Transcript Transition Metals - wellswaysciences

Week 27
•
Deduce the electron configurations of atoms and ions of the d-block
elements.
•
Describe the elements Ti–Cu as transition elements.
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Week 27
First row of the transition metal block
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Week 27
Energy-level diagram showing the overlap of the 3d and 4s sub-shells
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Transition Metals
• First Transition Series.
• ‘d block elements’ – that part of the periodic table
where the 3d orbitals are filling.
• Transition elements are defined as elements which
form at least one ion with a partly filled d shell.
• Why are Scandium (Sc) and Zinc (Zn) not considered
to be transition metals?
• Sc 1s22s22p63s23p6 3d1 4s2 forms only Sc3+ (d0)
• Zn 1s22s22p63s23p6 3d10 4s2 forms only Zn2+ (d10)
• N.B. The syllabus is clear that noble gas core notation
is unacceptable for transition metal electron
arrangements.
Properties
•
•
•
•
•
•
•
•
•
Shiny
Strong
Dense
High melting and boiling COMPARED WITH
main group metals – due to strength of
metallic bonding.
They also
have variable oxidation state
have coloured compounds
are catalysts
Shell Filling
• Predict electron arrangement for the
elements Sc to Zn. Z – Atomic number from
21 to 30. Use the Periodic Table NOT the
text book!
• The Aufbau Principle is followed (from the
lowest energy level out) as expected EXCEPT
• Cr 3d5 4s1 NOT 3d4 4s2
• Cu 3d10 4s1 NOT 3d9 4s2
Week 27
Electron configurations of the elements scandium to zinc
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Why?
• Cr -The maximum number of unpaired, parallel
spin electrons means that the former
arrangement is of LOWER energy than the
latter. The extra repulsion between paired
electrons outweighs the small energy
difference between 3d and 4s levels.
• Cu - 3d104s1 is more stable than 3d9 4s2 .
• (Exchange energy stabilises 3d10 .Electrons
have many places to go in the full 3d orbitals
and the electron distribution is spherical and
symmetrical so repulsion is less than in a full
4s orbital.)
Chemical Properties
• Close energies of the 3d and 4s shells means that ions
of similar stability can be formed by the loss of
different numbers of electrons.
• Stable electron arrangements might be expected
from loss of:
• a) all 3d and 4s electrons – giving noble gas
configurations.
• b) all 4s electrons leaving 3d untouched.
• c) the 4s and some 3d leaving 3d sub shell half filled.
• Other ions do exist which have varying stabilities.
• Stability of a particular cation depends to some
extent on the anion with which it combines. Small, not
easily polarised anions like F- and O2- stabilise high
oxidation states so high oxidation numbers are found
in oxidising agents containing many O atoms
e.g.KMnO4
Shell filling and emptying
• NOTE THAT although the 4s shell fills first when
dealing with electron arrangements the 4s electrons
are lost first when ions are formed.
• This is because of the shape of the orbitals.
• The 4s is spherical with a radius less than that the
radius of the 3d orbitals but once the d orbitals start
to fill they have significant electron density closer to
the nucleus than the 4s shells which are repelled
away.
• ON AVERAGE the radius of filling d orbitals are
LESS than the repelled 4s radius.
• This means that the 4s electrons are at higher
energy than the 3d and so are lost first when
ionisation occurs. This is shown by the great similarity
of 1st (and 2nd) ionisation energies across the d block
as the 4s electrons are lost and the shielding by d
electrons means that there is a constant effective
nuclear charge.
Week 27
Oxidation numbers and colours of the common d block metal ions
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Colour
• We see coloured objects as a result of light
striking them. When light falls on an object
some is reflected, some is absorbed and, if
the object is transparent, some may travel
through it.
• When white light passes through a solution
some of the radiation may be absorbed.
• If the wavelengths absorbed are in the visible
part of the spectrum the solution will appear
coloured.
• E.g. If a solution appears purple it is because
it absorbs green and yellow light and allows
red and blue light to pass through.
Why are most transition metal compounds
coloured?
• Transition metals have coloured ions because
the d energy levels are close enough together
to allow visible light to excite electrons from
a lower to a higher level. The theory is NOT
simple.
• Colour depends on the presence of partially
filled d shells.
• Transition metals have typical colours:
• Cu2+ blue
• Fe2+ green
• Fe3+ yellow etc.
Why are some transition metal compounds
white?
