Stoichiometric Calculations
1.
Review
of Fundamental Concepts
2.
Molarity
Calculations
3.
Normality
Calculations
4.
Density
Calculations
5.
Analytical
versus Equilibrium Concentration
7.
Expression
of Analytical Results, %, ppt, ppm, ppb
8.
Volumetric
Analysis Using Molarity
9.
Back-Titrations
10. Volumetric
Analysis using Normality
11. The
Titer Concept
Review of Fundamental Concepts
Formula Weight
It is assumed that you can calculate
the formula or molecular weights of compounds from respective atomic weights of
the elements forming these compounds. The formula weight (FWt) of a substance
is the sum of the atomic weights of the elements from which this substance is
formed from.
The formula weight of CaSO4.7H2O
is
|
Element
|
Atomic weight
|
|
Ca
|
40.08
|
|
S
|
32.06
|
|
11 O
|
11x16.00
|
|
14 H
|
14x1.00
|
|
FWt
|
S = 262.14
|
The Mole
The mole is the major word we will
use throughout the course. The mole is defined as gram molecular weight which
means that:
|
Mole
|
Grams
|
|
|
1 mol H2
|
2.00 g
|
|
|
1 mol O2
|
32.00 g
|
|
|
1mol O
|
16.00 g
|
|
|
1mol NaCl
|
58.5 g
|
|
|
1 mol Na2CO3
|
106.00 g
|
|
|
|
|
|
Assuming approximate atomic weights
of 1.00, 16.00, 23.00, 35.5, and 12.00 atomic mass unit for hydrogen, oxygen,
sodium, chlorine atom, and carbon, respectively.
The number of moles contained in a
specific mass of a substance can be calculated as:
mol = g substance/FWt
substance
The unit for the formula weight is
g/mol
In the same manner, the number of mmol
of a substance contained in a specific weight of the substance can be
calculated as
mmol = mol/1000
Or,
mmol = mg
substance/FWt substance
Look at this calculation
The number of mmol of Na2WO4 (FWt = 293.8 mg/mmol) present
in 500 mg of of Na2WO4 can be calculated as
? mmol of Na2WO4
= 500 mg/293.8 mg/mmol = 1.70 mmol
The number of mg contained in 0.25 mol of Fe2O3 (FWt =
159.7 mg/mmol) can be calculated as
? mg Fe2O3 =
0.25 mmol Fe2O3 x 159.7 mg/mmol = 39.9 mg
Therefore, either the number of mg
of a substance can be obtained from its mmols or vice versa.
Molarity of a solution can be
defined as the number of moles of solute dissolved n 1 L of solution. This
means that 1 mol of solute will be dissolved in some amount of water and the
volume will be adjusted to 1 L. The amount of water may be less than 1 L as the
final volume of solute and water is exactly 1 L.
Calculations Involving Molarity
These are very simple where
Molarity = mol/L = mmol/mL
This can be further formulated as
Number of moles =
Molarity X volume in Liters, or
mol = M (mol/L) x VL
Number of mmol =
Molarity x volume in mL , or
mmol = M (mmol/mL) x VmL
Look at the following calculations
involving molarity
Find the molarity of a solution
resulting from dissolving 1.26 g of AgNO3 (FWt = 169.9 g/mol) in a
total volume of 250 mL solution.
First find mol AgNO3 =
1.26 g AgNO3 / 169.9 g/mol = 7.42x10-3 mol
Also you should know that 250 mL =
0.25 L
Now one can calculate the molarity
directly from
Molarity = mol/L
M = 7.42x10-3 mol/0.25 L
= 0.0297 mol/L
We can find the molarity directly in
one step using dimensional analysis
? mol AgNO3
/ L = (1.26 g AgNO3 / 250 mL)
x ( mol AgNO3/169.9 g AgNO3)
x (1000 mL/1L) =0.0297 M
Let
us find the number of mg of NaCl per mL of a 0.25 M NaCl solution
First we should be able to recognize
the molarity as 0.25 mol/L or 0.25 mmol/mL. Of course, the second term offers
what we need directly
? mg NaCl in 1 mL = (0.25 mmol NaCl/mL) x (58.5
mg NaCl/mmol NaCl) = 14.6 mg NaCl/mL
A more practical problem
is to calculate the weight of a substance needed to prepare a specific volume
of a solution with predefined concentration. For example
Find the number of grams of Na2SO4
required to prepare 500 mL of 0.1 M solution.
First, we find mmoles needed from
the relation
mmol = M (mmol/mL)
x VmL
mmol Na2SO4 =
0.1 mmol/mL x 500 mL = 50 mmol
mmol = mg substance/FWt substance
? mg Na2SO4 =
50 mmol x 142 mg/mmol = 7100 mg or 7.1 g
One can use dimensional
analysis to find the answer in one step as
? g Na2SO4 = (0.1 mol Na2SO4/1000 mL) x
500 mL x (142
g Na2SO4/mol) = 7.1 g
When two or more solutions are
mixed, one can find the final concentration of each ion. However, you should
always remember that the number of moles ( or mmoles ) is additive. For example
Find the molarity of K+ after
mixing 100 mL of 0.25 M KCl with 200 mL of 0.1 M K2SO4.
