Transcript moles of Cu 2+ (aq)

AN INTRODUCTION TO
COPPER /
THIOSULPHATE
TITRATIONS
KNOCKHARDY PUBLISHING
2008
SPECIFICATIONS
KNOCKHARDY PUBLISHING
THIOSULPHATE TITRATIONS
INTRODUCTION
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COPPER(II) / THIOSULPHATE TITRATIONS
General theory
Copper(II) compounds can be analysed by a redox titration.
The general procedure is that excess potassium iodide solution is added to a
neutral solution of copper(II). This liberates iodine according to the equation
below and the amount of iodine is found by titration with sodium thiosulphate
solution. Just before the end-point, several drops of starch solution are added
and one continues the titration until the blue colour just disappears and an
off-white precipitate remains.
2Cu2+(aq) + 4I¯(aq) ——> 2CuI(s) + I2(aq)
2S2O32-(aq) + I2(aq) ——> S4O62-(aq) + 2I¯(aq)
therefore
moles of S2O32- = moles of Cu2+(aq)
COPPER(II) / THIOSULPHATE TITRATIONS
Practical details
A
A
Pipette a known volume of a solution of Cu2+ ions into a conical flask.
(alternatively dissolve a known mass of solid in water)
Neutralise the solution by adding sodium carbonate solution dropwise
until a feint precipitate starts to form.
COPPER(II) / THIOSULPHATE TITRATIONS
Practical details
A
B
B
Add excess potassium iodide solution to liberate iodine.
The copper(II) is reduced to copper(I) and half the iodide ions are
oxidised to iodine.
+2
2Cu2+(aq)
-1
+
4I¯(aq)
——>
+1 -1
2CuI(s)
off white solid
0
+
I2(aq)
COPPER(II) / THIOSULPHATE TITRATIONS
Practical details
A
C
B
C
Titrate with a standard solution of sodium thiosulphate until the
solution lightens. DO NOT ADD TOO MUCH.
The iodine is reduced back to iodide ions.
0
2S2O3
2-(aq)
+
I2(aq)
——>
-1
S 4 O6
2-(aq)
+
2I¯(aq)
COPPER(II) / THIOSULPHATE TITRATIONS
Practical details
A
D
B
C
D
Starch solution is added near the end point.
Starch gives a dark blue colouration in the presence of iodine.
COPPER(II) / THIOSULPHATE TITRATIONS
Practical details
A
E
B
C
D
Continue with the titration, adding the sodium thiosulphate
dropwise until the blue colour disappears at the end point. This
indicates that all the iodine has reacted.
Record the volume added and repeat to obtain concordant results.
E
TYPICAL CALCULATIONS
Percentage copper in a compound
1 titrate a known mass of copper compound or a known volume of a solution
2 calculate the moles of S2O32- needed
3 according to the equation… moles of Cu2+ = moles of S2O324 calculate the number of moles of Cu2+
5 calculate the mass of copper by multiplying the moles of copper by the
molar mass of copper.
6 divide the mass of copper by the mass of the weighed solid to find the
fraction and hence calculate the percentage of copper in the sample.