• Cu+ compounds are usually white. E.g. CuCl.
• Why?
• [Ar] 3d10 4s0 - no space in d orbitals for
electrons to be promoted into by absorption
of energy from white light.
• Why are Zn and Sc compounds all white?
• Zn – all 3d10 Sc – all 3d0
Transition Metals as Catalysts
• There are 2 types of catalyst.
• Heterogeneous Catalysts
• The catalyst and the reactants are in a
different phase (or state). The catalyst is
usually a solid.
• How many have we met?
• Iron in the Haber Process.
• Vanadium pentoxide or platinum in the
Contact Process.
• Nickel in the hydrogenation of alkenes.
• Manganese dioxide in the lab. prep. of oxygen
from hydrogen peroxide.
Action of Heterogeneous Catalysts
• In each of these the mechanism is a surface reaction
where reactant molecules are adsorbed onto the
catalyst surface using the available partially filled d
orbitals to form weak temporary bonds with the
metal.
• This weakens the original bonds between the reactant
molecules. New bonds form between adjacent atoms
on the surface and the product molecules diffuse
away, leaving the active site for more reaction.
• This is why catalysts are not used up in the reactions
they catalyse.
• This can be regarded as a lower activation energy
pathway
Homogeneous Catalysts
• We have not used these before.
• The catalyst and the reactants are in the
same phase or state.
• The action depends on the variable valency of
the transition metal.
• None are specifically mentioned in the
syllabus but a handy demonstration of Co2+ as
a catalyst can be demonstrated.
Precipitation Reactions
• Transition metal hydroxides are insoluble and
have distinctive colours which can be used to
identify the ions.
• Both sodium hydroxide and ammonium
hydroxide contain free hydroxide ions which
can react with the metal ion.
• Sodium hydroxide is a stronger alkali than
ammonium hydroxide since it is ionic.
• NaOH(s) + aq → Na+(aq) + OH-(aq)
Reactions with Ammonium Hydroxide
• ‘Ammonium hydroxide’ is actually ammonia
solution and contains a significant
concentration of ammonia gas.
• The alkali only forms as a result of the acidbase reaction with water:
• NH3(g) + H2O(l) → NH4+(aq) + OH-(aq)
•
• The equilibrium lies fairly LEFT since only
0.42% of ammonia molecules are reacted in
the above equilibrium.
• This means that it is an alkali, but a
significantly weaker one than sodium
hydroxide which is 100% dissociated in
solution.
Reactions with Sodium Hydroxide
Solution
• Simple precipitation occurs.
• e.g.
• Cu2+(aq) + 2OH-(aq) → Cu(OH)2(s)
•
Light blue ppt
Co2+(aq) + 2OH-(aq) → Co(OH)2(s)
•
light blue ppte →beige in air
• Fe2+(aq) + 2OH-(aq) → Fe(OH)2(s)
dark green gelatinous ppt → brown in
air
Week 28
•
Explain the term ligand in terms of coordinate bonding.
•
Describe and use the terms: complex ion and coordination number.
•
State and give examples of complexes with six-fold coordination with an
octahedral shape.
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Reactions with Ammonia
• Reactions with ammonia are the same to start
with, however ammonia can function as a
ligand and the initial precipitate is itself a
complex ion which redissolves to give a
solution of the complex ammine ion.
•
•
•
•
e.g. [Cu(OH)2(H2O)4](s) + 4NH3(aq) →
[Cu(NH3)4(H2O)2]2+(aq) + 2OH-(aq) + 2H2O(l)
Deep blue solution
This complexing takes place with many TM
ions.
Week 28
Complex ion [Fe(H2O)6]2+
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Week 28
[Co(H2O)6]2+ is an octahedral complex ion
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What is a Complex Ion?
• A complex ion is a transition metal ion bonded
to one or more ligands by coordinate (dative)
bonds.
• A ligand is a molecule or ion that can donate a
pair of electrons to the transition metal ion
to form a coordinate bond.
Shapes of Complexes
• 2 main shapes.
• Octahedral- 6 ligands at the points of an
octahedron.
• Tetrahedral- obvious.
• Square planar complexes are also important.
• The number of dative bonds from a ligand to
the metal cation is called the Coordination
Number. This determines the shape of the
complex. The most common coordination
numbers are 4 or 6.