The idea here is to calculate the
total mmol of K+ and divide it by volume in mL.
mmol K+ = mmol
K+ from KCl + mmol K+ fro K2SO4
= 0.25
mmol/ml x 100 mL + 2x0.1 mmol/mL x 200
mL = 65 mmol
Molarity = 65 mmol/(200 + 100) mL = 0.22 M
Note that the
concentration of K+ in 0.1M K2SO4 is 2x0.1M
(i.e. 0.2 M)
We have previously talked about
molarity as a method for expressing concentration. The second expression used
to describe concentration of a solution is the normality. Normality can be
defined as the number of equivalents of solute dissolved in 1 L of solution.
Therefore, it is important for us to define what we mean by the number of
equivalents, as well as the equivalent weight of a substance as a parallel term
to formula weight.
An equivalent is defined as the
weight of substance giving an Avogadro’s number of reacting units. Reacting
units are either protons (in acid base reactions) or electrons (in oxidation
reduction reactions). For example, HCl has one reacting unit (H+)
when reacting with a base like NaOH but sulfuric acid has two reacting units
(two protons) when reacting completely with a base. Therefore, we say that the
equivalent weight of HCl is equal to its
formula weight and the equivalent weight of sulfuric acid is one half its
formula weight. In the reaction where Mn(VII), in KMnO4, is reduced to Mn(II) five electrons are
involved and the equivalent weight of KMnO4 is equal to its formula
weight divided by 5.
Number of equivalents =
Normality x VL = (eq/L) x L
Number of milliequivalents
= Normality x VmL = (meq/mL)
x mL
Also, number of
equivalents = wt(g)/equivalent weight (g/eq)
Or, number of milliequivalents
= wt(mg)/equivalent weight (mg/meq)
Let us not be lost by the above
arguments and make concepts more practical. I think you just need to know the
following to answer any problem related to normality calculations:
Equivalent weight = FWt /n
Number of equivalents = n x number
of moles
Also, N = n M
Where n is the number of reacting
units ( protons or electrons ) and if you are forming factors always remember
that a mole contains n equivalents. The factor becomes (1 mol/n eq) or (n eq/1 mol).
One last thing to keep in mind is
that when dealing with normality problems always 1 eq of A reacts with 1 eq of
B regardless of the stoichiometry of the reaction since this stoichiometry was
used in the calculation of normalities.
Example
Find the equivalent weights of NH3
(FW = 17.03), H2C2O4
(FW = 90.04) in the reaction with excess
NaOH, and KMnO4 (FW = 158.04) when Mn(VII) is reduced to Mn(II).
Solution
Ammonia reacts with one proton only
Equivalent weights of NH3
= FW/1 = 17.03 g/eq
Two protons of oxalic acid react
with the base
Equivalent weights of H2C2O4
= FW/2 = 90.04/2 = 45.02 g/eq
Five electrons are
involved in the reduction of Mn(VII) to Mn(II)
Equivalent weights of KMnO4
= FW/5 = 158.04/5 = 31.608 g/eq
Example
Find the normality of the solution
containing 5.300 g/L of Na2CO3
(FW = 105.99), carbonate reacts with two protons.
Solution
Normality is the number of
equivalents per liter, therefore we first find the number of equivalents
eq wt = FW/2 = 105.99/2 = 53.00
eq = Wt/eq wt = 5.300/53.00 = 0.1000
N = eq/L = 0.1000 eq/1L = 0.1000 N
The problem can be worked out simply
as below
? eq Na2CO3 /L = (5.300
g Na2CO3 /L )
x (1 mol Na2CO3 /105.99 g Na2CO3
) x (2 eq Na2CO3
/1 mol Na2CO3) = 0.1 N
The other choice is to find the molarity first and the
convert it to normality using the relation
N = n M
No of mol = 5.300 g/(105.99 g/mol)
M = mol/L = [5.300
g/(105.99 g/mol)]/ 1L
N = n M = 2 x [5.300 g/(105.99 g/mol)]/ 1L = 0.1000
A further option is to find the number of moles first
followed by multiplying the result by 2 to obtain the number of equivalents.
Example
Find the normality of the solution
containing 5.267 g/L K2Cr2O7 (FW = 294.19) if
Cr6+ is reduced to Cr3+.
Solution
The same as the previous example
N = eq/L , therefore
we should find the number of eq where eq = wt/eq wt , therefore we should find the equivalent
weight where eq wt = FW/n. Here each Contributes three electrons and
since the dichromate contains two Cr atoms we have 6 reacting units
Eq wt = (294.19 g/mol)/(6 eq/mol)
Eq = 5.267 g/ (294.19 g/mol)/(6 eq/mol)
N = eq/L = (294.19 g/mol)/(6 eq/mol)/1L = 0.1074 eq/L
Using the dimensional analysis we
may write
? eq
K2Cr2O7 /L = (5.267 g K2Cr2O7
/L) x (mol K2Cr2O7
/294.19 g K2Cr2O7 ) x (6 eq K2Cr2O7 /mol K2Cr2O7
) = 0.0174 eq/L
Again one can choose
to calculate the molarity then convert it to normality
mol = 5.267 g/(294.19 g/mol)
M = mol/L = [5.267 g/(294.19
g/mol)]/L
N = n M
N = (6 eq/mol)x [5.267 g/(294.19 g/mol)]/L = 0.1074 eq/L
In this section, you will learn how
to find the molarity of solution from two pieces of information ( density and
percentage). Usually the calculation is simple and can be done using several
procedures. Look at the examples below:
Example
What volume of concentrated HCl (FW
= 36.5g/mol, 32%, density = 1.1g/mL) are required to prepare 500 mL of 2.0 M
solution.
