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AN INTRODUCTION TO COPPER / THIOSULPHATE TITRATIONS KNOCKHARDY PUBLISHING 2008 SPECIFICATIONS

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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... Navigation is achieved by... either clicking on the grey arrows at the foot of each page orusing the left and right arrow keys on the keyboard KNOCKHARDY PUBLISHING THIOSULPHATE TITRATIONS

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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. 2Cu 2+ (aq) + 4I¯(aq) ——> 2CuI(s) + I 2 (aq) 2S 2 O 3 2- (aq) + I 2 (aq) ——> S 4 O 6 2- (aq) + 2I¯(aq) therefore moles of S 2 O 3 2- = moles of Cu 2+ (aq)

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COPPER(II) / THIOSULPHATE TITRATIONS Practical details Pipette a known volume of a solution of Cu 2+ 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

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COPPER(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. 2Cu 2+ (aq) + 4I¯(aq) ——> 2CuI(s) + I 2 (aq) off white solid ABAB B

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COPPER(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. 2S 2 O 3 2- (aq) + I 2 (aq) ——> S 4 O 6 2- (aq) + 2I¯(aq) AB C C 0 -1

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COPPER(II) / THIOSULPHATE TITRATIONS Practical details Starch solution is added near the end point. Starch gives a dark blue colouration in the presence of iodine. AB C D D

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COPPER(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. AB C D E E

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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 S 2 O 3 2- needed 3 according to the equation… moles of Cu 2+ = moles of S 2 O calculate the number of moles of Cu 2+ 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.

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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 S 2 O 3 2- needed 3 according to the equation… moles of Cu 2+ = moles of S 2 O calculate the number of moles of Cu 2+ 5 calculate the number of moles of CuSO 4 moles of CuSO 4 = moles of Cu 2+ (there is one Cu 2+ in every CuSO 4 ) 6 calculate the mass of copper sulphate by multiplying the moles of copper sulphate by the molar mass of copper sulphate (CuSO 4 ) 7 calculate mass of water ( = mass of CuSO 4.xH 2 O - mass of CuSO 4 ) 8 divide the mass of water by 18 to find the number of moles of water 9Compare the ration of moles of… water : moles of CuSO 4 to find x

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CALCULATIONS – Example 1 25cm 3 of a solution of CuSO 4.xH 2 O of concentration g dm -3 was placed in a conical flask and an excess of KI added cm 3 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

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CALCULATIONS – Example 1 25cm 3 of a solution of CuSO 4.xH 2 O of concentration g dm -3 was placed in a conical flask and an excess of KI added cm 3 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 equations2Cu 2+ (aq) + 4I¯(aq) —> 2CuI(s) + I 2 (aq) 2S 2 O 3 2- (aq) + I 2 (aq) —> S 4 O 6 2- (aq) + 2I¯(aq) you get moles of S 2 O 3 2- = moles of Cu 2+ (aq)

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CALCULATIONS – Example 1 25cm 3 of a solution of CuSO 4.xH 2 O of concentration g dm -3 was placed in a conical flask and an excess of KI added cm 3 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 equations2Cu 2+ (aq) + 4I¯(aq) —> 2CuI(s) + I 2 (aq) 2S 2 O 3 2- (aq) + I 2 (aq) —> S 4 O 6 2- (aq) + 2I¯(aq) you get moles of S 2 O 3 2- = moles of Cu 2+ (aq) moles of S 2 O 3 2- = 0.12 x / 1000= 2.67 x moles of Cu 2+ in 25cm 3 = moles of S 2 O 3 2- = 2.67 x 10 -3

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CALCULATIONS – Example 1 25cm 3 of a solution of CuSO 4.xH 2 O of concentration g dm -3 was placed in a conical flask and an excess of KI added cm 3 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 equations2Cu 2+ (aq) + 4I¯(aq) —> 2CuI(s) + I 2 (aq) 2S 2 O 3 2- (aq) + I 2 (aq) —> S 4 O 6 2- (aq) + 2I¯(aq) you get moles of S 2 O 3 2- = moles of Cu 2+ (aq) moles of S 2 O 3 2- = 0.12 x / 1000= 2.67 x moles of Cu 2+ in 25cm 3 = moles of S 2 O 3 2- = 2.67 x moles of Cu 2+ in 1dm 3 = 2.67 x x 40= 0.107