Shapes of Complexes
Coordination Shape
number
Example
2
linear
[Ag(NH3)2]+
4
tetrahedral
[CuCl4]2-
6
octahedral
[Co(NH3)6]2+
Structure
Ligands
• The number of coordinate bonds a ligand can
form is called its DENTITION. Each
coordinate bond is formed from a pair of
electrons.
Electron
pairs
One pair
Dentition
Example
mondentate
Cl-, H2O, NH3,
Two pairs
bidentate
Six pairs
‘en’ ethane-1,2diamine
hexadentate EDTA
Formulae of Complex Ions
• In the formula:
• Use square brackets to group the species making up
the complex ion.
• The overall charge is shown outside the brackets.
• Complexes can be positive, neutral or negative overall,
depending on the charges on the ligands in the
complex.
• In diagrams of complexes, wedges are used as with
organic isomers to indicate bonds coming out of the
plane, hatched wedges to show bonds going into the
paper and solid lines are in the plane of the paper.
Naming Complex Ions (NOS)
• 1. Number the ligands using Greek prefixes
(mono,di,tri etc.).
• 2. Identify the ligands using names ending in ‘o’ e.g.
ammino (NH3), cyano (CN-), chloro (Cl- ), hydroxo
(OH-), aquo (H2O).
• 3.Name the cation:
• English if positively charged complex. Latinised,
ending in suffix ‘ate’ if negatively charged complex.
e.g. aluminate, cuprate, ferrate etc..
• 4. Indicate the oxidation number of the central
cation by using Roman Numerals. ( I, II, III etc.)
Try These
Formula of
complex ion
1
2
3
4
[Cu(NH3)4]2+
tetra
ammino
copper
II
chloro
cuprate
II
[CuCl4]2- tetra
[Fe(CN)6]3-
hexa
cyano
ferrate
III
[Cu(NH3)2]+
di
ammino
copper
I
[Al(OH)4][Zn(NH3)4]2
+
[Fe(CN)6]4-
1
Isomerism in Transition Metal
Complexes
• Stereoisomerism
• Molecules or complexes with the same structural
formula but with a different spatial arrangement of
these atoms.
• Cis-Trans isomerism
• This occurs in square planar or octahedral complexes.
• Square Planar Complexes
• In square planar complexes there must be 2 of each
ligand present for this to occur.
• In cis isomers ligands are at 90o to one another.
• In trans isomers the ligands are at 180o to one
another.
• The example on p.211, [NiCl2(NH3)2] is the example
required by the syllabus.
• It is essential to draw the isomers with the triangular
shaped bonds to highlight the 3D.
• Cis-Platin
• An effective anti cancer drug. Platin binds to the
DNA in fast growing cells. When the cells divide in
the presence of platin the daughter cells have
incorrectly copied DNA. As a result the cells die, and
so the reproduction of the cancer cells stops.
• Cis platin binds more effectively than trans platin.
• Unpleasant side effects are listed on the following
slide, courtesy of Chemocare.com.
Week 28
Cis–trans isomerism in [NiCl2(NH3)2]
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Week 28
Structure of cis-platin
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Cisplatin
What Cisplatin Is Used For:
* Used to treat testicular, ovarian, bladder, head and neck, esophageal,
small and non-small cell lung, breast, cervical, stomach and prostate cancers.
Also to treat Hodgkin's and non-Hodgkin's lymphomas, neuroblastoma,
sarcomas, multiple myeloma, melanoma, and mesothelioma.
The following Cisplatin side effects are common (occurring in greater than
30%) for patients receiving Cisplatin:
* Nausea and vomiting. Nausea may last up to 1 week after therapy. Antinausea medication is given before the infusion, and a prescription is also
given for use after.
* Kidney toxicity. Effects on kidney function are dose related, observed 1020 days after therapy, and are generally reversible.
* Blood test abnormalities (low magnesium, low calcium, low potassium)
* Low white blood cells (this may put you at increased risk for infection)
* Low red blood cells (anemia)
Other Side Effects
* Peripheral neuropathy: Although less common, a serious side effect of
decreased sensation and paresthesia (numbness and tingling of the extremities)
may be noted. Sensory loss, numbness and tingling, and difficulty in walking may
last for at least as long as therapy is continued. These side effects may become
progressively more severe with continued treatment, and your doctor may decide
to decrease your dose. Neurologic effects may be irreversible.
* High frequency hearing loss. Ringing in the ears.
* Loss of appetite
* Taste changes, metallic taste
* Increases in blood tests measuring liver function. These return to normal
once treatment is discontinued. (see liver problems).