Solution
Always start with the density and
find how many grams of solute in each mL of solution. Remember that only a
percentage of the solution is solute .
g HCl/ml = 1.1 x 0.32 g HCl / mL
The problem is now simple as it
requires conversion of grams HCl to mmol since the molarity is mmol per mL
mmol HCl = 1.1x0.32 x103
mg HCl/(36.5 mg/mmol) = 9.64 mmol
M = 9.64 mmol/mL
= 9.64 M
Now, we can calculate the volume required from the relation
MiVi
(before dilution) = MfVf (after dilution)
9.64 x VmL= 2.0 x 500mL
VmL = 104 mL
This means that 104 mL of the
concentrated HCl should be added to distilled water and the volume should then
be adjusted to 500 mL
Example
How many mL of concentrated H2SO4
(FW = 98.1 g/mol, 94%, d = 1.831 g/mL)
are required to prepare 1 L of 0.100 M solution?
Solution
g H2SO4 / mL =
1.831 x 0.94 g /mL
Now find the mmol acid present
mmol H2SO4 =
(1.831 x 0.94 x 1000 mg) / (98.1 mg/mmol)
The molarity can then be calculated
as
M = mmol/mL = [(1.831 x 0.94 x 1000
mg) / (98.1 mg/mmol)] / mL = 17.5 M
To find the volume required to
prepare the solution
MiVi
(before dilution) = MfVf (after dilution)
17.5 x VmL = 0.100 x 1000
mL
VmL = 5.71 mL which
should be added to distilled water and then adjusted to 1 L.
When we prepare a solution by
weighing a specific amount of solute and dissolve it in a specific volume of
solution, we get a solution with specific concentration. This concentration is
referred to as analytical concentration.
However, the concentration in solution may be
different from the analytical concentration, especially when partially
dissociating substances are used. An example would be clear if we
consider preparing 0.1 M acetic acid (weak acid) by dissolving 0.1 mol of the
acid in 1 L solution. Now, we have an analytical concentration of acetic acid (HOAc)
equals 0.1 M. But what is the actual equilibrium concentration of HOAc?
We have
HOAc = H+ + OAc-
The analytical
concentration ( CHOAc ) = 0.1 M
CHOAc = [HOAc]undissociated
+ [OAc-]
The equilibrium
concentration = [HOAc]undissociated.
For good electrolytes which are 100%
dissociated in water the analytical and equilibrium concentrations can be
calculated for the ions, rather than the whole species. For example, a 1.0 M CaCl2 in water results
in 0 M CaCl2, 1.0 M Ca2+, and 2.0 M Cl- since
all calcium chloride dissociates in solution. For species x we express the analytical
concentration as Cx and the
equilibrium concentration as [x]. All concentrations we use in calculations
using equilibrium constants are equilibrium concentrations.
In many cases, a dilution step or
steps are involved in analytical procedures. One should always remember that in
any dilution the number of mmoles of the initial (concentrated) solution is
equal to the number of mmoles of the diluted solution. This means
MiVi (concentrated) = MfVf
(dilute)
Example
Prepare 200 mL of 0.12 M KNO3
solution from 0.48 M solution.
Solution
MiVi
(concentrated) = MfVf (dilute)
0.48 x VmL = 0.12 x 200
VmL = (0.12 x 200)/0.48 =
50 mL
Therefore, 50 mL of 0.48
M KNO3 should be diluted to 200 mL to obtain 0.12 M solution
Example
A 5.0 g Mn sample was dissolved in
100 mL water. If the percentage of Mn (At wt = 55 g/mol) in the sample is about
5%. What volume is needed to prepare 100 mL of approximately 3.0x10-3
M solution.
Solution
First we find approximate mol Mn in
the sample = 5.0 x (5/100) g Mn/(55 g/mol) = 4.5x10-3 mol
Molarity of Mn solution = (4.5x10-3
x 103 mmol)/100 mL = 4.5x10-2 M
The problem can now be solved easily
using the dilution relation
MiVi
(concentrated) = MfVf (dilute)
4.5x10-2 x VmL
= 3.0x10-3 x 100
VmL = (3.0x10-3
x 100)/4.5x10-2 = 6.7 mL
Therefore, about 6.7 mL
of the Mn sample should be diluted to obtain an approximate concentration of
3.0x10-3 M solution.
A result can be reported as a
percentage, part per thousand (ppt), part per million (ppm), part per billion (ppb),
etc.. Solutes can be solids , liquids,
or gases. The concentration can be expressed as weight solute per weight sample
(w/w), weight solute per volume solution (w/v), or volume solute per volume
solution (v/v).
For Solid Solutes
% (w/w) = [weight solute (g)/weight
sample (g)] x 100
ppt (w/w) = [weight solute
(g)/weight sample (g)] x 1000
ppm (w/w) = [weight solute
(g)/weight sample (g)] x 106
ppb (w/w) = [weight solute
(g)/weight sample (g)] x 109
A ppm can be
represented by several terms like the one above, (mg solute/kg sample), ( g
solute/106g sample), etc..
If the solute is
dissolved in solution we have
% (w/v) = [weight solute (g)/volume
sample (mL)] x 100
ppt (w/v) = [weight solute
(g)/volume sample (mL)] x 1000
ppm (w/v) = [weight solute
(g)/volume sample (mL)] x 106
ppb (w/v) = [weight solute
(g)/volume sample (mL)] x 109
Also a ppm can be
expressed as above or as (g solute/106 mL solution), (mg solute/L
solution), or (mg/mL), etc..