TYPICAL CALCULATIONS
Number of water molecules of crystallisation
1 titrate a known mass of copper compound or a known volume of a solution
2 calculate the moles of S2O32- needed
3 according to the equation… moles of Cu2+ = moles of S2O324 calculate the number of moles of Cu2+
5 calculate the number of moles of CuSO4
moles of CuSO4 = moles of Cu2+ (there is one Cu2+ in every CuSO4)
6 calculate the mass of copper sulphate by multiplying the moles of copper
sulphate by the molar mass of copper sulphate (CuSO4)
7 calculate mass of water ( = mass of CuSO4.xH2O - mass of CuSO4)
8 divide the mass of water by 18 to find the number of moles of water
9
Compare the ration of moles of… water : moles of CuSO4 to find x
CALCULATIONS – Example 1
25cm3 of a solution of CuSO4.xH2O of concentration 26.50 g dm-3 was placed in a
conical flask and an excess of KI added. 22.25cm3 of a 0.12 mol dm-3 solution of sodium
thiosulphate was required to reduce the iodine produced. Calculate…
a) the percentage of copper in copper(II) sulphate solution
b) the molar mass of the copper(II) sulphate
c) the number of water molecules of crystallisation (x) in the formula
CALCULATIONS – Example 1
25cm3 of a solution of CuSO4.xH2O of concentration 26.50 g dm-3 was placed in a
conical flask and an excess of KI added. 22.25cm3 of a 0.12 mol dm-3 solution of sodium
thiosulphate was required to reduce the iodine produced. Calculate…
a) the percentage of copper in copper(II) sulphate solution
b) the molar mass of the copper(II) sulphate
c) the number of water molecules of crystallisation (x) in the formula
From these two equations
you get
2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)
2S2O32-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)
moles of S2O32- = moles of Cu2+(aq)
CALCULATIONS – Example 1
25cm3 of a solution of CuSO4.xH2O of concentration 26.50 g dm-3 was placed in a
conical flask and an excess of KI added. 22.25cm3 of a 0.12 mol dm-3 solution of sodium
thiosulphate was required to reduce the iodine produced. Calculate…
a) the percentage of copper in copper(II) sulphate solution
b) the molar mass of the copper(II) sulphate
c) the number of water molecules of crystallisation (x) in the formula
From these two equations
you get
moles of S2O32moles of Cu2+ in 25cm3
2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)
2S2O32-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)
moles of S2O32- = moles of Cu2+(aq)
= 0.12 x 22.25 / 1000
= moles of S2O32-
= 2.67 x 10-3
= 2.67 x 10-3
CALCULATIONS – Example 1
25cm3 of a solution of CuSO4.xH2O of concentration 26.50 g dm-3 was placed in a
conical flask and an excess of KI added. 22.25cm3 of a 0.12 mol dm-3 solution of sodium
thiosulphate was required to reduce the iodine produced. Calculate…
a) the percentage of copper in copper(II) sulphate solution
b) the molar mass of the copper(II) sulphate
c) the number of water molecules of crystallisation (x) in the formula
From these two equations
you get
moles of S2O32moles of Cu2+ in 25cm3
moles of Cu2+ in 1dm3
2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)
2S2O32-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)
moles of S2O32- = moles of Cu2+(aq)
= 0.12 x 22.25 / 1000
= 2.67 x 10-3
= moles of S2O32= 2.67 x 10-3
= 2.67 x 10-3 x 40= 0.107
CALCULATIONS – Example 1
25cm3 of a solution of CuSO4.xH2O of concentration 26.50 g dm-3 was placed in a
conical flask and an excess of KI added. 22.25cm3 of a 0.12 mol dm-3 solution of sodium
thiosulphate was required to reduce the iodine produced. Calculate…
a) the percentage of copper in copper(II) sulphate solution
b) the molar mass of the copper(II) sulphate
c) the number of water molecules of crystallisation (x) in the formula
From these two equations
you get