* Hair loss
* Your fertility, meaning your ability to conceive or father a child, may be
affected by Cisplatin. Please discuss this issue with your health care provider.
Not all Cisplatin side effects are listed above, some that are rare (occurring in
less than 10% of patients) are not listed here.
Week 28
Complex ion [Fe(H2O)6]2+
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Week 28
[Co(H2O)6]2+ is an octahedral complex ion
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Cis trans isomerism in octahedral
complexes
• In an octahedral complex all bond angles are 90o, even
though it does not always look that way.
• Most aqueous ions are actually octahedral hexa-aquo
complexes, though only in transition metal chemistry
is this mentioned.
• For cis-trans isomerism of octahedral complexes
there must be 4 identical ligands and 2 of one other
ligand.
• In the cis isomer the 2 identical ligands are at 90o to
one another.
• In the trans form the identical ligands are at 180o to
one another.
Drawing cis-trans isomers
•
•
•
•
•
Draw cis trans isomers of the following:
[Co(NH3)4Cl2]+
[Pt(NH3)4Br2]
[Fe(CN)2(CH3NC)4]
[NiCl2(NH3)2]
Week 28
Cis–trans isomerism in [Co(NH3)4Cl2]+
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trans-[Co(NH3)4 Cl2]+ andcis-[Co(NH3)4 Cl2]+
Just cos they’re pretty
Bi and Multi-dentate ligands
• Bidentate ligands such as ‘en’ ethane-1,2-diamine or
ethanedioate ions form one dative bond from each N
in the molecule so are bidentate.
• These can form octahedral complexes with cis-trans
isomers as long as 2 bidentate ligands are present
with 2 monodentate ligands, e.g. [CoCl2(en)2] or
[Cr(C2O4)2(H2O)2]• EDTA is a chelating hexadentate ligand – chelos is
Latin for claw- so the ligand is forming a pincer-like
grip on the central metal ion.
• In general polydentate ligands are more powerful and
stable than simple unidentate ligands
Week 28
Cis and trans isomers of [Cr(C2O4)2(H2O)2]–
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Week 28
Complex ion [Ni(en)3]2+
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Week 28
Structure of the ethanedioate ligand, C2O42–
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Week 28
EDTA4– – note the six pairs of electrons, each of which can form a coordinate
bond
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Optical Isomerism
• As with organic chemistry optical isomers are non
superimposable mirror images of each other which
rotate plane polarised light through identical angles
but in opposite directions.
• Optical isomerism occurs when:
• 3 molecules or ions of a bidentate ligand form an
octahedral complex
• 2 molecules or ions of a bidentate ligand and 2
molecules or ions of a monodentate ligand forma an
octahedral complex
• An octahedral complex forms with one hexadentate
ligand. See pics.
Week 28
Optical isomers of [Ni(NH2CH2CH2NH2)3]2+
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Ligand Substitution in Complexes
• Well done on p. 214/5 in text book.
• A ligand substitution reaction is one in which a
ligand in a complex ion is replaced by another
ligand.
• These are usually seen when an aqueous ion
containing the hexaaquo complex ion is
reacted with a solution containing another
ligand.
• E.g. when ammonia solution is added to an
aqueous solution of a copper salt.
• Pale blue → pale blue gelatinous precipitate→
dark blue solution.
Changes?
• [Cu(H2O)6]2+(aq) +4NH3(aq)D
[Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l)
• Note that the intermediate precipitate is not
included in this equation.
• (The precipitate is actually [Cu(OH)2(H2O)4](s)
a neutral complex which is why it
precipitates.)
• See others on the page.
The biochemical importance of iron in
haemoglobin
Haem
Haemoglobin
• Haem is the name given to the iron complex on
the previous slide where the Fe2+ uses 4 of its
6 coordination sites in coordinate bonds to
the nitrogen lone pairs of the porphyrin ring.
• A fifth site is used to bind to the protein
globin.
• The sixth site is used to bind reversibly with
oxygen to form RED oxyhaemoglobin.
• When oxygen is removed it is replaced by
water to form BLUE deoxyhaemoglobin (which
is why veins look blue as they carry
deoxygenated blood back to the lungs).
Carrying carbon dioxide
• When oxygenated blood reaches the cells and
tissues carbon dioxide, present in high
concentration displaces the oxygen, forming
carbaminohaemoglobin.