For Liquid Solutes
% (v/v) = [volume solute (mL)/volume
sample (mL)] x 100
ppt (v/v) = [volume solute
(mL)/volume sample (mL)] x 1000
ppm (v/v) = [volume solute (mL)/volume sample (mL)] x 106
ppb (v/v) = [volume solute (mL)/volume sample (mL)] x 109
A ppm can be expressed as above or
as (mL/L), (mL/103 L), etc..
It is very important
here to recognize the similarity between the abovementioned relations for easy
remembrance of these terms when needed.
Example
A 2.6 g sample was analyzed and
found to contain 3.6 mg zinc. Find the concentration of zinc in ppm and ppb.
Solution
A ppm is microgram solute
per gram sample, therefore
Ppm Zn = 3.6 mg
Zn/2.6 g sample = 1.4 ppm
A ppb is nanogram
solute/gram sample, therefore
Ppb Zn = 2.6 x103 ng
Zn/2.6 g sample = 1400 ppb
Example
A 25.0 mL
sample was found to contain 26.7 mg glucose. Express the concentration
as ppm and mg/dL glucose.
Solution
A ppm is defined as mg/mL, therefore
Ppm = 26.7 mg/(25.0x10-3
mL) = 1.07x103 ppm
Or one can use dimensional analysis
considering always a ppm as mg/L as below
? mg/L glucose = (26.7 mg/25.0 mL) x (10-3
mg/mg) x (106 mL/L)
= 1.07x103 ppm
Now let us find mg glucose per
deciliter
?mg glucose/dL = = (26.7 mg/25.0 mL) x (10-3 mg/mg) x (106 mL/L)
x (L/10dL) = 107 mg/dL
The other important type
of problems is to change from ppm or ppb into molarity and vice versa
Example
Find the molar concentration of a
1.00 ppm Li (at wt = 6.94 g/mol) and Pb (at wt = 207 g/mol).
Solution
A 1.00 ppm is 1.00 mg/L,
therefore change this 1.00 mg into moles to obtain molarity.
? mol Li/L = (1.00x10-3 g
Li/L) x ( 1 mol Li/6.94 g Li) = 1.44x10-4 M
? mol Pb/L = (1.00x10-3 g
Pb/L) x ( 1 mol Pb/207 g Li) = 4.83x10-6 M
Example
Find the number of mg Na2CO3
(FW = 106 g/mol) required to prepare 500 mL of 9.20 ppm Na solution.
Solution
The idea is to find the of mg sodium
( 23.0 mg/mmol) required and then get the mmoles sodium and relate it to mmoles
sodium carbonate followed by calculation of the weight of sodium carbonate.
? mg Na = 9.20 mg/L x 0.5 L = 4.6 mg
Na
mmol Na = 4.60 mg Na/23.0 mg/mmol
mmol Na2CO3
= 1/2 mmol Na = 4.60/46.0 mmol = 0.100
? mg Na2CO3 =
(4.60/46.0) mmol Na2CO3 x (106
mg Na2CO3/ mmol Na2CO3) =
10.6 mg
One can work such a
problem in one step as below
? mg Na2CO3 = (9.2 mg Na/1000mL) x 500 mL x (1mmol Na/23.0 mg Na) x (1
mmol Na2CO3/2 mmol Na) x (106
mg Na2CO3/1 mmol Na2CO3)
= 10.6 mg
In this section we look at calculations involved in
titration processes as well as general quantitative reactions. In a volumetric
titration, an analyte of unknown concentration is titrated with a standard in
presence of a suitable indicator. For a reaction to be used in titration the
following characteristics should be satisfied:
1. The stoichiometry of the reaction should be exactly known.
This means that we should know the number of moles of A reacting with 1 mole of
B.
- The reaction should be rapid and reaction between A and B should occur immediately and instantly after addition of each drop of titrant (the solution in the burette).
- There should be no side reactions. A reacts with B only.
- The reaction should be quantitative. A reacts completely with B.
- There should exist a suitable indicator which has distinct color change.
- There should be very good agreement between the equivalence point (theoretical) and the end point (experimental). This means that Both points should occur at the same volume of titrant or at most a very close volume. Three reasons exist for the disagreement between the equivalence and end points. The first is whether the suitable indicator was selected, the second is related to concentration of reactants, and the third is related to the value of the equilibrium constant. These factors will be discussed in details later in the course.
Standard solutions
A standard solution is a solution of known and exactly
defined concentration. Usually standards are classified as either primary standards or secondary standards. There
are not too many secondary standards available to analysts and standardization
of other substances is necessary to prepare secondary standards. A primary standard
should have the following properties:
- Should have a purity of at least 99.98%
- Stable to drying, a necessary step to expel adsorbed water molecules before weighing
- Should have high formula weight as the uncertainty in weight is decreased when weight is increased
- Should be non hygroscopic
- Should possess the same properties as that required for a titration
NaOH and HCl are not primary standards and therefore should be standardized using a primary or
secondary standard. NaOH absorbs CO2 from air, highly hygroscopic,
and usually of low purity. HCl and other acid in solution are not standards as
the percentage written on the reagent bottle is a claimed value and should not
be taken as guaranteed.
Molarity Volumetric
Calculations
Volumetric calculations involving molarity are rather
simple. The way this information is presented in the text is not very helpful.
Therefore, disregard and forget about all equations and relations listed in
rectangles in the text, you will not need it. What you really need is to use
the stoichiometry of the reaction to find how many mmol of A as compared to the number of mmoles of B.
Example
A 0.4671 g sample containing NaHCO3 (FW = 84.01
mg/mmol) was dissolved and titrated with 0.1067 M HCl requiring 40.72 mL. Find
the percentage of bicarbonate in the sample.
