Transcript Example

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.
The equation should be the first thing to formulate
Na2CO3 +2 HCl g 2NaCl + H2CO3
mmol Na2CO3 = ½ mmol HCl
(mg Na2CO3/FW) = ½ x ( MHCl x VmL (HCl) )
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(mg Na2CO3/FW) = ½ x ( MHCl x VmL (HCl) )
(mg Na2CO3/106) = ½ x 0.1067 x 40.72
mg Na2CO3 = 106 * ½ x 0.1067 x 40.72 = 230
% Na2CO3 = (230 x 10-3 g/0.4671 g ) x 100 =
49.3 %
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Example
How many mL of 0.25 M NaOH will react with
5.0 mL of 0.10 M H2SO4.
H2SO4 + 2 NaOH g Na2SO4 + 2 H2O
mmol NaOH = 2 mmol H2SO4
MNaOH x VmL(NaOH) = 2 {M(H2SO4) x VmL(H2SO4)}
0.25 x VmL = 2 x 0.10 x 5.0
VmL = 4.0 mL
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We can also calculate the volume in one
step using dimensional analysis:
? mL NaOH = (mL NaOH/0.25 mmol NaOH) x (2
mmol NaOH/mmol H2SO4) x (0.10 mmol H2SO4
/ mL H2SO4) x 5.0 mL H2SO4 = 4.0 mL
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Example
A 0.1876 g of pure 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 g 2 NaCl + H2CO3
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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
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MHCl x 35.86 = 2 * 187.6 mg/ (106 mg/mmol)
MHCl = 0.09872 M
The same result can be obtained
dimensional analysis in one single step:
using
? 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
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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+ g Mn2+ + 5 Fe3+ + 4 H2O
mmol Fe = 5 mmol KMnO4
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(1)
Now substitute for mmol 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
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To calculate the mg Fe2O3 we set the following
2Fe g Fe2O3
mmol Fe = 2 mmol Fe2O3
Substitute for mmol Fe in equation 1
2* [mg Fe2O3/ (159.7 mg/mmol)] = 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 Fe/mmol KMnO4) x (mmol Fe2O3 /2 mmol Fe) x (
159.7 mg Fe2O3/mmol Fe2O3) = 331 mg
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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
Al3+ + EDTA g Al-EDTA
mmol Al = mmol EDTA
(1)
However, 2Al g Al2O3
mmol Al = 2 mmol Al2O3 , substitute in equation (1)
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2*mmol Al2O3 = mmol EDTA
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 CaCl2
Molarity x VmL(EDTA) = Molarity x VmL (CaCl2)
MEDTA x 30.0 = 0.100 x 25.0
MEDTA = 0.0833 M
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Now we can solve the problem using relation (1)
2*[mg Al2O3/ ( 101.96 mg/mmol)] = 0.0833 x 20.5
? mg Al2O3 = 87.1 mg
% Al2O3 = (87.1 mg/1000 mg) x 100 = 8.71%
Another approach:
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 Al/mmol EDTA) x (mmol Al2O3/2mmol Al) x
(101.96 mg Al2O3/mmol Al2O3) = 87.1 mg
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We have seen in previous sections that correct
solution of any problem involving reactions
between two substances requires setting up two
important relations:
1.
2.
3.
Writing a balanced chemical equation representing
stoichiometric relationships.
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:
mmol = M x VmL
mmol = mg/FW
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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+ g 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
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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
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