• This is carried back to the lungs where
oxygen once again displaces the carbon
dioxide to form oxyhaemoglobin.
• These reactions are ligand substitution
reactions where oxygen, water, carbon
dioxide and carbon monoxide all displace one
another
Carrying Carbon Monoxide
• Carbon monoxide binds 200 times more effectively
with haemoglobin than oxygen and the reaction is
irreversible.
• High concentrations of CO use up the binding sites
and so asphyxiation results – although victims look
healthy because carboxyhaemoglobin is a very bright
cherry red.
When inspired air contains CO levels as low as 0.02%,
headache and nausea occur. If the CO concentration
is increased to 0.1%, unconsciousness will follow. In
heavy smokers, up to 20% of the oxygen active sites
can be blocked by CO.
Stability Constants Kstab
• The stability constant, Kstab of a complex ion is the
equilibrium constant for the formation of a complex
ion in a solvent from its constituent ions.
• Kstab = prods
as before.
•
reactants
• For the reaction:
• [Co(H2O)6]2+(aq) + 4Cl-(aq) D [CoCl4]2-(aq) + 6H2O(l)
• Kstab = prods
•
reactants
= [CoCl42- ]
[[Co(H2O)6]2+][Cl-]4
• Water not included because large excess and concentration
virtually constant.
• Units also need to be calculated as before.
The value of Kstab
• Large value indicates equilibium lies RIGHTmore products than reactants.
• Complexes with high Kstab are more stable
than those with lower values.
• A large value of Kstab shows the ion is easily
formed.
• Since stability constants are often very large
they can be logged to make the numbers more
manageable.
Some values:
Complex ion
Colour
[CuCl4]2-
yellow
Stability
constant log Kstab
5.6
[Cu(NH3)4(H2O)2]2+
Deep blue
13.1
[Cu(edta)]2-
Pale blue
18.8
[Ni(NH3)6]2+
purple
8.0
[Fe(CN)6]3-
yellow
31.0
• Each of the previous values were compared
with the hexaaquo complex and so can be
thought of as that many times more stable
than the hexaaquo complex.
• Predict the order that you would add
ammonia, edta and HCl to a pale blue solution
of copper sulphate to prove the stability of
the relevant complexes listed.
Redox Titrations
• Examples include:
• Potassium permanganate with iron II,
• Potassium permanganate and hydrogen
peroxide
• Sodium thiosulphate and iodine
• Estimating copper using iodine and sodium
thiosulphate
Reagents
Solution in
burette
Start colour Colour at
end point
KMnO4/Fe2+
KMnO4
Pale green
First hint of
pink
KMnO4/H2O2 KMnO4
Colourless
First hint of
pale pink
I2/Na2S2O3
Thio
orange
colourless
Cu2+/I2/thio
thio
Orange in
white ppte
Colourless
with
unchanged
white ppte
Permanganate Titrations
•
•
•
•
•
•
Permanganate is the oxidising agent.
Fe2+ is oxidised to Fe3+.
Half equations:
MnO4- + 8H+ + 5e → Mn2+ + 4H2O
Fe2+ →Fe3+ + e
Need 5 Fe2+ to supply the 5e needed to reduce the
MnO4• Dilute sulphuric acid is added to the conical flask
containing the permanganate to supply the H+ needed
to balance the equation.
• Without the acid the reduction stops at a brown
precipitate of MnO2 and the reaction is incomplete.
Hydrogen Peroxide Titration
• Permanganate is the oxidising agent.
• H2O2 → 2H+ + 2e- + O2
• MnO4- + 8H+ + 5e → Mn2+ + 4H2O
• Purple
pale pink
• As before the sulphuric must be added to the permanganate to
supply the H+ needed to complete the reduction of the MnO4-
• Balanced equation: (by combining the 2 half equations)
• 5H2O2 + 2MnO4- + 6H+ → + 2Mn2+ +8H2O + 5O2
Iodine/Thio Titration
•
I2(aq) + 2S2O32-(aq) → 2I-(aq) + S4O62-(aq)
• The iodine is liberated from KI by an oxidising agent e.g.
permanganate.
• The iodine is then reduced back to iodide by the thiosulphate
• The orange iodine solution is generated in a conical flask and
thio run in from a burette until the solution is straw yellow.
• At this point starch is added to form a blue solution. (Care must
be taken so that blobs of iodine trapped in the starch micelles
are not formed so the starch must not be added too soon.)
