Transcript Volumetric analysis 2

Volumetric Analysis Redox Honours
Oxidation / Reduction
Volumetric Analysis Redox Honours
What you will learn
The use of KMnO4 as an oxidising agent in titrations.
Ammonium iron(II) sulfate as primary standard
Solving redox volumetic problems.
Analysis of iron tablets.
The use if Na2S2O3 as reducing agent in titrations
Analysis of household bleach.
Experiments
–
Standard ammonium iron sulfate solution
–
Standardise KMnO4 solution
–
Determine amount of iron in an iron tablet
–
Standardise Na2S2O3 with iodine solution
Volumetric Analysis Redox Honours
Potassium Permanganate (KMnO4) Oxidising
Purple coloured, not a primary standard, must be standardised.
Agent
Unstable, decomposes with heat and sunlight. Therefore its concentration
therefore varies over time.
Acts as an oxidising agent in acidic solution only. (learn equation)
MnO4- + 8H+ +5e-  Mn2+ + 4H2O
H2SO4 is used to acidify KMnO4 as nitric is an oxidising agent itself so can't
be used nor can HCl as the Cl- ions would be oxidised to Cl2
In neutral or basic solution the above reaction does not happen, instead the
Mn (+7) is oxidised to Mn (+4) instead of Mn2+ forming
MnO4- + 2H2O + 3e-  MnO2 + 4OHKMnO4 acts as its own indicator, purple to colourless in acidic solution.
Volumetric Analysis Redox Honours
Potassium Permanganate (KMnO4) Oxidising
Because of the purple colour the bottom of the meniscus is difficult to see so
we
read from the top of the meniscus on the burette.
Agent
Because of its intense colour, low solubility and its good oxidising ability
KMnO4 is used fairly dilute. e.g around 0.02 M
Volumetric Analysis Redox Honours
Potassium Permanganate (KMnO4) reaction
KMnO4 oxidises Fe2+ to Fe3+ ions. As shown by the following half equations
with Fe2+ ions
Oxidising Agent: MnO4- + 8H+ +5e-  Mn2+ + 4H2O
Reducing Agent: Fe2+  Fe3+ + esince MnO4- can accept 5 electrons and Iron can only lose 1 it takes 5Fe to
react with 1 MnO4The full equation then is
MnO4- + 8H+ + 5Fe2+  5Fe3+ + Mn2+ + 4H2O
Note the ratio of MnO4- : Fe2+ IS 1:5
The Fe2+ reducing agent is used to standardise KMnO4
Volumetric Analysis Redox Honours
Potassium Permanganate (KMnO4) reaction
FeSO4.7H2O (Iron (II) sulfate) cannot be used as primary standard because
with Fe2+ ions
(1)The crystals oxidise in air.
(2)Lose their water of crystallisation spontaneously to the air
(efflourescence)
(3)The anhydrous FeSO4 absorbs moisture from the air (deliquescence).
Ammonium Iron(II) Sulfate ((NH4)2SO4.FeSO4.7H2O) is used as primary
standard because
(1)It is very stable even though it is hydrated.
(2)The high Mr = 392 ensure low %errors in weighing inaccuracies.
(3)NOTE Must be dissolved in water with dil. H2SO4 to stop the oxidation
of Fe2+ to Fe3+ by the air.
Volumetric Analysis Redox Honours
Solving Volumetric Problems for Redox
The following formula is used in solving volumetric
Reactions
problems.
Vol of
oxidising
agent (cm3)
no. of
moles of
0xidising
agent
Molarity
of
oxidising
agent
Vol of reducing agent (cm3)
V0 x Mo = Vr x Mred
no
nr
Molarity of
reducing agent
no. of moles of
reducing agent
Volumetric Analysis Redox Honours
Example 1 Standardising KMnO4
A 0.12 M (NH4)2SO4.FeSO4.7H2O was made up and used to standardise KMnO4.