moles of S2O32moles of Cu2+ in 25cm3
moles of Cu2+ in 1dm3
mass of Cu2+ in 1dm3
2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)
2S2O32-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)
moles of S2O32- = moles of Cu2+(aq)
=
=
=
=
0.12 x 22.25 / 1000
= 2.67 x 10-3
moles of S2O32= 2.67 x 10-3
2.67 x 10-3 x 40= 0.107
0.107 x 63.5
= 0.679g
CALCULATIONS – Example 1
25cm3 of a solution of CuSO4.xH2O of concentration 26.50 g dm-3 was placed in a
conical flask and an excess of KI added. 22.25cm3 of a 0.12 mol dm-3 solution of sodium
thiosulphate was required to reduce the iodine produced. Calculate…
a) the percentage of copper in copper(II) sulphate solution
b) the molar mass of the copper(II) sulphate
c) the number of water molecules of crystallisation (x) in the formula
From these two equations
you get
2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)
2S2O32-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)
moles of S2O32- = moles of Cu2+(aq)
moles of S2O32moles of Cu2+ in 25cm3
moles of Cu2+ in 1dm3
mass of Cu2+ in 1dm3
=
=
=
=
0.12 x 22.25 / 1000
= 2.67 x 10-3
moles of S2O32= 2.67 x 10-3
2.67 x 10-3 x 40= 0.107
0.107 x 63.5
= 0.679g
% of Cu2+ in compound
= 0.679 / 26.50 x 100
= 25.64%
CALCULATIONS – Example 1
25cm3 of a solution of CuSO4.xH2O of concentration 26.50 g dm-3 was placed in a
conical flask and an excess of KI added. 22.25cm3 of a 0.12 mol dm-3 solution of sodium
thiosulphate was required to reduce the iodine produced. Calculate…
a) the percentage of copper in copper(II) sulphate solution
b) the molar mass of the copper(II) sulphate
c) the number of water molecules of crystallisation (x) in the formula
From these two equations
you get
moles of S2O32moles of Cu2+ in 25cm3
moles of Cu2+ in 1dm3
mass of Cu2+ in 1dm3
2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)
2S2O32-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)
moles of S2O32- = moles of Cu2+(aq)
=
=
=
=
0.12 x 22.25 / 1000
= 2.67 x 10-3
moles of S2O32= 2.67 x 10-3
2.67 x 10-3 x 40= 0.107
0.107 x 63.5
= 0.679g
CALCULATIONS – Example 1
25cm3 of a solution of CuSO4.xH2O of concentration 26.50 g dm-3 was placed in a
conical flask and an excess of KI added. 22.25cm3 of a 0.12 mol dm-3 solution of sodium
thiosulphate was required to reduce the iodine produced. Calculate…
a) the percentage of copper in copper(II) sulphate solution
b) the molar mass of the copper(II) sulphate
c) the number of water molecules of crystallisation (x) in the formula
From these two equations
you get
2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)
2S2O32-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)
moles of S2O32- = moles of Cu2+(aq)
moles of S2O32moles of Cu2+ in 25cm3
moles of Cu2+ in 1dm3
mass of Cu2+ in 1dm3
=
=
=
=
0.12 x 22.25 / 1000
= 2.67 x 10-3
moles of S2O32= 2.67 x 10-3
2.67 x 10-3 x 40= 0.107
0.107 x 63.5
= 0.679g
molar mass of compound
= mass/moles
= 26.50 / 0.107
= 247.66
CALCULATIONS – Example 1
25cm3 of a solution of CuSO4.xH2O of concentration 26.50 g dm-3 was placed in a
conical flask and an excess of KI added. 22.25cm3 of a 0.12 mol dm-3 solution of sodium
thiosulphate was required to reduce the iodine produced. Calculate…
a) the percentage of copper in copper(II) sulphate solution
b) the molar mass of the copper(II) sulphate
c) the number of water molecules of crystallisation (x) in the formula
From these two equations
you get
2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)
2S2O32-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)
moles of S2O32- = moles of Cu2+(aq)
moles of S2O32moles of Cu2+ in 25cm3
moles of Cu2+ in 1dm3
mass of Cu2+ in 1dm3
=
=
=
=
% of Cu2+ in compound
= 0.679 / 26.50 x 100
molar mass of compound
mass of water
=
=
=
=
moles of water (x)