• The blue colour should then be discharged by the addition of
only one drop more of thio.
• Since 1 mol I2(aq) ≡ 2mol thiosulphate we can work back to find
out the concentration of the original oxiding agent.
Analysing Copper
• The analysis is done by adding KI(aq) to a solution of
Cu2+ ions.
• This precipitates all the copper as CuI, a white solid.
• 2Cu2+ + 4I- → 2CuI + I2
• The iodine produced can be found by titrating the
mixture with sodium thiosulphate solution of known
concentration.
• The thiosulphate ions reduce the iodine back to
iodide ions
• 2S2O32- + I2 → S4O62- + 2I• As the end point is neared, the brown colour due to
the iodine becomes faint and starch solution is added
to turn the mixture blue. The disappearance of the
blue is easier to judge than the loss of the pale
yellow/brown colour.
Analysing results
• Since 2Cu2+ ≡ I2 ≡ 2S2O32• Then the number of moles of thiosulphate
added is the same as the number of moles of
copper originally present.
• From this the original mass of copper can be
calculated.
The maths!
• Standardising an approximately 0.02
moldm-3 solution of potassium permanganate
using an iron II salt.
• RMM iron ammonium sulphate = 392
• Mass iron ammonium sulphate = 9.829g
• Volume KMnO4 used in titration = 25.70 cm3
• Equations:MnO4- + 8H+ + 5e → Mn2+ + 4H2O
• Fe2+ →Fe3+ + e
• The solid was made up to 250 cm3 in a
volumetric flask.
• The titration was carried out using 25cm3
portions of the iron II salt.
• 1. Work out the balanced equation for the reaction.
• MnO4- + 8H+ + 5 Fe2+ → Mn2+ + 4H2O + 5 Fe3+
• 2. Calculate the number of moles of iron II salt
weighed out.
• 9.829 = 0.025mols
• 392
• 3. Calculate the concentration of the iron II salt
solution in the flask
Conc = mols
= 0.025 = 0.1003 moldm-3
•
volume (litres)
0.25
• 4. How many mols of iron II salt were used in each
titration?
• 0.1003/1000 x 25 = 0.002508 mols
• 5. From the equation for the reaction how many moles
of permanganate reacted with the iron II salt in the
titration.
• 5Fe2+ ≡ 1MnO4• MnO4- = 0.002508/5 = 0.00050 mol in 25.7cm3
• 6. What was the accurate concentration for the
permanganate supplied?
• Concentration KMnO4 = 0.00050 x 1000
•
25.7
•
= 0.0195 mol dm-3
To standardise a solution of hydrogen
peroxide using potassium permanganate
• 5.5cm3 of hydrogen peroxide solution were made up
to 250cm3 in a volumetric flask.
• 25cm3 of this solution was titrated with standardised
potassium permanganate solution.
• 10cm3 of dilute sulphuric acid was added to each
titration to provide the H+ ions needed to complete
the reduction of the permanganate.
• 19.38cm3 of 0.0194 mol dm-3 potassium permanganate
were required to complete the oxidation of the
hydrogen peroxide.
• Equations:
• H2O2 → 2H+ + 2e- + O2
• MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
Calculations
• 1. What is the balanced equation for the reaction?
• 2MnO4- + 6H+ + 5H2O2 → 5O2 + 2Mn2+ +8H2O
• 2.How many mols of MnO4- were present at the end
point for this reaction?
• 19.38 cm3 of 0.0194 mol dm-3 potassium permanganate
were present at the end point.
•  moles = conc x vol (dm3)
•
= 0.0194 x 19.38 = 0.0003759 moles
•
1000
• 3.How many moles of hydrogen peroxide must have
been present?
• 2MnO4- ≡ 5H2O2
• peroxide = 0.0003759 x 5
•
2
•
= 0.00093975 mols peroxide
• 4. Calculate the concentration of the hydrogen
peroxide in the conical flask.
• 25cm3 peroxide contain 0.00093975 mols
•  = 0.00093975 = 0.03759 mol dm-3
•
0.025
• 5. Calculate the original concentration of the
peroxide used to make up the solution in the conical
flask.
• [Peroxide] = 0.03759 mol dm-3
• This concentration was for the solution in the
volumetric flask.
•  For the moles in 250cm3 volumetric flask
• = 0.03759 = 0.009398 moles.
