Transcript File - Science at St. Dominics

Potassium permanganate
Titrations
A potassium manganate(VII)/ammonium iron(II)
sulfate titration
• Potassium manganate (VII) (potassium permanganate,
KMnO4) solution can be standardised by titration against a
standard solution of ammonium iron(II) sulfate solution.
MnO4- + 8H+ + 5Fe+2  Mn+2 + 5Fe+3 + 4H2O
Safety
Potassium manganate(VII) solution is an oxidising agent and can be a skin irritant.
If it is washed off, it may leave a brown stain that will slowly disappear.
Ammonium iron(II) sulfate is harmful if ingested in quantity, and is an eye irritant.
Dilute sulfuric acid
i
is harmful to eyes and an irritant to skin.
Once the oxidation numbers
are balanced,
Make sure the overall
equation still balances...
•Each Mn goes down 5 in number (reduction)
•RIG – Each Mn is gaining 5 electrons.
•It is the oxidising agent
MnO4 ―+5 Fe+2 +8H+
( + 7 ) ( - 2)
( +2)
( +1)
Mn+2 +5 Fe+3 + 4 H2O
( +2)
( +3)
( +1 ) ( - 2)
x + 4(-2) = -1
x – 8 = -1
x = -1+ 8
x= 7
•Each Fe goes down up 1 in number ( oxidation)
•OIL – Each Fe is losing 1 electron
•It is the reducing agent
•Altogether ( 5 Fe) are losing 5 electrons
Why is ammonium iron(II) sulfate
suitable as a primary standard?
Because it is stable and available in a highly
pure form.
Adding acid to the iron(II) sulfate
solution
• Iron(II) is very susceptible to air
oxidation, forming iron(III), under neutral or
alkaline conditions but this oxidation is
inhibited in the presence of acids.
Which acid is used for the reaction?
• The normal source of acid used is dilute sulfuric acid.
Sulfuric acid is a good source of H+, and the SO4-2 ions
are not reactive.
• Hydrochloric acid is not suitable as the Cl- ions would
be oxidised by the KMnO4 (instead of the Fe+2 ions
being oxidised)
• Nitric acid is not suitable as the NO3- ion is reduced
instead of the Mn+7.
•
Adding more acid before the reaction
• The iron(II) solution is measured using a
pipette and placed in the conical flask
• The titration is carried out under acidic
conditions, so about 10 cm3 of dilute sulfuric
acid is added to the Fe+2 solution before the
titration
Why is acid needed in the reaction?
• because in neutral or alkaline conditions Mn+7 is reduced only
as far as Mn+4 which is brown and would make it difficult to
determine when the end point of the titration happens.
• Acidic conditions ensure that Mn+7 is fully
reduced to Mn+2
Read from the top of the meniscus
• the potassium manganate(VII) solution is
placed in a burette.
• Read from the top of meniscus because the
very dark colour of the manganate(VII)
solution makes the meniscus difficult to see.
No indicator is needed
• No indicator is needed, as the manganate(VII)
ions are decolourised in the reaction until the
end-point, when a pale pink colour persists.
Autocatalysis
• The reaction is catalysed by a product of the
reaction - Mn2+ ions.
• The first droplet of MnO4- added decolourises
slowly.
• As soon as some Mn2+ is made it acts as a
catalyst and speeds up the next reaction – so
the next drops of MnO4 decolourise quickly
246. A solution of potassium permanganate is standardised
against a 0.11 M iron(II) sulfate solution. 25 cm3 of the iron(II)
sulfate solution required 28.5 cm3 of the permanganate
solution.
Calculate the concentration of the potassium permanganate
(KMnO4) solution in (a) moles per litre (b) grams per litre
The equation for the reaction is:
MnO4 ―+ 5Fe+2 + 8H+
Mn+2 + 5 Fe+3 + 4H2O
(a) Find concentration of potassium permanganate
solution in moles per litre
MnO4 ―+ 5Fe+2 + 8H+
•
Mn+2 + 5 Fe+3 + 4H2O
V1 X M1 = V2 x M2
n1
n2
Solution 1 – MnO4 -
Solution 2 – Fe+2
V1 = 28.5cm3
V2 = 25cm3
M1 = ?