Solution
We should write the equation in order to identify the stoichiometry
NaHCO3
+ HCl = NaCl + H2CO3
Now it is clear that the number of mmol of bicarbonate is equal to the number
of mmol HCl
Mmol NaHCO3 = mmol HCl
Mmol = M x VmL
Mmol NaHCO3 = ( 0.1067 mmol/ml ) x 40.72 mL =
4.345 mmol
Now get mg bicarbonate by multiplying mmol times FW
Mg NaHCO3 = 4.345 mmol x (84.01 mg/mmol) = 365.01
% NaHCO3 = (365.01 x 10-3
g/0.4671 g) x 100 = 78.14%
We can use dimensional analysis to calculate the mg NaHCO3
directly then get the percentage as
above.
? mg NaHCO3 = (0.1067
mmol HCl/ml) x 40.72 mL x (mmol NaHCO3/mmol
HCl) x (84.01 mg NaHCO3/ mmol
NaHCO3) = 365.0 mg
Example
A 0.4671 g sample containing Na2CO3
(FW = 106mg/mmol) was dissolved and titrated with 0.1067 M HCl requiring 40.72
mL. Find the percentage of carbonate in the sample.
Solution
The equation should be the first thing to formulate
Na2CO3
+2 HCl = 2NaCl + H2CO3
Mmol Na2CO3 = ½ mmol HCl
Now get the number of mmol Na2CO3
= ½ x ( MHCl x VmL (HCl) )
Now get the number of mmol Na2CO3 = ½
x 0.1067 x 40.72 = 2.172 mmol
Now get mg Na2CO3 = mmol x FW = 2.172 x 106 = 230 mg
% Na2CO3 = (230 x 10-3
g/0.4671 g ) x 100 = 49.3 %
Example
How many mL of 0.25 M NaOH will react with 5.0 mL of 0.10 M
H2SO4.
H2SO4
+ 2 NaOH = Na2SO4
+ 2 H2O
Solution
Mmol NaOH = 2 mmol H2SO4
Mmol NaOH = 2 {M (H2SO4)
x VmL (H2SO4)}
Mmol NaOH = 2 x 0.10 x 5.0 = 1.0 mmol
Mmol NaOH = MNaOH x VmL (NaOH)
1.0 = 0.25 x VmL
VmL = 4.0 mL
We can also calculate the volume in one step using
dimensional analysis
? mL NaOH = (mL NaOH/0.25 mmol NaOH)
x (2 mmolNaOH/mmol H2SO4)
x (0.10 mmol H2SO4 / mL H2SO4)
x 5.0 mL = 4.0 mL
In a previous lecture you were
introduced to calculations involved in titrimetric reactions. We agreed that
you do not have to memorize relations listed in the text and the only relation
you need to remember is the mmol. A mmol is mg/FW or molarity times volume
(mL). We have also agreed that writing the equation of the reaction is
essential and is the first step in solving any problem where a reaction takes
place. Let us look at more problems and develop a logic solution using the simple
mmol concept.
Example
A 0.1876 g of sodium carbonate (FW = 106 mg/mmol) was
titrated with approximately 0.1 M HCl requiring 35.86 mL. Find the molarity of HCl.
Solution
The first thing to do is to write
the equation. For the reaction. You should remember that carbonate reacts with
two protons
Na2CO3
+ 2 HCl = 2 NaCl + H2CO3
The second step is to
relate the number of mmol HCl to mmol carbonate, Where it is clear from the
equation that we have 2 mmol HCl and 1 mmol carbonate. This is translated to
the following
Mmol HCl = 2 mmol Na2CO3
Now let us substitute for mmol HCl by MHCl X VmL,
and substitute for mmol carbonate by mg carbonate/FW carbonate. This gives
MHCl x 35.86 = 187.6 mg/ (106 mg/mmol)
MHCl = 0.09872 M
The same result can be obtained using dimensional analysis
in one single step:
? mmol HCl/mL = (187.6 mg Na2CO3
/35.86 mL HCl) x ( mmol Na2CO3/106
mg Na2CO3) x (2
mmol HCl/mmol Na2CO3) = 0.09872 M
Example
An acidified and reduced iron sample required 40.2 mL of
0.0206 M KMnO4. Find mg Fe (at wt = 55.8) and mg Fe2O3
(FW = 159.7 mg/mmol).
Solution
The first step is to write the
chemical equation
MnO4-
+ 5 Fe2+ + 8 H+ = Mn2+ + 5 Fe3+ + 4
H2O
MnO4-
+ 2.5 Fe22+ + 8 H+ = Mn2+ + 5 Fe3+
+ 4 H2O
Mmol Fe = 5 mmol KMnO4
Now substitute for mf Fe by mg Fe/at wt Fe and substitute
for mmol KMnO4 my molarity of permanganate times volume, we then get
[mg Fe/(55.8 mg/mmol)] = 5 x (0.0206 mmol/mL) x 40.2 mL
mg Fe = 231 mg
This can also be done in a single step as follows:
? mg Fe = (0.0206 mmol KMnO4
/mL) x 40.2 mL x (5 mmol Fe/mmol KMnO4)
x ( 55.8 mg Fe/mmol Fe) = 231 mg
To calculate the mg Fe2O3 we set the
following
Mmol Fe2O3 = 2.5 mmol KMnO4
[mg Fe2O3/ (159.7 mg/mmol)] = 2.5 x
(0.0206 mmol/mL) x 40.2 mL
mg Fe2O3 = 331 mg
The last step can also be done using dimensional analysis as
follows:
? mg Fe2O3 = = (0.0206 mmol KMnO4 /mL) x 40.2 mL x (5 mmol Fe2O3 /mmol KMnO4)
x ( 55.8 mg Fe2O3/mmol Fe)
= 331 mg
Example
A 1.00 g Al sample required 20.5 mL EDTA. Find the % Al2O3
(FW = 101.96) in the sample if 30.0 mL EDTA required 25.0 mL of 0.100 M CaCl2.