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CALCULATIONS – Example 1 25cm 3 of a solution of CuSO 4.xH 2 O of concentration g dm -3 was placed in a conical flask and an excess of KI added cm 3 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 equations2Cu 2+ (aq) + 4I¯(aq) —> 2CuI(s) + I 2 (aq) 2S 2 O 3 2- (aq) + I 2 (aq) —> S 4 O 6 2- (aq) + 2I¯(aq) you get moles of S 2 O 3 2- = moles of Cu 2+ (aq) moles of S 2 O 3 2- = 0.12 x / 1000= 2.67 x moles of Cu 2+ in 25cm 3 = moles of S 2 O 3 2- = 2.67 x moles of Cu 2+ in 1dm 3 = 2.67 x x 40= mass of Cu 2+ in 1dm 3 = x 63.5= 0.679g

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CALCULATIONS – Example 1 25cm 3 of a solution of CuSO 4.xH 2 O of concentration g dm -3 was placed in a conical flask and an excess of KI added cm 3 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 equations2Cu 2+ (aq) + 4I¯(aq) —> 2CuI(s) + I 2 (aq) 2S 2 O 3 2- (aq) + I 2 (aq) —> S 4 O 6 2- (aq) + 2I¯(aq) you get moles of S 2 O 3 2- = moles of Cu 2+ (aq) moles of S 2 O 3 2- = 0.12 x / 1000= 2.67 x moles of Cu 2+ in 25cm 3 = moles of S 2 O 3 2- = 2.67 x moles of Cu 2+ in 1dm 3 = 2.67 x x 40= mass of Cu 2+ in 1dm 3 = x 63.5= 6.79g % of Cu 2+ in compound = 6.79 / x 100= 25.64%

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CALCULATIONS – Example 1 25cm 3 of a solution of CuSO 4.xH 2 O of concentration g dm -3 was placed in a conical flask and an excess of KI added cm 3 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 equations2Cu 2+ (aq) + 4I¯(aq) —> 2CuI(s) + I 2 (aq) 2S 2 O 3 2- (aq) + I 2 (aq) —> S 4 O 6 2- (aq) + 2I¯(aq) you get moles of S 2 O 3 2- = moles of Cu 2+ (aq) moles of S 2 O 3 2- = 0.12 x / 1000= 2.67 x moles of Cu 2+ in 25cm 3 = moles of S 2 O 3 2- = 2.67 x moles of Cu 2+ in 1dm 3 = 2.67 x x 40= mass of Cu 2+ in 1dm 3 = x 63.5= 6.79g

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CALCULATIONS – Example 1 25cm 3 of a solution of CuSO 4.xH 2 O of concentration g dm -3 was placed in a conical flask and an excess of KI added cm 3 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 equations2Cu 2+ (aq) + 4I¯(aq) —> 2CuI(s) + I 2 (aq) 2S 2 O 3 2- (aq) + I 2 (aq) —> S 4 O 6 2- (aq) + 2I¯(aq) you get moles of S 2 O 3 2- = moles of Cu 2+ (aq) moles of S 2 O 3 2- = 0.12 x / 1000= 2.67 x moles of Cu 2+ in 25cm 3 = moles of S 2 O 3 2- = 2.67 x moles of Cu 2+ in 1dm 3 = 2.67 x x 40= mass of Cu 2+ in 1dm 3 = x 63.5= 6.79g molar mass of compound= mass/moles = / =

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CALCULATIONS – Example 1 25cm 3 of a solution of CuSO 4.xH 2 O of concentration g dm -3 was placed in a conical flask and an excess of KI added cm 3 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 equations2Cu 2+ (aq) + 4I¯(aq) —> 2CuI(s) + I 2 (aq) 2S 2 O 3 2- (aq) + I 2 (aq) —> S 4 O 6 2- (aq) + 2I¯(aq) you get moles of S 2 O 3 2- = moles of Cu 2+ (aq) moles of S 2 O 3 2- = 0.12 x / 1000= 2.67 x moles of Cu 2+ in 25cm 3 = moles of S 2 O 3 2- = 2.67 x moles of Cu 2+ in 1dm 3 = 2.67 x x 40= mass of Cu 2+ in 1dm 3 = x 63.5= 6.79g % of Cu 2+ in compound = 6.79 / x 100= 25.64% molar mass of compound= mass/moles = / = mass of water= mass of CuSO 4.xH 2 O - mass of CuSO 4 = = moles of water (x)= mass / molar mass= / 18= 4.9 (5)

