AN INTRODUCTION TOAN INTRODUCTION TO

COPPER / COPPER / THIOSULPHATE THIOSULPHATE

TITRATIONSTITRATIONS

KNOCKHARDY PUBLISHINGKNOCKHARDY PUBLISHING2008 2008

SPECIFICATIONSSPECIFICATIONS

INTRODUCTION

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THIOSULPHATE TITRATIONSTHIOSULPHATE TITRATIONS

COPPER(II) / THIOSULPHATE TITRATIONSCOPPER(II) / THIOSULPHATE TITRATIONS

General theoryGeneral 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) ——> S4O6

2-(aq) + 2I¯(aq)

therefore moles of S2O32- = moles of Cu2+(aq)

COPPER(II) / THIOSULPHATE TITRATIONSCOPPER(II) / THIOSULPHATE TITRATIONS

Practical details

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.

A

A

COPPER(II) / THIOSULPHATE TITRATIONSCOPPER(II) / THIOSULPHATE TITRATIONS

Practical details

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.

2Cu2+(aq) + 4I¯(aq) ——> 2CuI(s) + I2(aq)off white solid

A B

B

+2 -1 +1 -1 0

COPPER(II) / THIOSULPHATE TITRATIONSCOPPER(II) / THIOSULPHATE TITRATIONS

Practical details

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.

2S2O32-(aq) + I2(aq) ——> S4O6

2-(aq) + 2I¯(aq)

A B C

C

0 -1

COPPER(II) / THIOSULPHATE TITRATIONSCOPPER(II) / THIOSULPHATE TITRATIONS

Practical details

Starch solution is added near the end point.

Starch gives a dark blue colouration in the presence of iodine.

A B C D

D

COPPER(II) / THIOSULPHATE TITRATIONSCOPPER(II) / THIOSULPHATE TITRATIONS

Practical details

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.

A B C D E

E

TYPICAL CALCULATIONSTYPICAL 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 S2O32-

4 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 CALCULATIONSTYPICAL 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 S2O32-

4 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 1CALCULATIONS – 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 solutionb) the molar mass of the copper(II) sulphatec) the number of water molecules of crystallisation (x) in the formula

CALCULATIONS – Example 1CALCULATIONS – 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 solutionb) the molar mass of the copper(II) sulphatec) the number of water molecules of crystallisation (x) in the formula

From these two equations 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)2S2O3

2-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)

you get moles of S2O32- = moles of Cu2+(aq)

CALCULATIONS – Example 1CALCULATIONS – 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 solutionb) the molar mass of the copper(II) sulphatec) the number of water molecules of crystallisation (x) in the formula

From these two equations 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)2S2O3

2-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)

you get moles of S2O32- = moles of Cu2+(aq)

moles of S2O32- = 0.12 x 22.25 / 1000 = 2.67 x 10-3

moles of Cu2+ in 25cm3 = moles of S2O32- = 2.67 x 10-3

CALCULATIONS – Example 1CALCULATIONS – 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 solutionb) the molar mass of the copper(II) sulphatec) the number of water molecules of crystallisation (x) in the formula

From these two equations 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)2S2O3

2-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)

you get moles of S2O32- = moles of Cu2+(aq)

moles of S2O32- = 0.12 x 22.25 / 1000 = 2.67 x 10-3

moles of Cu2+ in 25cm3 = moles of S2O32- = 2.67 x 10-3

moles of Cu2+ in 1dm3 = 2.67 x 10-3 x 40= 0.107

CALCULATIONS – Example 1CALCULATIONS – 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 solutionb) the molar mass of the copper(II) sulphatec) the number of water molecules of crystallisation (x) in the formula

From these two equations 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)2S2O3

2-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)

you get moles of S2O32- = moles of Cu2+(aq)

moles of S2O32- = 0.12 x 22.25 / 1000 = 2.67 x 10-3

moles of Cu2+ in 25cm3 = moles of S2O32- = 2.67 x 10-3

moles of Cu2+ in 1dm3 = 2.67 x 10-3 x 40= 0.107mass of Cu2+ in 1dm3 = 0.107 x 63.5 = 0.679g

CALCULATIONS – Example 1CALCULATIONS – 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 solutionb) the molar mass of the copper(II) sulphatec) the number of water molecules of crystallisation (x) in the formula

From these two equations 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)2S2O3

2-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)

you get moles of S2O32- = moles of Cu2+(aq)

moles of S2O32- = 0.12 x 22.25 / 1000 = 2.67 x 10-3

moles of Cu2+ in 25cm3 = moles of S2O32- = 2.67 x 10-3

moles of Cu2+ in 1dm3 = 2.67 x 10-3 x 40= 0.107mass of Cu2+ in 1dm3 = 0.107 x 63.5 = 6.79g

% of Cu2+ in compound = 6.79 / 26.50 x 100 = 25.64%

CALCULATIONS – Example 1CALCULATIONS – 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 solutionb) the molar mass of the copper(II) sulphatec) the number of water molecules of crystallisation (x) in the formula

From these two equations 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)2S2O3

2-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)

you get moles of S2O32- = moles of Cu2+(aq)

moles of S2O32- = 0.12 x 22.25 / 1000 = 2.67 x 10-3

moles of Cu2+ in 25cm3 = moles of S2O32- = 2.67 x 10-3

moles of Cu2+ in 1dm3 = 2.67 x 10-3 x 40= 0.107mass of Cu2+ in 1dm3 = 0.107 x 63.5 = 6.79g