Solution
We should know that EDTA reacts in a
1:1 ratio with metal ions Therefore, the equation is
Al
+ EDTA = Al-EDTA
AL2O3
+ 2 EDTA = 2 Al-EDTA
Therefore, 1 mmol Al2O3
reacts with 2 mmoles EDTA since it
contains 2 mmol AL
Mmol Al2O3 = ½ mmol
EDTA (1)
The same procedure above is repeated in the calculation.
First we substitute mg aluminum oxide/FW for the mmol of aluminum oxide and
substitute mmol EDTA by molarity times volume. But we do not have the molarity
of EDTA, therefore,let us calculate the molarity of EDTA.
Mmol EDTA = mmol Ca
Molarity x VmL(EDTA)
= Molarity x VmL (Ca)
MEDTA x 30.0 = 0.100 x 25.0
MEDTA = 0.0833 M
Now we can solve the problem using relation (1)
[mg Al2O3/ ( 101.96 mg/mmol)]
= ½ x 0.833 x 20.5
? mg Al2O3 = 87.1 mg
% Al2O3 = (87.1 mg/1000 mg) x 100 =
8.71%
After calculation of the molarity of EDTA, one can do the
rest of the calculation in a single step as follows:
? mg Al2O3
= (0.0833 mmol EDTA/mL) x 20.5 mL x (mmol Al2O3/2mmol EDTA) x (101.96 mg Al2O3/mmol Al2O3)
= 87.1 mg
We have indicated earlier that the key to solve any dilution
problem is to remember that the number of mmol is the same in both initial and
final solutions. Dilution decreases the concentration but the volume increases
where the mmol remains constant.
Example
What volume of 0.4 M Ba(OH)2 should be added to
50 mL of 0.30 M NaOH in order to obtain a solution that is 0.5 M in OH-.
Solution
We have to be able to see that the mmol OH-
coming from Ba(OH)2 and NaOH will equal the number of mmol of OH-
in the final solution, which is
Mmol OH- from Ba(OH)2 + mmol
OH- from NaOH = mmo OH- in final solution
The mmol OH- from Ba(OH)2
is molarity of OH- times volume and so are other terms. Molarity
of OH- from Ba(OH)2
is 0.8 M (twice the concentration of Ba(OH)2, and its volume is x
mL. Now performing the substitution we
get
0.8 x x + 0.30 x 50 = 0.5 x (x + 50)
x = 33 mL
We have seen in previous sections
that correct solution of any problem involving reactions between two substances
requires setting up two important relations:
1. Writing a balanced chemical equation
representing stoichiometric relationships.
2. Formulating a relationship between the number of mmol
of substance A and mmol of substance B.
The last step in the calculation is to substitute for the mmol
A or B by either one of the following according to given information:
1. mmol = M x VmL
2. mmol = mg/FW
Only one unknown will be encountered and should be easily
determined following the abovementioned steps.
Example
Find the volume of 0.100 M KMnO4 that will react
with 50.0 mL of 0.200 M H2O2 according to the following
equation:
5 H2O2 + 2 KMnO4 + 6 H+ = 2 Mn2+
+ 5 O2 + 8 H2O
Solution
We have the equation ready therefore the following step is
to formulate the relationship between the number of moles of the two reactants.
We always start with the one we want to calculate, that is
mmol KMnO4 = 2/5 mmol H2O2
It is clear that we should substitute M x VmL for
mmol in both substances as this information is given to us.
0.100 x VmL = 2/5 x 0.200 x 50.0
VmL = 40.0 mL KMnO4
Example
Find the volume of 0.100 M KMnO4 that will react
with 50.0 mL of 0.200 M MnSO4 according to the following equation:
3 MnSO4 + 2 KMnO4 + 4 OH- = 2 MnO2
+ 2 H2O + 2 K+ + SO42-
Solution
Again, we have the equation ready therefore we turn to
formulating a relationship for the number of mmol for both substances.
mmol KMnO4 = 2/3 mmol MnSO4
It is clear that we should substitute M x VmL for
mmol in both substances as this information is given to us.
0.100 x VmL = 2/3 x 0.200 x 50.0
VmL = 66.7 mL KMnO4
Example
Find the volume of 0.100 M KMnO4 that will react
with 500 mg of H2C2O4 (FW = 90.0mg/mmol)
according to the following equation:
5 H2C2O4 + 2 KMnO4 + 6 H+ = 10 CO2
+ 2 Mn2++ 8 H2O + 2 K+
Solution
We have the equation ready therefore we turn to formulating
a relationship for the number of mmol for both substances.
mmol KMnO4 = 2/5 mmol H2C2O4
It is clear that we should substitute M x VmL for
mmol permanganate while substitute for mmol oxalic acid by mg/FW. This gives:
0.100 x VmL = 2/5 x 500 mg / (90.0 mg/mmol)
VmL = 22.2 mL KMnO4
In this technique, an accurately known amount of a reagent
is added to analyte in such a way that some excess of the added reagent is
left. This excess is then titrated to determine its amount and thus:
mmol reagent taken = mmol reagent reacted with analyte + mmol
reagent titrated
Therefore, the analyte can be determined since we know mmol
reagent added and mmol reagent titrated.
mmol reagent reacted = mmol reagent taken – mmol
reagent titrated
Finally, the number of
mmol reagent reacted can be related to the number of mmol analyte from
the stoichiometry of the reaction between the two substances.