In a titration 25 cm3 of the (NH4)2SO4.FeSO4.7H2O took an average of 23.75 cm3
of KMnO4 . Find the (a) molarity of the KMnO4 and (b) conc. of KMnO4 in
g/L. The equation for the reaction is:
MnO4- + 8H+ + 5Fe2+  5Fe3+ + Mn2+ + 4H2O
Answer
M of KMnO =39+55 +4(16)=158 g
r
vo=23.55
Mo= ?
no = 1
Mred=0.12
Vr=25
nr= 5
Vo x Mo = Vr x Mred
n0
nr
23.75 x Mo = 25 x 0.12
1
5
Ma = 25 x 0.12 x 1
23.75 x 5
= 0.0253 M
KMnO
4
g/L = molarity x Mr
= 0.0253 x 158
= 3.997 g/L
Volumetric Analysis Redox Honours
Example 2 Finding x (water of crystallisation) in FeSO4.xH2O
6.42 g FeSO4.xH2O crystals were dissolved in deionised water with a little
dil. H2SO4 and made up to 250 cm3. In a titration 25 cm3 of the FeSO4.xH2O
took an average of 25.5 cm3 of 0.018 M KMnO4 . Find the value of x in the
formula FeSO4.xH2O. The equation for the reaction is:
MnO4- + 8H+ + 5Fe2+  5Fe3+ + Mn2+ + 4H2O
Answer
0.9184 = 0.023 mol/250cm3
V xM =V xM
o
o
r
red
vo=25.5
n0
nr
Mo= 0.018
This means that 6.42 g of
25.5 x 0.018 = 25 x Mred crystals is 0.023 moles.
no = 1
1
5
Mred=?
Mred = 25.5 x 0.018 x 5
Vr=25
25 x 1
nr= 5
= 0.0918 M FeSO4xH2O
Why divide by 4?
Volumetric Analysis Redox Honours
Answer cont.
Rem no. of moles = mass
Mr
Now 0.023 moles of FeSO4xH2O = 6.42 g
1 mole (Mr) of FeSO4xH2O = 6.42
0.023
= 279.13 g
i.e 1 mole of FeSO4xH2O = 279.13
 152 + 18x = 279.13
Mr of FeSO4 =56+32 +4(16)=152 g

18x = 127.13

x =7
 the formula is FeSO47H2O
Volumetric Analysis Redox Honours
The Need for Iron in Our Bodies
Iron is need in our red blood cells to carry oxygen for respiration and
hence energy.
Haemoglobin is the molecule that carries oxygen in the blood and iron is
part of this molecule.
If we become deficient in iron it can be remedied with a course of iron
tablets. The active ingredient is FeSO4 .
Iron deficiency is called anaemia.
We will analyse iron tablets for the quantity of Iron (II) sulfate.
Volumetric Analysis Redox Honours
Example 3 Finding the amount of iron in an iron tablet.
To analyse the amount of iron (II) sulfate in an iron tablet 5
tablets of mass 1.2 g were dissolved in deionise water with a
little dil. H2SO4 added and made up to 250 cm 3 in a
volumetric flask. 25 cm3 portions of this solution were titrated
against 0.15 M KMnO4 taking ón average 5.75 cm3.
Find (a) the mass of FeSO4 in each tablet, (b) the mass of
iron in each tablet and (c) the % FeSO4 in each tablet.
The equation for the reaction is.
MnO4- + 8H+ + 5Fe2+  5Fe3+ + Mn2+ + 4H2O
Volumetric Analysis Redox Honours
Answer
vo=5.75
Mo= 0.015
no = 1
Mred=?
Vr=25
nr = 5
Mr of FeSO4 =56+32 +4(16)=152 g
Vo x Mo = Vr x Mred
n0
nr
5.75 x 0.015 = 25 x Mred
1
5
Mred = 5.75 x 0.015 x 5
25 x 1
= 0.0173 M FeSO4
g/L = molarity x Mr
= 0.0173 x 152
= 2.63 g/L
Volumetric Analysis Redox Honours
Answer Cont.
Mr of FeSO4 =56+32 +4(16)=152 g
In 250 cm3 there would be
2.634 = 0.658 g FeSO4 in 5 tablets
 in 1 tablet there is 0.658=0.132 g
5
This means that 0.132 g of
FeSO4 in each tablet.
% Fe in FeSO4 = mass Fe x100
Mass FeSO4
= 56 x100
152
= 36.84%
Mass of Fe/tablet =36.84% of
0.132 = 0.049 g
Why divide by 4 and then 5?
Rem:
mass 1 tablet = total mass of tablets
5
= 0.24 g
%FeSO4/tablet = mass FeSO4
Mass 1 tablet
= 0.132 x100
0.24
= 55%
%FeSO4/tablet = 55%
Volumetric Analysis Redox Honours
Sodium Thiosulfate (Na2S2O3) reaction with
Na2S2O3 is an important reducing agent. It is used in analysing them amount
Iodine
of oxygen and chlorine in water, extracting gold, as an antidote to cyanide
poisoning and developing film photographs.
