KNOCKHARDY PUBLISHING AN INTRODUCTION TO COPPER / THIOSULPHATE TITRATIONS 2008 SPECIFICATIONS KNOCKHARDY PUBLISHING

KNOCKHARDY PUBLISHING THIOSULPHATE TITRATIONS INTRODUCTION This Powerpoint show is one of several produced to help students understand selected topics at AS and A2 level Chemistry. It is based on the requirements of the AQA and OCR specifications but is suitable for other examination boards. Individual students may use the material at home for revision purposes or it may be used for classroom teaching if an interactive white board is available. Accompanying notes on this, and the full range of AS and A2 topics, are available from the KNOCKHARDY SCIENCE WEBSITE at... www.knockhardy.org.uk/sci.htm Navigation is achieved by... either clicking on the grey arrows at the foot of each page or using the left and right arrow keys on the keyboard

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. 2Cu2+(aq) + 4I¯(aq) ——> 2CuI(s) + I2(aq) off white solid +2 -1 +1 -1 0

COPPER(II) / THIOSULPHATE TITRATIONS Practical details A B C 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. 2S2O32-(aq) + I2(aq) ——> S4O62-(aq) + 2I¯(aq) 0 -1

COPPER(II) / THIOSULPHATE TITRATIONS Practical details A B C D 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 B C D E E 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.

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 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 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 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 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq) 2S2O32-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq) you get 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 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq) 2S2O32-(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 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 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq) 2S2O32-(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 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 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq) 2S2O32-(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 mass of Cu2+ in 1dm3 = 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 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq) 2S2O32-(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 mass of Cu2+ in 1dm3 = 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 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq) 2S2O32-(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 mass of Cu2+ in 1dm3 = 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 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq) 2S2O32-(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 mass of Cu2+ in 1dm3 = 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 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq) 2S2O32-(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 mass of Cu2+ in 1dm3 = 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 = 247.66 mass of water = mass of CuSO4.xH2O - mass of CuSO4 = 247.66 - 159.50 = 88.16 moles of water (x) = mass / molar mass = 88.16 / 18 = 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 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq) 2S2O32-(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 mass of Cu2+ in 1dm3 = 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 = 247.66 mass of water = mass of CuSO4.xH2O - mass of CuSO4 = 88.16 moles of water (x) = mass / molar mass = 88.16 / 18 = 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 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq) 2S2O32-(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 mass of Cu2+ in 1dm3 = 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 = 247.66 mass of water = mass of CuSO4.xH2O - mass of CuSO4 = 88.16 moles of water (x) = mass / molar mass = 88.16 / 18 = 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 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq) 2S2O32-(aq) + I2(aq) —> S4O62-(aq) + 2I¯(aq) you get 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 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq) 2S2O32-(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 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 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq) 2S2O32-(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 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 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq) 2S2O32-(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 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 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq) 2S2O32-(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 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 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq) 2S2O32-(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 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 2Cu2+(aq) + 4I¯(aq) —> 2CuI(s) + I2(aq) 2S2O32-(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%

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