Why Do We Use Back-Titrations?
Back-titrations are important especially in some situations
like:
1. When the titration reaction is slow. Addition of an excess
reagent will force the reaction to proceed faster.
2. When the titration reaction lacks a good indicator. We will
see details of this later.
3. When the analyte is not very stable. Addition of excess
reagent will finish the analyte instantly thus overcoming stability problems.
Example
A 2.63 g Cr(III) sample was dissolved and analyzed by
addition of 5.00 mL of 0.0103 M EDTA. The excess EDTA required 1.32 mL of
0.0112 M Zn(II). Calculate % CrCl3 (FW=158.4 mg/mmol) in the sample.
Solution
We should remember that EDTA reacts in a 1:1 ratio. The
first step in the calculation is to find the mmol EDTA reacted from the
relation:
mmol EDTA reacted = mmol EDTA taken – mmol EDTA titrated
We substitute for both mmol EDTA taken and mmol EDTA
titrated by M x VmL for each. Also since EDTA reacts in a 1:1 ratio,
we can state the following:
1. mmol EDTA reacted = mmol CrCl3.
2. mmol EDTA titrated
= mmol Zn(II).
Therefore, we now have the following reformulated relation:
mmol CrCl3 = mmol EDTA taken – mmol Zn(II)
Substitution gives:
mmol CrCl3 = 0.0103 x 5.00 –
0.0112 x 1.32
mmol CrCl3 = 0.0367 mmol
We can then find the number of mg CrCl3 by
multiplying mmol times FW.
Mg CrCl3 = 0.0367 mmol x 158.4 mg/mmol = 5.81 mg
% CrCl3 = (5.81 mg/2.63x103 mg) x 100
= 0.221%
Example
A 0.500 g sample containing sodium carbonate (FW=106 mg/mmol)
was dissolved and analyzed by addition of 50.0 mL of 0.100 M HCl solution. The
excess HCl required 5.6 mL of 0.05 M NaOH solution. Find the percentage of Na2CO3
n the sample.
Solution
First, write the chemical equations involved
Na2CO3
+ 2 HCl = 2 NaCl + H2CO3
HCl
+ NaOH = NaCl + H2O
mmol HCl reacted = mmol HCl taken – mmol
HCl titrated
1. mmol HCl reacted = 2 mmol Na2CO3
2. mmol HCl titrated = mmol NaOH
Now we can reformulate the above relation to read
2
mmol Na2CO3 = mmol
HCl taken – mmol NaOH
mmol Na2CO3 = 1/2 ( 0.100 x
50.0 – 0.050 x 5.6) = 2.36 mmol
? mg Na2CO3 = 2.36 mmol x 106 mg/mmol
= 250 mg
% Na2CO3 = (250 mg/0.500 x103
mg) x 100 = 50.0%
Example
A 0.200 g sample containing MnO2 was dissolved
and analyzed by addition of 50.0 mL of
0.100 M Fe2+ to drive the reaction
2 Fe2+ + MnO2 +
4 H+ = 2 Fe3+ + Mn2+ + 2 H2O
The excess Fe2+ required 15.0 mL of 0.0200 M KMnO4.
Find % Mn3O4 (FW= 228.8 mg/mmol) in the sample.
5
Fe2+ + MnO4- + 8 H+ = 5 Fe3+
+ Mn2+ + 4 H2O
Solution
First, calculate mmol MnO2 then change it to mmol
Mn3O4 . We have the relation
mmol Fe2+ reacted = mmol Fe+ taken – mmol Fe+ titrated
Now we should perform the following substitutions:
1. mmol Fe2+ reacted
= 2
mmol MnO2
2. mmol Fe2+ titrated = 5 mmol KMnO4
Now reformulate the above relation:
2
mmol MnO2 = mmol Fe2+
taken – 5 mmol KMnO4
Substitution gives:
mmol MnO2 = 1/2( 0.100 x
50.0 – 5 x 0.0200 x 15.0) = 1.75 mmol
3 MnO2 = Mn3O4 + O2
mmol Mn3O4 = 1/3 mmol MnO2
mmol Mn3O4 = 1/3 x 1.75 mmol = 0.5833
?mg Mn3O4 = 0.5833 mmol x 228.8 mg/mmol
= 133.5 mg
% Mn3O4 = (133.5 mg/200 mg)
x 100 = 66.7%
We have seen previously that solving volumetric problems
required setting up a relation between the number of mmoles or reacting
species. In case of normality, calculation is easier as we always have the
number of milli equivalents (meq) of substance A is equal to meq of substance
B, regardless of the stoichiometry in the chemical equation. Of course this is
because the number of moles involved in the reaction is accounted for in the
calculation of meqs.
Therefore, the first step in a calculation using normalities
is to write down the relation
meq A = meq B
The next step is to substitute for meq by one of the
following relations
1. meq = N x VmL
2. meq = mg/eq wt
3.
meq
= n x mmol
We should remember from previous lectures that:
1. eq wt = FW/n
2. N = n x M
Therefore, change from molarity or number of
moles to normality or number of equivalents using the above relations.