M2 = 0.11
n1 = 1
n2 = 5
Question
235(h)(i)
V1 X M1 = V2 x M2
n1
n2
(28.5)X (M1) = (25) x (0.11)
1
5
M1 = (25) x (0.11) x (1)
(5) x (28.5)
M1 = .0193
The concentration of potassium permanganate solution is 0.0193 M
(moles per litre)
(ii) What is the concentration in grams per litre?
• Moles PER LITRE
X RMM
Grams PER LITRE
0.0193 x rmm = grams per litre
0.0193 x 158g = 3.0494
There are 3.0494g of KMn04 in one litre.
Q247.
22.5 cm3 of a solution of 0.02 M KMnO4 solution
reacted completely with 25 cm3 of a solution of
ammonium iron(II) sulfate. Calculate the
concentration of the ammonium iron(II) sulfate
solution in
(a) moles per litre
(b) grams of ammonium iron(II) sulfate,
(NH4)2SO4.FeSO4.6H2O, per litre.
The equation for the reaction is:
MnO4 ―+ 5Fe+2 + 8H+
Mn+2 + 5 Fe+3 + 4H2O
(a) Find concentration of potassium permanganate
solution in moles per litre
MnO4 ―+ 5Fe+2 + 8H+
•
Mn+2 + 5 Fe+3 + 4H2O
V1 X M1 = V2 x M2
n1
n2
Solution 1 – MnO4 -
Solution 2 – Fe+2
V1 = 22.5cm3
V2 = 25cm3
M1 = 0.02M
M2 = ?
n1 = 1
n2 = 5
V1 X M1 = V2 x M2
n1
(22.5)X (0.02) = (25) x (M2)
1
5
(22.5)X (0.02)X (5) = M2
(1) x (25)
0.09 = M2
The concentration of the iron (II) sulfate solution was
0.09M (moles per litre)
n2
(ii) What is the concentration in grams per litre?
• Know Moles PER LITRE
Find Grams PER LITRE
0.09 moles x RMM = Grams per litre
0.09 mole x 392 = 35.28
There are 35.28 g of (NH4)2SO4.FeSO4.6H2O in one
litre.
248
• A solution of ammonium iron(II) sulfate,
(NH4)2SO4.FeSO4.6H2O was made up by dissolving
9.80 g of this crystalline solid in 250 cm3 of
acidified solution.
• 25.0 cm3 of this solution completely reacted with
24.65 cm3 of a potassium permanganate
solution.
• Calculate the concentration of the potassium
permanganate (KMnO4) solution in
• (a) moles per litre (b) grams per litre
• The equation for the reaction is:
MnO4 ―+ 5Fe+2 + 8H+
Mn+2 + 5 Fe+3 + 4H2O
(a) Find concentration of potassium permanganate
solution in moles per litre
MnO4 ―+ 5Fe+2 + 8H+
•
Mn+2 + 5 Fe+3 + 4H2O
V1 X M1 = V2 x M2
n1
n2
Solution 1 – MnO4 -
Solution 2 – Fe+2
V1 = 24.65cm3
V2 = 25cm3
M1 = ?
M2 = ??
n1 = 1
n2 = 5
Finding molarity of Ammonium Iron(II)
sulfate solution
(i) Finding grams per litre
1.
2.
We know that 9.8g are in 250cm3 of solution
In one litre of solution there would be (9.8x4) = 39.2g
(ii) Finding moles per litre
Grams per litre
Moles per litre
/ RMM
39.2g / 392 = 0.1 Moles per litre
There are 0.1 moles of Ammonium Iron (II) sulfate in 1 litre.
This is the molarity.