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CALCULATIONS – Example 1 25cm 3 of a solution of CuSO 4.xH 2 O of concentration g dm -3 was placed in a conical flask and an excess of KI added cm 3 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 equations2Cu 2+ (aq) + 4I¯(aq) —> 2CuI(s) + I 2 (aq) 2S 2 O 3 2- (aq) + I 2 (aq) —> S 4 O 6 2- (aq) + 2I¯(aq) you get moles of S 2 O 3 2- = moles of Cu 2+ (aq) moles of S 2 O 3 2- = 0.12 x / 1000= 2.67 x moles of Cu 2+ in 25cm 3 = moles of S 2 O 3 2- = 2.67 x moles of Cu 2+ in 1dm 3 = 2.67 x x 40= mass of Cu 2+ in 1dm 3 = x 63.5= 6.79g % of Cu 2+ in compound = 6.79 / x 100= 25.64% molar mass of compound= mass/moles = / = mass of water= mass of CuSO 4.xH 2 O - mass of CuSO 4 = moles of water (x)= mass / molar mass= / 18= 4.9 (5)

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CALCULATIONS – Example 2 25cm 3 of a solution of CuSO 4.xH 2 O of concentration g dm -3 was placed in a conical flask and an excess of KI added cm 3 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 equations2Cu 2+ (aq) + 4I¯(aq) —> 2CuI(s) + I 2 (aq) 2S 2 O 3 2- (aq) + I 2 (aq) —> S 4 O 6 2- (aq) + 2I¯(aq) you get moles of S 2 O 3 2- = moles of Cu 2+ (aq) moles of S 2 O 3 2- = 0.12 x / 1000= 2.67 x moles of Cu 2+ in 25cm 3 = moles of S 2 O 3 2- = 2.67 x moles of Cu 2+ in 1dm 3 = 2.67 x x 40= mass of Cu 2+ in 1dm 3 = x 63.5= 6.79g % of Cu 2+ in compound = 6.79 / x 100= 25.64% molar mass of compound= mass/moles = / = mass of water= mass of CuSO 4.xH 2 O - mass of CuSO 4 = moles of water (x)= mass / molar mass= / 18= 4.9 (5)

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CALCULATIONS – Example g of a copper alloy was dissolved in acid and the solution made up to 250cm 3. 25cm 3 was then pipetted into a conical flask and an excess of KI added. It was found 25cm 3 of a 0.100M sodium thiosulphate solution Calculate the percentage of copper in the alloy.

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CALCULATIONS – Example g of a copper alloy was dissolved in acid and the solution made up to 250cm 3. 25cm 3 was then pipetted into a conical flask and an excess of KI added. It was found 25cm 3 of a 0.100M sodium thiosulphate solution Calculate the percentage of copper in the alloy. From these two equations2Cu 2+ (aq) + 4I¯(aq) —> 2CuI(s) + I 2 (aq) 2S 2 O 3 2- (aq) + I 2 (aq) —> S 4 O 6 2- (aq) + 2I¯(aq) you get moles of S 2 O 3 2- = moles of Cu 2+ (aq)

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CALCULATIONS – Example g of a copper alloy was dissolved in acid and the solution made up to 250cm 3. 25cm 3 was then pipetted into a conical flask and an excess of KI added. It was found 25cm 3 of a 0.100M sodium thiosulphate solution Calculate the percentage of copper in the alloy. From these two equations2Cu 2+ (aq) + 4I¯(aq) —> 2CuI(s) + I 2 (aq) 2S 2 O 3 2- (aq) + I 2 (aq) —> S 4 O 6 2- (aq) + 2I¯(aq) you get moles of S 2 O 3 2- = moles of Cu 2+ (aq) moles of S 2 O 3 2- = x / 1000= 2.50 x 10 -3

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CALCULATIONS – Example g of a copper alloy was dissolved in acid and the solution made up to 250cm 3. 25cm 3 was then pipetted into a conical flask and an excess of KI added. It was found 25cm 3 of a 0.100M sodium thiosulphate solution Calculate the percentage of copper in the alloy. From these two equations2Cu 2+ (aq) + 4I¯(aq) —> 2CuI(s) + I 2 (aq) 2S 2 O 3 2- (aq) + I 2 (aq) —> S 4 O 6 2- (aq) + 2I¯(aq) you get moles of S 2 O 3 2- = moles of Cu 2+ (aq) moles of S 2 O 3 2- = x / 1000= 2.50 x moles of Cu 2+ in 25cm 3 = moles of S 2 O 3 2- = 2.50 x 10 -3