•
4
• This came originally from 5.5 cm3 of concentrated
peroxide.
• Original peroxide contained 0.009398 moles in 5.5
cm3
Concentration of original, undiluted
hydrogen peroxide
• Original peroxide contained 0.009398 moles in 5.5
cm3
•  Original concentration = moles
volume in dm3
= 0.009398
0.0055
= 1.709 mol dm-3
6 Hydrogen peroxide is usually sold as a solution with an
oxygen volume concentration. What is the oxygen
volume concentration for this solution?
e.g. 20 volume hydrogen peroxide produces 20 volumes
of oxygen for 1 volume of aqueous peroxide.
2H2O2 → H2O + O2
• e.g. 20 volume hydrogen peroxide produces 20
volumes of oxygen for 1 volume of aqueous peroxide.
• 2H2O2 → H2O + O2
• 2 moles peroxide → 24 litres O2 at RTP
• 1mole
12 litres
• Our solution is 1.709 moles per litre
•  Oxygen volume concentration = 1.709 x12 = 20.508
vol
To standardise an approximately 0.1 mol dm-3 solution of
sodium thiosulphate using standardised potassium
permanganate solution
• 1g KI solid was added to acidified distilled water and 25 cm3
standardised (0.0194 mol dm-3) permanganate was added.
• Sodium thiosulphate solution was run in from a burette until the
solution was straw yellow. At this point 2 drops of starch
solution was added and the titration continued until the blue
starch colour was discharged.
• Volume of thio at end point = 23.95 cm3
• The KI was oxidised to iodine by the permanganate and then
reduced again by the thiosulphate.
• The KI was in excess to ensure that all the permanganate had
been used up in the oxidation.
Equations
• 2MnO4- + 16H+ + 10I- → 2Mn2+ + 8H2O +5I2
• 2S2O32- + I2→ S4O62- +2I• 1. What is the molar ratio for thiosulphate and permanganate
ions?
• Since 2 MnO4 ≡ 5I2
• And 2 thio ≡ 1 I2
• Then 5 I2
=10 thio = 2MnO4
• 5 mol thiosulphate ≡ 1 mol permanganate
• 2. Calculate the number of mols of permanganate present in the
conical flask.
• 25.00 cm3 of 0.0194 mol dm-3 permanganate present in flask.
• Moles = conc x vol in dm3 = 0.0194 x 0.025 =0.000485 moles
•
Use the previous value to calculate the number of moles
of thiosulphate present at the end point.
• 0.000485 moles permanganate ≡ 5 x 0.000485 moles thio
•
= 0.002425 moles thio.
• 3. Use the number of moles of thio needed at the
endpoint to calculate the concentration of the
thiosulphate solution.
• 0.002425 moles in 23.95cm3 solution.
• Concentration = moles
•
volume (dm3)
•
= 0.002425
•
0.02395
•
= 0.10125 = 0.1013 mol dm-3
Analysing Copper
• A 1.00g piece of brass is dissolved in nitric acid to produce a
mixture of copper (II) nitrate and zinc nitrate solutions.
• An excess of potassium iodide solution is added, causing copper
(I) iodide and iodine to form.
• The iodine formed reacts with 47.8 cm3 of 0.200 mol dm-3
sodium thiosulphate solution.
•
I2(aq) + 2S2O32-(aq) → 2I-(aq) + S4O62-(aq)
• 2Cu2+(aq) + 4I-(aq) → 2CuI(s) + I2(aq)
• 1. What is the molar ratio for thiosulphate and copper ions?
• 2. How many moles of sodium thiosulphate were needed to react
with all of the iodine?
• 3. How many moles Cu2+ were present before the excess of
potassium iodide solution was added?
• 4.What mass of copper is this? (Ar copper 63.5)
• 5. What was the percentage by mass of copper in the original
piece of brass?
Ans
•
•
•
•
•
•
•
•
•
•
1. 2Cu2+(aq) ≡ I2(aq) ≡ 2S2O32-(aq)
2. 47.8 cm3 of 0.200 mol dm-3 solution.
Moles = conc x vol in dm3
= 0.200 x 0.0478
= 0.00956 mol thiosulphate
3.Moles thio = moles Cu2+ originally present
 moles Cu2+ = 0.00956 mol
4. Mass copper = 0.00956 x 63.5
= 0.60706g
5.% by mass copper = copper/total
= 0.60706 x 100
1
•
= 60.7% ( 3 sf)