(a) Find concentration of potassium permanganate
solution in moles per litre
MnO4 ―+ 5Fe+2 + 8H+
•
Mn+2 + 5 Fe+3 + 4H2O
V1 X M1 = V2 x M2
n1
n2
Solution 1 – MnO4 -
Solution 2 – Fe+2
V1 = 24.65cm3
V2 = 25cm3
M1 = ?
M2 = 0.1
n1 = 1
n2 = 5
V1 X M1 = V2 x M2
n1
n2
(24.65)X (M1) = (25) x (0.1)
1
5
M1 = (25) x (0.1) x (1)
(5) x (24.65)
M1 =0.0203
The concentration of potassium permanganate solution is 0.0203 M
(moles per litre)
(ii) What is the concentration in grams per litre?
• Moles PER LITRE
X RMM
Grams PER LITRE
0.0203 x rmm = grams per litre
0.0203 x 158g = 3.2074
There are 3.2074g of KMn04 in one litre.
249
A standard solution of ammonium iron(II) sulfate,
(NH4)2SO4.FeSO4.6H2O, was
prepared by dissolving 11.76 g of the crystals in dilute sulfuric acid and
making up
the solution to 250 cm3 with deionised water from a washbottle.
(b) Calculate the molarity (moles per litre) of this solution.
25cm3 of this solution was taken and further acidified with dilute
sulfuric acid, It
required 33.3 cm3 of a potassium permanganate solution for complete
reaction
according to the equation.
(e) Calculate the molarity of the potassium permanganate solution.
MnO4 ―+ 5Fe+2 + 8H+
Mn+2 + 5 Fe+3 + 4H2O
b. Finding molarity of Ammonium
Iron(II) sulfate solution
(i) Finding grams per litre
1.
2.
We know that 11.76g are in 250cm3 of solution
In one litre of solution there would be (11.76x4) = 47.04g
(ii) Finding moles per litre
Grams per litre/ RMM = moles per litre
47.04/ 392g = 0.12
There are 0.12 moles of Ammonium Iron (II) sulfate
in 1 litre. This is the molarity.
•Each Mn goes down 5 in number (reduction)
•RIG – Each Mn is gaining 5 electrons.
•It is the oxidising agent
MnO4 ―+5 Fe+2 +8H+
( - 7 ) ( - 2)
( +2)
( +1)
Once the oxidation numbers
are balanced,
Make sure the overall
equation still balances...
Mn+2 +5 Fe+3 + 4 H2O
( +2)
( +3)
( +1 ) ( - 2)
x + 4(-2) = -1
x – 8 = -1
x = -1+ 8
x= 7
•Each Fe goes down up 1 in number ( oxidation)
•OIL – Each Fe is losing 1 electron
•It is the reducing agent
•Altogether ( 5 Fe) are losing 5 electrons
(a) Find concentration of potassium permanganate
solution in moles per litre
MnO4 ―+ 5Fe+2 + 8H+
•
Mn+2 + 5 Fe+3 + 4H2O
V1 X M1 = V2 x M2
n1
n2
Solution 1 – MnO4 -
Solution 2 – Fe+2
V1 = 33.3cm3
V2 = 25cm3
M1 = ?
M2 = 0.12
n1 = 1
n2 = 5
V1 X M1 = V2 x M2
n1
n2
(33.3)X (M1) = (25) x ()
1
5
M1 = (25) x (0.12) x (1)
(5) x (33.3)
M1 = 0.0180
The concentration of potassium permanganate solution is 0.0180M
(moles per litre)
• 250. A solution of potassium manganate(VII) was
prepared by a group of students. They then
wished to standardise this solution against a 0.1
M standard solution of ammonium iron(II)
sulfate,(NH4)2SO4.FeSO4.6H2O solution. The
potassium manganate(VII) solution (KMnO4) was
placed into a burette and titrated against 25 cm3
volumes of ammonium iron(II) sulfate.