0.12 x 22.25 / 1000
= 2.67 x 10-3
moles of S2O32= 2.67 x 10-3
2.67 x 10-3 x 40= 0.107
0.107 x 63.5
= 0.679g
= 25.64%
mass/moles
= 26.50 / 0.107
mass of CuSO4.xH2O - mass of CuSO4
247.66 - 159.50
mass / molar mass
= 88.16 / 18
= 247.66
= 88.16
= 4.9 (5)
CALCULATIONS – Example 1
25cm3 of a solution of CuSO4.xH2O of concentration 26.50 g dm-3 was placed in a
conical flask and an excess of KI added. 22.25cm3 of a 0.12 mol dm-3 solution of sodium
thiosulphate was required to reduce the iodine produced. Calculate…
a) the percentage of copper in copper(II) sulphate solution
b) the molar mass of the copper(II) sulphate
c) the number of water molecules of crystallisation (x) in the formula
From these two equations
you get
2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)
2S2O32-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)
moles of S2O32- = moles of Cu2+(aq)
moles of S2O32moles of Cu2+ in 25cm3
moles of Cu2+ in 1dm3
mass of Cu2+ in 1dm3
=
=
=
=
0.12 x 22.25 / 1000
= 2.67 x 10-3
moles of S2O32= 2.67 x 10-3
2.67 x 10-3 x 40= 0.107
0.107 x 63.5
= 0.679g
% of Cu2+ in compound
= 0.679 / 26.50 x 100
= 25.64%
molar mass of compound
= mass/moles
= 26.50 / 0.107
mass of water
moles of water (x)
= mass of CuSO4.xH2O - mass of CuSO4
= mass / molar mass
= 88.16 / 18
= 247.66
= 88.16
= 4.9 (5)
CALCULATIONS – Example 2
25cm3 of a solution of CuSO4.xH2O of concentration 26.50 g dm-3 was placed in a
conical flask and an excess of KI added. 22.25cm3 of a 0.12 mol dm-3 solution of sodium
thiosulphate was required to reduce the iodine produced. Calculate…
a) the percentage of copper in copper(II) sulphate solution
b) the molar mass of the copper(II) sulphate
c) the number of water molecules of crystallisation (x) in the formula
From these two equations
you get
2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)
2S2O32-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)
moles of S2O32- = moles of Cu2+(aq)
moles of S2O32moles of Cu2+ in 25cm3
moles of Cu2+ in 1dm3
mass of Cu2+ in 1dm3
=
=
=
=
0.12 x 22.25 / 1000
= 2.67 x 10-3
moles of S2O32= 2.67 x 10-3
2.67 x 10-3 x 40= 0.107
0.107 x 63.5
= 0.679g
% of Cu2+ in compound
= 0.679 / 26.50 x 100
= 25.64%
molar mass of compound
= mass/moles
= 26.50 / 0.107
mass of water
moles of water (x)
= mass of CuSO4.xH2O - mass of CuSO4
= mass / molar mass
= 88.16 / 18
= 247.66
= 88.16
= 4.9 (5)
CALCULATIONS – Example 2
3.00g of a copper alloy was dissolved in acid and the solution made up to 250cm3.
25cm3 was then pipetted into a conical flask and an excess of KI added. It was found
25cm3 of a 0.100M sodium thiosulphate solution
Calculate the percentage of copper in the alloy.
CALCULATIONS – Example 2
3.00g of a copper alloy was dissolved in acid and the solution made up to 250cm3.
25cm3 was then pipetted into a conical flask and an excess of KI added. It was found
25cm3 of a 0.100M sodium thiosulphate solution
Calculate the percentage of copper in the alloy.
From these two equations
you get
2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)
2S2O32-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)
moles of S2O32- = moles of Cu2+(aq)
CALCULATIONS – Example 2
3.00g of a copper alloy was dissolved in acid and the solution made up to 250cm3.
25cm3 was then pipetted into a conical flask and an excess of KI added. It was found
25cm3 of a 0.100M sodium thiosulphate solution
Calculate the percentage of copper in the alloy.
From these two equations
you get
moles of S2O32-
2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)
2S2O32-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)
moles of S2O32- = moles of Cu2+(aq)
= 0.100 x 25.00 / 1000
= 2.50 x 10-3
CALCULATIONS – Example 2
3.00g of a copper alloy was dissolved in acid and the solution made up to 250cm3.