Again, I believe you do not have to remember the long
relations presented in the text and keep things simple by following the steps I
suggest for you to solve any normality volumetric problem. The most direct way
for understanding these concepts is to look at the following examples:
Example
A 0.4671 g sample containing sodium bicarbonate was titrated
with HCl requiring 40.72 mL. The acid was standardized by titrating 0.1876 g of
sodium carbonate (FW = 106 mg/mmol) requiring 37.86 mL of the acid. Find the
percentage of NaHCO3 (FW=84.0 mg/mmol) in the sample.
Solution
This problem was solved previously using molarity. The point
here is to use normalities in order to practice and learn how to use the meq
concept. First, let us find the normality of the acid from reaction with carbonate
(reacts with two protons as you know).
Eq wt Na2CO3 = FW/2 =
53 and eq wt NaHCO3 = FW/1 =
84.0
meq HCl = meq Na2CO3
Now substitute for meq
as mentioned above:
Normality x volume (mL) = wt (mg)/ eq wt
N x 37.86
= 0.4671 x 1000 mg/ (53 mg/meq)
NHCl= 0.0935 eq/L
Now we can find meq
NaHCO3 where
meq NaHCO3 = meq HCl
mg NaHCO3 / eq wt = N x VmL
mg NaHCO3 /84.0 = 0.0935 x 40.72
mg NaHCO3 =
84.0 x 0.0935 x 40.72 mg
% NaHCO3 = (84.0 x 0.0935 x 40.72 mg/476.1 mg) x
100 = 67.2%
Example
Use normalities to calculate how many mL of a 0.10 M H2SO4
will react with 20 mL of 0.25 M NaOH.
Solution
We can first convert molarities to normalities:
N = n x M
N (H2SO4) = 2 x 0.10 = 0.20
N (NaOH) = 1 x 0.25 = 0.25
meq H2SO4 = meq NaOH
Substitute for meq as usual ( either NVmL or mg/eq
wt)
0.20 x VmL = 0.25 x 20
VmL = 25 mL
Example
Find the normality of sodium carbonate (FW = 106) in a
solution containing 0.212 g carbonate in 100 mL solution if :
a.
The carbonate is used as a monobasic
base.
b. The carbonate is used as a dibasic base.
Solution
If carbonate is a monobasic base then eq wt =
FW/1 = 106/1 = 106 mg/meq
To find the normality of the solution we find the weight per
mL and then convert the weight per mL to meq/mL. We have 212 mg/100 mL which
means 2.12 mg/mL. Now the point is how many meq per 2.12 mg sodium carbonate.
meq = mg/eq wt = 2.12/106 = 0.02 meq
Then the normality is 0.02 N
If carbonate is to be used as a dibasic salt then the eq wt
= FW/2 = 106/2 = 53 mg/meq.
To find the normality of the solution we find the weight per
mL and then convert the weight per mL to meq/mL. We have 212 mg/100 mL which
means 2.12 mg/mL. Now the point is how many meq per 2.12 mg sodium carbonate.
meq = mg/eq wt = 2.12/53 = 0.04 meq
Then normality will be 0.04 N
Example
How many mg of I2 (FW = 254 mg/mmol) should you
weigh to prepare 250 mL of 0.100 N solution in the reaction:
I2
+ 2e = 2 I-
Solution
To find the number of mg I2 to be weighed and
dissolved we get the meq required. We have:
meq = N x VmL
meq = 0.100 x 250 = 25.0 meq
mg I2 = meq x eq wt = meq x FW/2 =
25.0 x 254/2 = 3.18x103 mg
Example
Find the normality of the solution containing 0.25 g/L H2C2O4
(FW = 90.0 mg/mmol). Oxalic acid reacts as a diacidic substance.
Solution
Since oxalic acid acts as a diacidic substance
then its eq wt = FW/2 = 90.0/2 = 45.0 mg/meq
We convert the mg acid per mL to meq acid per mL We have
0.25 mg acid/mL
meq acid = 0.25 mg/45.0 mg/meq = 0.0056 meq
Therefore, the normality is 0.0056 meq/mL
In many situations where routine titrations are carried out
and to avoid wasting time in performing calculations, one can calculate the
weight of analyte in mg equivalent to 1 mL of titrant. The obtained value is
called the titer of the titrant. For example, an EDTA bottle is labeled as
having a titer of 2.345 mg CaCO3. This means that each mL of EDTA
consumed in a titration of calcium carbonate corresponds to 2.345 mg CaCO3.
If the titration required 6.75 mL EDTA then we have in solution 2.345 x 6.75 mg
of calcium carbonate.
Example
What is the titer of a 5.442g/L K2Cr2O7
(FW = 294.2 mg/mmol) in terms of Fe2O3 (FW = 159.7 mg/mmol).
The equation is:
6
Fe2+ + Cr2O72- + 14 H+ =
6 Fe3+ + 2 Cr3+ + 7 H2O
Solution
1 mL of K2Cr2O7 contains
5.442 mg K2Cr2O7 per mL. Therefore let us find
how many mg Fe2O3 corresponds to this value of K2Cr2O7.
mmol K2Cr2O7
per mL = 5.442 mg/294.2 mg/mmol = 0.0185
mmol Fe2O3 = 3 mmol K2Cr2O7
mmol Fe2O3 = 3 x 0.0185
= 0.0555 mmol
mg = mmol x FW
mg Fe2O3 = 0.0555 mmol x (159.7 mg/mmol)
= 8.86 mg
Therefore, the titer of K2Cr2O7
in terms of Fe2O3 is 8.86 mg Fe2O3
per mL K2Cr2O7
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