It is a reducing agent reacting with iodine in the following manner
I2
+
2S2O32-

S4O62- +
2I-
Iodine + thiosulfate ion  tetrathionate ion + iodide ion
oxidising agent
reducing agent
Sodium thiosulfate is colourless crystalline solid Na2S2O3.5H2O
Thiosulfate ion S2O32-
tetrathionate ion S4O62-
Volumetric Analysis Redox Honours
Sodium Thiosulfate (Na2S2O3) reaction with
Na2S2O3 is not a primary standard, its crystals are efflorescent (lose water of
Iodine
crystallisation to the air).
Sodium thiosulfate is standardised by titrating it against standard I 2 solution.
I2 + 2S2O32-  S4O62- + 2Ioxidation numbers of the sulfur
+2
+2.5
oxidation number of the sulfur increases so it is oxidised  reducing agent
A standard solution of iodine cannot be made up directly, as iodine sublimes
and is almost completely insoluble in water.
Volumetric Analysis Redox Honours
Making a Standard Iodine Solution
A standard iodine solution is made by reacting standardised KMnO4 solution
with excess potassium iodide.
Sodium thiosulfate is standardised by titrating it against standard I 2 solution
and reacts according to the following equation.
2MnO4- + 10I- + 16H+  2Mn2+ + 8H2O + 5I2
Excess potassium is added for the following reasons
(1)Having the I- ions in excess means all the KMnO4 is used up
for 2KMnO4 : 5I2
so the exact amount of Iodine formed can be calculated.
(2)The presence of I- keeps the I2 in solution by the reaction below making it
charged. (remember I2 is non-polar)
I + I-  I -
Volumetric Analysis Redox Honours
Detecting the End Point in an Iodine /
As the thiosulfate is added to the iodine solution the reddish brown colour
Thiosulfate
gradually turnsTitration
yellow, then paler yellow until the solution goes colourless.
I2 + 2S2O32-  S4O62- + 2Ireddish brown
colourless
There is no sudden colour change so an indicator is needed.
Starch indicator is added which goes blue-black with Iodine. When all the
Iodine has reacted the blue-black colour suddenly disappears.
The starch is only added when the iodine is pale yellow near end point as I 2
strongly adsorbs to starch so adding earlier makes it less accurate.
Permanganate reacts with iodine as in the equation below.
2MnO4- + 10I- + 16H+  2Mn2+ + 8H2O + 5I2
The reacting ratios are KMnO4  5I2  10Na2S2O3
Volumetric Analysis Redox Honours
Example 4 Iodine/ thiosulfate titration calculation
25 cm3 of a 0.045 M Iodine solution was titrate against sodium thiosulfate. The
average titre was 27.45 cm3. Find the (a) molarity of the Na2S2O3 and (b)
conc. of Na2S2O3 in g/L. The equation for the reaction is:
I2 + 2S2O32-  S4O62- + 2IAnswer
M of Na S O .5H O=2(23)+2(32) +3(16)+ 5(18) =248 g
r
2 2
3
2
Vo x Mo = Vr x Mred
vo=25
n0
nr
Mo= 0.045
25 x 0.045 = 27.45 x Mred
no = 1
1
2
Mred=?
Vr=27.45 Mred = 25 x 0.045 x 2
27.45 x 1
nr= 2
= 0.082 M Na2S2O3.5H2O
g/L = molarity x Mr
= 0.082 x 248
= 20.336 g/L
Volumetric Analysis Redox Honours
Example 5 Finding the Na2S2O3.5H2O using KMnO4
25 cm3 of 0.02 M KMnO4 was pipetted into a conical flask.
Dil. H2SO4 and excess KI solution were added. The liberated
I2 was titrated against Na2S2O3 solution the average titre being
18.7 cm3.
Find the (a) molarity of the Na2S2O35H2O and (b) conc. of
Na2S2O35H2O in g/L. The equation for the reactions are:
I2 + 2S2O32-  S4O62- + 2I2MnO4- +16H+ + 10I-  5I2 + Mn2+ + 8H2O
the reacting ratios are
● KMnO  5I  10Na S O
4
2
2 2 3
 n0=2 and nr =10
Volumetric Analysis Redox Honours
Answe
r
Mr of Na2S2O3.5H2O=2(23)+2(32) +3(16)+ 5(18) =248 g
Vo x Mo = Vr x Mred
vo=25
n0
nr
Mo= 0.02
25 x 0.02 = 18.7 x Mred
no = 2
2
10
Mred=?