25cm3 was then pipetted into a conical flask and an excess of KI added. It was found
25cm3 of a 0.100M sodium thiosulphate solution
Calculate the percentage of copper in the alloy.
From these two equations
you get
2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)
2S2O32-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)
moles of S2O32- = moles of Cu2+(aq)
moles of S2O32-
= 0.100 x 25.00 / 1000
= 2.50 x 10-3
moles of Cu2+ in 25cm3
= moles of S2O32-
= 2.50 x 10-3
CALCULATIONS – Example 2
3.00g of a copper alloy was dissolved in acid and the solution made up to 250cm3.
25cm3 was then pipetted into a conical flask and an excess of KI added. It was found
25cm3 of a 0.100M sodium thiosulphate solution
Calculate the percentage of copper in the alloy.
From these two equations
you get
2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)
2S2O32-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)
moles of S2O32- = moles of Cu2+(aq)
moles of S2O32-
= 0.100 x 25.00 / 1000
= 2.50 x 10-3
moles of Cu2+ in 25cm3
= moles of S2O32-
= 2.50 x 10-3
moles of Cu2+ in 250cm3
= 2.50 x 10-3 x 10= 0.025
CALCULATIONS – Example 2
3.00g of a copper alloy was dissolved in acid and the solution made up to 250cm3.
25cm3 was then pipetted into a conical flask and an excess of KI added. It was found
25cm3 of a 0.100M sodium thiosulphate solution
Calculate the percentage of copper in the alloy.
From these two equations
you get
2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)
2S2O32-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)
moles of S2O32- = moles of Cu2+(aq)
moles of S2O32-
= 0.100 x 25.00 / 1000
= 2.50 x 10-3
moles of Cu2+ in 25cm3
= moles of S2O32-
= 2.50 x 10-3
moles of Cu2+ in 250cm3
= 2.50 x 10-3 x 10= 0.025
mass of Cu2+ in 250cm3
= 0.025 x 63.5
= 1.588g
CALCULATIONS – Example 2
3.00g of a copper alloy was dissolved in acid and the solution made up to 250cm3.
25cm3 was then pipetted into a conical flask and an excess of KI added. It was found
25cm3 of a 0.100M sodium thiosulphate solution
Calculate the percentage of copper in the alloy.
From these two equations
you get
2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)
2S2O32-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)
moles of S2O32- = moles of Cu2+(aq)
moles of S2O32-
= 0.100 x 25.00 / 1000
= 2.50 x 10-3
moles of Cu2+ in 25cm3
= moles of S2O32-
= 2.50 x 10-3
moles of Cu2+ in 250cm3
= 2.50 x 10-3 x 10= 0.025
mass of Cu2+ in 250cm3
= 0.025 x 63.5
= 1.588g
% of Cu in the alloy
= 1.588 / 3.00 x 100
= 52.91%
CALCULATIONS – Example 2
3.00g of a copper alloy was dissolved in acid and the solution made up to 250cm3.
25cm3 was then pipetted into a conical flask and an excess of KI added. It was found
25cm3 of a 0.100M sodium thiosulphate solution
Calculate the percentage of copper in the alloy.
From these two equations
you get
2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)
2S2O32-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)
moles of S2O32- = moles of Cu2+(aq)
moles of S2O32-
= 0.100 x 25.00 / 1000
= 2.50 x 10-3
moles of Cu2+ in 25cm3
= moles of S2O32-
= 2.50 x 10-3
moles of Cu2+ in 250cm3
= 2.50 x 10-3 x 10= 0.025
mass of Cu2+ in 250cm3
= 0.025 x 63.5
= 1.588g
% of Cu in the alloy
= 1.588 / 3.00 x 100
= 52.91%
AN INTRODUCTION TO
COPPER /
THIOSULPHATE
TITRATIONS
THE END
© 2010 JONATHAN HOPTON & KNOCKHARDY PUBLISHING