CALCULATIONS – Example 1CALCULATIONS – 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 solutionb) the molar mass of the copper(II) sulphatec) the number of water molecules of crystallisation (x) in the formula

From these two equations 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)2S2O3

2-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)

you get moles of S2O32- = moles of Cu2+(aq)

moles of S2O32- = 0.12 x 22.25 / 1000 = 2.67 x 10-3

moles of Cu2+ in 25cm3 = moles of S2O32- = 2.67 x 10-3

moles of Cu2+ in 1dm3 = 2.67 x 10-3 x 40= 0.107mass of Cu2+ in 1dm3 = 0.107 x 63.5 = 6.79g

molar mass of compound = mass/moles = 26.50 / 0.107 = 247.66

CALCULATIONS – Example 1CALCULATIONS – 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 solutionb) the molar mass of the copper(II) sulphatec) the number of water molecules of crystallisation (x) in the formula

From these two equations 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)2S2O3

2-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)

you get moles of S2O32- = moles of Cu2+(aq)

moles of S2O32- = 0.12 x 22.25 / 1000 = 2.67 x 10-3

moles of Cu2+ in 25cm3 = moles of S2O32- = 2.67 x 10-3

moles of Cu2+ in 1dm3 = 2.67 x 10-3 x 40= 0.107mass of Cu2+ in 1dm3 = 0.107 x 63.5 = 6.79g

% of Cu2+ in compound = 6.79 / 26.50 x 100 = 25.64%

molar mass of compound = mass/moles = 26.50 / 0.107 = 247.66mass of water = mass of CuSO4.xH2O - mass of CuSO4

= 247.66 - 159.50 = 88.16moles of water (x) = mass / molar mass = 88.16 / 18 = 4.9 (5)

CALCULATIONS – Example 1CALCULATIONS – 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 solutionb) the molar mass of the copper(II) sulphatec) the number of water molecules of crystallisation (x) in the formula

From these two equations 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)2S2O3

2-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)

you get moles of S2O32- = moles of Cu2+(aq)

moles of S2O32- = 0.12 x 22.25 / 1000 = 2.67 x 10-3

moles of Cu2+ in 25cm3 = moles of S2O32- = 2.67 x 10-3

moles of Cu2+ in 1dm3 = 2.67 x 10-3 x 40= 0.107mass of Cu2+ in 1dm3 = 0.107 x 63.5 = 6.79g

% of Cu2+ in compound = 6.79 / 26.50 x 100 = 25.64%

molar mass of compound = mass/moles = 26.50 / 0.107 = 247.66

mass of water = mass of CuSO4.xH2O - mass of CuSO4 = 88.16moles of water (x) = mass / molar mass = 88.16 / 18 = 4.9 (5)

CALCULATIONS – Example 2CALCULATIONS – 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 solutionb) the molar mass of the copper(II) sulphatec) the number of water molecules of crystallisation (x) in the formula

From these two equations 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)2S2O3

2-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)

you get moles of S2O32- = moles of Cu2+(aq)

moles of S2O32- = 0.12 x 22.25 / 1000 = 2.67 x 10-3

moles of Cu2+ in 25cm3 = moles of S2O32- = 2.67 x 10-3

moles of Cu2+ in 1dm3 = 2.67 x 10-3 x 40= 0.107mass of Cu2+ in 1dm3 = 0.107 x 63.5 = 6.79g

% of Cu2+ in compound = 6.79 / 26.50 x 100 = 25.64%

molar mass of compound = mass/moles = 26.50 / 0.107 = 247.66

mass of water = mass of CuSO4.xH2O - mass of CuSO4 = 88.16moles of water (x) = mass / molar mass = 88.16 / 18 = 4.9 (5)

CALCULATIONS – Example 2CALCULATIONS – 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 solutionCalculate the percentage of copper in the alloy.

CALCULATIONS – Example 2CALCULATIONS – 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 solutionCalculate the percentage of copper in the alloy.

From these two equations 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)2S2O3

2-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)

you get moles of S2O32- = moles of Cu2+(aq)

CALCULATIONS – Example 2CALCULATIONS – 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 solutionCalculate the percentage of copper in the alloy.

From these two equations 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)2S2O3

2-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)

you get moles of S2O32- = moles of Cu2+(aq)

moles of S2O32- = 0.100 x 25.00 / 1000 = 2.50 x 10-3

CALCULATIONS – Example 2CALCULATIONS – 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 solutionCalculate the percentage of copper in the alloy.

From these two equations 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)2S2O3

2-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)

you get 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 2CALCULATIONS – 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 solutionCalculate the percentage of copper in the alloy.

From these two equations 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)2S2O3

2-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)

you get 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 2CALCULATIONS – 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 solutionCalculate the percentage of copper in the alloy.

From these two equations 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)2S2O3

2-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)

you get 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 2CALCULATIONS – 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 solutionCalculate the percentage of copper in the alloy.

From these two equations 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)2S2O3

2-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)

you get 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 2CALCULATIONS – 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 solutionCalculate the percentage of copper in the alloy.

From these two equations 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq)2S2O3

2-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq)

you get 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%

© 2010 JONATHAN HOPTON & KNOCKHARDY PUBLISHING© 2010 JONATHAN HOPTON & KNOCKHARDY PUBLISHING

THE ENDTHE END

AN INTRODUCTION TOAN INTRODUCTION TO

COPPER / COPPER / THIOSULPHATE THIOSULPHATE

TITRATIONSTITRATIONS