Vr=18.7 Mred = 25 x 0.02 x 10
18.7 x 2
nr= 10
= 0.1337 M Na2S2O3.5H2O
g/L = molarity x Mr
= 0.1337 x 248
= 33.158 g/L
Volumetric Analysis Redox Honours
Example 6 Finding the KMnO4 conc. using the Na2S2O3.5H2O
When excess potassium iodide was added to a solution of acidied
KMnO4 it required 22.5 cm3 of 0.125 M Na2S2O3.5H2O to reduce
the iodine liberated.
Find the mass of KMnO4 in 25 cm3 of solution. The equation
for the reactions are:
I2 + 2S2O32-  S4O62- + 2I2MnO4- +16H+ + 10I-  5I2 + Mn2+ + 8H2O
the reacting ratios are
● KMnO  5I  10Na S O
4
2
2 2 3
 n0=2 and nr =10
Volumetric Analysis Redox Honours
Answe
r
Mr of KMnO4=39+55 +4(16) =158 g
Vo x Mo = Vr x Mred
vo=25
n0
nr
Mo= 0.02
25 x Mo = 22.5 x 0.125
no = 2
2
10
Mred=?
Vr=22.5 Mo = 22.5 x 0.125 x 2
25 x 10
nr= 10
= 0.0225 M KMnO4
g/L = molarity x Mr
= 0.0225 x 158
= 3.555 g/L
= 0.09 g/25 cm3
Divide by 40 why?
Volumetric Analysis Redox Honours
Example 7 Iodine/ thiosulfate titration calculation
Na2S2O3.5H2O was prepared by adding about 33 g if impure
crystals to water and making up to 1 litre. A burette was filled with
this solution using a funnel. The funnel was remove and the tap
open and closed briefly before recording the reading. 25 cm3 of
0.05 M iodine solution was put in a conical flask to be
titrated against the thiosulfate. When the end point was near
starch indicator was added. The average titre was 20 cm 3.
The equation for the reaction is:
I2 + 2S2O32-  S4O62- + 2Ithe reacting ratios is
● 1I  2Na S O therefore  n =1 and n =2
2
2 2 3
0
r
Volumetric Analysis Redox Honours
(i) Why is it not necessary to know the exact mass of thiosulfate crystals?
answer 7(i) sodium thiosulfate is not primary standard, the exact
mass can be calculated from the titration results.
(ii) Why was (a) the funnel removed and (b) the tap open and closed briefly
before recording the reading on the burette?
answer 7(ii) (a) drops of solution could stick to the funnel and
drop down later. (b) to ensure the space below the burette tap is
filled.
(iii) What is the difference in procedure regarding rinsing the conical flask
as opposed to the burette and pipette?
answer 7(iii) (the pipette and burette are always rinsed with the
solutions they are to contain but conical flask with deionised water
only. This is because an exact volume is added to the conical flask
so rinsing with the solution would add extra volume. Adding
deionised water does not change the amount of dissolved material.
Volumetric Analysis Redox Honours
(iv) mention two steps of the procedure (other than adding starch
indicator) carried out with the conical flask and its contents.
answer 7(iv) The flask should be continuously swirled. Any drops
on the side should be washed down with deionised water. A white
tile underneath the flask makes the colour change easier to see.
(v) How can you tell when the end point is near? What was observed (a)
when the starch indicator was added and (b) when the end point was
reached?
answer 7(v) The end point is near when the iodine in the flask
turns pale yellow. (a) The solution turned blue black when starch
was added. (b) At the end point the solution goes colourless (not
clear why?)
Volumetric Analysis Redox Honours
(vi) find the conc. Of the thiosulfate solution in (a) moles / litre (b) g/L
crystalline of Na2S2O3.5H2O
I2 + 2S2O32-  S4O62- + 2IAnswer 7(vi)
Mr of Na2S2O3.5H2O=2(23)+2(32) +3(16)+ 5(18) =248 g
Vo x Mo = Vr x Mred
vo=25
n0
nr
Mo= 0.05
25 x 0.05 = 20 x Mred
no = 1
1
2
Mred=?
Mred = 25 x 0.05 x 2
Vr=20
20 x 1
nr= 2
= 0.125 M Na2S2O3.5H2O
g/L = molarity x Mr
= 0.125 x 248
= 31 g/L