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CALCULATIONS – Example g of a copper alloy was dissolved in acid and the solution made up to 250cm 3. 25cm 3 was then pipetted into a conical flask and an excess of KI added. It was found 25cm 3 of a 0.100M sodium thiosulphate solution Calculate the percentage of copper in the alloy. From these two equations2Cu 2+ (aq) + 4I¯(aq) —> 2CuI(s) + I 2 (aq) 2S 2 O 3 2- (aq) + I 2 (aq) —> S 4 O 6 2- (aq) + 2I¯(aq) you get moles of S 2 O 3 2- = moles of Cu 2+ (aq) moles of S 2 O 3 2- = x / 1000= 2.50 x moles of Cu 2+ in 25cm 3 = moles of S 2 O 3 2- = 2.50 x moles of Cu 2+ in 250cm 3 = 2.50 x x 10= 0.025

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CALCULATIONS – Example g of a copper alloy was dissolved in acid and the solution made up to 250cm 3. 25cm 3 was then pipetted into a conical flask and an excess of KI added. It was found 25cm 3 of a 0.100M sodium thiosulphate solution Calculate the percentage of copper in the alloy. From these two equations2Cu 2+ (aq) + 4I¯(aq) —> 2CuI(s) + I 2 (aq) 2S 2 O 3 2- (aq) + I 2 (aq) —> S 4 O 6 2- (aq) + 2I¯(aq) you get moles of S 2 O 3 2- = moles of Cu 2+ (aq) moles of S 2 O 3 2- = x / 1000= 2.50 x moles of Cu 2+ in 25cm 3 = moles of S 2 O 3 2- = 2.50 x moles of Cu 2+ in 250cm 3 = 2.50 x x 10= mass of Cu 2+ in 250cm 3 = x 63.5= 1.588g

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CALCULATIONS – Example g of a copper alloy was dissolved in acid and the solution made up to 250cm 3. 25cm 3 was then pipetted into a conical flask and an excess of KI added. It was found 25cm 3 of a 0.100M sodium thiosulphate solution Calculate the percentage of copper in the alloy. From these two equations2Cu 2+ (aq) + 4I¯(aq) —> 2CuI(s) + I 2 (aq) 2S 2 O 3 2- (aq) + I 2 (aq) —> S 4 O 6 2- (aq) + 2I¯(aq) you get moles of S 2 O 3 2- = moles of Cu 2+ (aq) moles of S 2 O 3 2- = x / 1000= 2.50 x moles of Cu 2+ in 25cm 3 = moles of S 2 O 3 2- = 2.50 x moles of Cu 2+ in 250cm 3 = 2.50 x x 10= mass of Cu 2+ in 250cm 3 = x 63.5= 1.588g % of Cu in the alloy = / 3.00 x 100= 52.91%

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CALCULATIONS – Example g of a copper alloy was dissolved in acid and the solution made up to 250cm 3. 25cm 3 was then pipetted into a conical flask and an excess of KI added. It was found 25cm 3 of a 0.100M sodium thiosulphate solution Calculate the percentage of copper in the alloy. From these two equations2Cu 2+ (aq) + 4I¯(aq) —> 2CuI(s) + I 2 (aq) 2S 2 O 3 2- (aq) + I 2 (aq) —> S 4 O 6 2- (aq) + 2I¯(aq) you get moles of S 2 O 3 2- = moles of Cu 2+ (aq) moles of S 2 O 3 2- = x / 1000= 2.50 x moles of Cu 2+ in 25cm 3 = moles of S 2 O 3 2- = 2.50 x moles of Cu 2+ in 250cm 3 = 2.50 x x 10= mass of Cu 2+ in 250cm 3 = x 63.5= 1.588g % of Cu in the alloy = / 3.00 x 100= 52.91%

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© 2010 JONATHAN HOPTON & KNOCKHARDY PUBLISHING THE END AN INTRODUCTION TO COPPER / THIOSULPHATE TITRATIONS

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