• The titration results were 22.8 cm3, 22.4 cm3 and
22.5 cm3
Calculate the molarity of the potassium
permanganate solutionand its concentration in
grams per litre.
(f) Find concentration of potassium permanganate
solution in moles per litre
MnO4 ―+ 5Fe+2 + 8H+
•
Mn+2 + 5 Fe+3 + 4H2O
V1 X M1 = V2 x M2
n1
n2
Solution 1 – MnO4 -
Solution 2 – Fe+2
V1 = 22.45cm3 ( average)
V2 = 25cm3
M1 = ?
M2 = 0.1
n1 = 1
n2 = 5
Question 250 f
V1 X M1 = V2 x M2
n1
n2
(22.45)X (M1) = (25) x (0.1)
1
5
M1 = (25) x (0.1) x (1)
(5) x (22.45)
M1 = 0.0223
The concentration of potassium permanganate solution is 0.0223M
(moles per litre)
(ii) What is the concentration in grams per litre?
• Moles PER LITRE
X RMM
Grams PER LITRE
0.0223 x rmm = grams per litre
0.0223 x 158g = 3.6814
There are 3.6814 g of KMn04 in one litre.
Mandatory experiment
• Determination of the amount of iron in an iron
tablet by titration against a standardised
potassium permanganate solution
LC Past paper questions on this
titration
Why are iron tablets sometimes
prescribed?
• To treat anaemia (iron deficiency)
Why must potassium permanganate
solutions be standardised?
• Potassium permanganate is unstable and pure
and so is not a primary standard
Why should a potassium
permanganate solution be
standardised immediately before use?
• It is unstable and will start to breakdown in
the presence of light
What reagent must be used to
standardise potassium permangante
solution?
• Ammonium Iron(II) sulfate
Why is it important to use the dilute
sulfuric acid as well as the deionised
water when making up the solution
from the tablets?
Sulfuric acid prevents Iron (II) being oxidised to
Iron(III) by the air
Describe the procedure for making up
the 250cm3 solution from the tablets
•
•
•
•
•
•
•
•
Weigh out on a clock glass
Transfer to a beaker of dilute sulfuric acid
Add rinsings to beaker
Stir to dissolve
Add to a volumetric flask with a funnel
Add rinsings from rod and beaker
Fill with deionised water near to the calibration mark
Add deionised water dropwise until the bottom of the
meniscus reaches the calibration mark
• Read at eye level/ on a vertical surface
• Invert 20 times to make a homogenous mixture
Explain why dilute sulfuric acid must
be added to the titration flask before
the starting?
• Its stops Mn(7) being converted to Mn(4)
which is brown and makes it difficult to see
colour changes in the reaction. This may give
an inaccurate titre result
How was the end point detected?
• A permanent pink colour in the conical flask
252.
Six iron tablets of total mass 1.47 g (consisting of iron(II)
sulfate) were
dissolved in dilute sulfuric acid and deionised water and the
solution was made up to 250 cm3 in a volumetric flask. 20
cm3 portions of the resulting solution were titrated against
0.015 M potassium permanganate. The average titration
figure was 5.47 cm3.
(b) Calculate:
(i) the mass of anhydrous FeSO4 in each tablet,
(ii) the mass of iron in each tablet and
(iii) the percentage of FeSO4 in each tablet.
The equation for the reaction is:
MnO4 ―+ 5Fe+2 + 8H+
Mn+2 + 5 Fe+3 + 4H2O
(a) Find concentration of iron sulfate solution in moles
per litre
MnO4 ―+ 5Fe+2 + 8H+
•
Mn+2 + 5 Fe+3 + 4H2O
V1 X M1 = V2 x M2
n1
n2
Solution 1 – MnO4 -
Solution 2 – Fe+2
V1 = 5.47cm3 ( average)
V2 = 20cm3
M1 = 0.015
M2 = ?
n1 = 1
n2 = 5
V1 X M1 = V2 x M2
n1
n2
(5.47)X (0.015) = (20) x (M2)
1
5
0.8205 =
(20) x (M2)
(5)
M1 = 0.0205
The concentration of iron(II) sulfate solution is 0.0205 M (moles per
litre)
Moles of Iron (II) sulfate in 250cm3?
