202 g was used to standardize a potassium
permanganate solution in a redox titration in an acidic solution. The potassium
permanganate solution was in turn used to find the percent iron in a 2.012 g sample.
32.41 mL of the solution was used in the standardization step and 41.51 mL of potassium
permanganate solution was used when reacting with iron. The molar mass of sodium
oxalate is 134.00 g/mol and the redox reaction involves two electrons from the sodium
oxalate. The iron reacted by changing from iron(II) to iron(III). What was the normality of
the potassium permanganate solution and what is the percent iron in the iron sample? Give
your answer to the proper number of significant digits.
Titration, Normality and Equivalency Exercise
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For the First Titration:
A) BALANCED REACTION
B)
Calculations for the normality of KMnO4:
ANS
N
For the second titration
C)
BALANCED REACTION:
D) Calculations to calculate the percent of Fe in the sample
ANS
% Fe
It is very important that you complete the previous questions before you obtain a quiz.
You will need the information before the next lab period.
If you have completed all 20 question, btain a quiz.