• 0.0205 moles of Iron(II) sulfate per litre
• 0.0205 /4 = .0051 moles per 250cm3
(ii) What is the mass of FeSO4 in each tablet?
• Moles PER LITRE
X RMM
Grams PER LITRE
0.0051 x rmm = grams per litre
0.0051 x 152g = 0.7752
There are 0.7752 g of Iron(II) Sulfate in 250cm3
So in each tablet…
There are 0.7752 / 6= 0.1292g of Iron(II) Sulfate in
each tablet
(ii) What is the mass of iron in each tablet?
• Moles PER LITRE
X RMM
Grams PER LITRE
0.0051 x rmm = grams per litre
0.0051 x 56g = 0.2856g
There are 0.2856g of Iron in 250cm3
So in each tablet…
There are 0.2856 / 6 = 0.0476g of Iron in each tablet
% of iron(II) Sulfate in each tablet?
Mass of iron(II) Sulfate per tablet x 100
Total mass per tablet
1
0.1292g x 100
0.245g
1
52.7347%
253
In an analysis of iron tablets to determine the amount of iron sulfate per
tablet, five tablets whose total mass was 1.20 g were dissolved in dilute
sulfuric acid and the solution made up to 250 cm3 in a volumetric flask.
25 cm3 portions of the resulting solution were titrated against 0.015 M
potassium permanganate. The average titration figure was 5.75 cm3.
(a) What would be observed if the student forgot to add the dilute
sulfuric acid?
(b) Calculate:
(i) the mass of anhydrous FeSO4 in each tablet,
(ii) the mass of iron in each tablet and
(iii) the percentage of FeSO4 in each tablet.
• The equation for the reaction is:
MnO4 ―+ 5Fe+2 + 8H+
Mn+2 + 5 Fe+3 + 4H2O
(a) Find concentration of potassium permanganate
solution in moles per litre
MnO4 ―+ 5Fe+2 + 8H+
•
Mn+2 + 5 Fe+3 + 4H2O
V1 X M1 = V2 x M2
n1
n2
Solution 1 – MnO4 -
Solution 2 – Fe+2
V1 = 5.75cm3 ( average)
V2 = 25cm3
M1 = 0.015
M2 = ?
n1 = 1
n2 = 5
V1 X M1 = V2 x M2
n1
n2
(5.75)X (0.015) = (25) x (M2)
1
5
0.862 =
(25) x (M2)
(5)
M1 = 0.01725
The concentration of iron solution is 0.0173 M (moles per litre)
Moles of Iron (II) sulfate in 250cm3
• 0.0173 moles of Iron(II) sulfate per litre
• 0.0173/4 = .0043 moles per 250cm3
(ii) What is the mass of FeSO4 in each tablet?
• Moles
X RMM
Grams
0.0043 x rmm = grams
0.0043 x 152g = 0.6536
There are 0.6536 g of Iron(II) Sulfate in 250cm3
So in each tablet…
There are 0.6536 / 5= 0.1307g of Iron(II) Sulfate in
each tablet
(ii) What is the mass of iron in each tablet?
• Moles PER LITRE
X RMM
Grams PER LITRE
0.0043 x rmm = grams per litre
0.0043 x 56g = 0.2408g
There are 0.2408g of Iron in 250cm3
So in each tablet…
There are 0.2408 / 5 = 0.0482g of Iron in each tablet
% of iron(II) Sulfate in each tablet?
Mass of iron(II) Sulfate per tablet x 100
Total mass per tablet
1
0.0482g x 100
0.24g
1
20.0833%
Iron tablets may be used in the treatment of anaemia.
To analyse the iron(II) content of commercially available iron tablets a student
used four tablets, each of mass 0.360 g, to make up 250 cm3 of solution in a
volumetric flask using dilute sulfuric acid and deionised water.
About 15 cm3 of dilute sulfuric acid was added to 25 cm3 portions of this
iron(II) solution and the mixture then titrated with a 0.010 M solution of
potassium manganate(VII), KMnO4.The titration reaction is described by
the equation
MnO4 ― + 5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H2O
(e) In the titrations the 25 cm3 portions of the iron(II) solution made from the
tablets required 13.9 cm3 of the 0.010 M KMnO4 solution.
Calculate
(i) the concentration of the iron(II) solution in moles per litre
(ii) the mass of iron(II) in one tablet
(iii) the percentage by mass of iron(II) in each tablet.
(a) Find concentration of potassium permanganate
solution in moles per litre
MnO4 ―+ 5Fe+2 + 8H+
•
Mn+2 + 5 Fe+3 + 4H2O
V1 X M1 = V2 x M2
n1
n2
Solution 1 – MnO4 -
Solution 2 – Fe+2
V1 = 13.9 cm3 ( average)
V2 = 25cm3
M1 = 0.01
M2 = ?
n1 = 1
n2 = 5
V1 X M1 = V2 x M2
n1
n2
(13.9)X (0.01) = (25) x (M2)
1
5
0.139 =
(25) x (M2)
(5)
M1 = 0.0278
The concentration of iron(II) sulfate solution is 0.0278M (moles per
litre)
Moles of Iron in 250cm3?
• 0.0278 moles of Iron(II) sulfate per litre
0.0278/4 = .0070 moles per 250cm3
(ii) What is the mass of iron in each tablet?
• Moles PER LITRE
X RMM
Grams PER LITRE
0.0070 x rmm = grams per litre
0.0070 x 56g = 0.392g
There are 0.392g of Iron in 250cm3
So in each tablet…
There are 0.392/ 4 = 0.098g of Iron in each tablet
% of iron in each tablet?
Mass of iron per tablet x 100
Total mass per tablet
0.098g x 100
0.360g
1
27.2222%
1
• The following experiment was carried out to find the mass of iron in
an
• iron tablet. A 250 cm3 volume of solution containing five tablets
• dissolved in dilute sulfuric acid was carefully made up in a
volumetric
• flask. The molarity of this solution was found by titrating it in 25
cm3
• volumes against a 0.005 M solution of potassium manganate(VII)
which
• had been previously standardised. The potassium manganate(VII)
• solution was put in the burette and a number of titrations were
carried
• out. The average titration figure was 17.5 cm3.
d) Assuming that the tablets are exactly equal in mass, calculate the
mass of iron in each tablet.
(a) Find concentration of potassium permanganate
solution in moles per litre
MnO4 ―+ 5Fe+2 + 8H+
•
Mn+2 + 5 Fe+3 + 4H2O
V1 X M1 = V2 x M2
n1
n2
Solution 1 – MnO4 -
Solution 2 – Fe+2
V1 = 17.5 cm3 ( average)
V2 = 25cm3
M1 = 0.005
M2 = ?
n1 = 1
n2 = 5
V1 X M1 = V2 x M2
n1
n2
(17.5)X (0.005) = (25) x (M2)
1
5
0.0875
(M2)
=
(25) x
(5)
M1 = 0.0175
The concentration of iron(II) sulfate solution is 0.0175 M (moles per
litre)
Moles of Iron in 250cm3?
• 0.0175 moles of Iron per litre
0.0175/4 = .0044 moles per 250cm3
(ii) What is the mass of iron in each tablet?
• Moles PER LITRE
X RMM
Grams PER LITRE
0.0044 x rmm = grams per litre
0.0044 x 56g = 0.2464g
There are 0.2464g of Iron in 250cm3
So in each tablet…
There are 0.2464/ 4 = 0.0493g of Iron in each tablet