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S  0 ``  X*  0 `   Z*  0X `   Z*z  bA޽h @ ?Parchment ̙33 Default Design 0 zr@ ( )   0H% P    P*    0-     R*  d  c $ ?    0D{  @  RClick to edit Master text styles Second level Third level Fourth level Fifth level!     S  6{ `P   P*    6 1 `  { R*  H  0޽h ? ̙3380___PPT10. 0X(   $+ X X 0@ P    X*  X 0D     Z*  X 6J `P   X*  X 6pN `   Z* H X 0޽h ? ̙3380___PPT10.1 0L0 P/(    0 3AN INTRODUCTION TO COPPER / THIOSULPHATE TITRATIONSD4(2 f!0f<f + <̙? x @i 0 0 , <̙? x @e 0 0 - 0x m OKNOCKHARDY PUBLISHING(2 fJ . C "A KTRE  / 0"`>B 2008 SPECIFICATIONS<G$G   B  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0   0 (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0c   0PT0Pa 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 (2 2(2 2  75, dB   <D?     0<,{0P( OKNOCKHARDY PUBLISHING(2    09X eTHIOSULPHATE TITRATIONS2(2$f<fB  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0   0dX (  ddB d <D)P0Q d 0Pp@ 0 0dB d@ <D)PXP d 0X0@ 0 0 d 0vq~ r$COPPER(II) / THIOSULPHATE TITRATIONS2%(2$ff d 0}"`0H  :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)!*![!Z $           4    t   !  o   u   f B d s *޽h ? ̙33y___PPT10Y+D=' = @B +w  0L0  ~  $ (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0襋q~ r$COPPER(II) / THIOSULPHATE TITRATIONS2%(2$ff   0D"`0 ]Practical details!    0D"` { oPipette 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.6* &  + P   C (Acuthiog  ! VAParchmentS" ?0 " VAParchmentS" ?m0H  # 0d 8K  cA 0 2f!   $ 0   OA 2f  B  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0   pt (  tdB t <D)P0Q t 0Pp@ 0 0dB t@ <D)PXP t 0X0@ 0 0 t 0@q~ r$COPPER(II) / THIOSULPHATE TITRATIONS2%(2$ff t 0"`0 ]Practical details!  0 t 0"` H 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! : !@>      P  t C (Acuthiog   t VAParchmentS" ?80  t VAParchmentS" ?b=   t 0h:d 8K  vA BB 2!f    t 0p:   OB 2f   t 0 : Q+2 -1 +1 -1 0 R 2R! R B t s *޽h ? ̙33y___PPT10Y+D=' = @B +1  0L0 @ 8  (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0:q~ r$COPPER(II) / THIOSULPHATE TITRATIONS2%(2$ff  0:"`0 ]Practical details!  v  0:"` H{ 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)U!5       &+    P   C (Acuthiog    VAParchmentS" ? 0   VAParchmentS" ? 8s    0,*:d 8K  hA B C0 2f     0x-:   OC 2f    03:@ f 0 -1 g 2g! g B  s *޽h ? ̙33y___PPT10Y+D=' = @B +?  0L0 NF |(  |dB | <D)P0Q | 0Pp@ 0 0dB |@ <D)PXP | 0X0@ 0 0 | 0<:q~ r$COPPER(II) / THIOSULPHATE TITRATIONS2%(2$ff | 0?:"`0 ]Practical details!  C | 0C:"` H! mStarch solution is added near the end point. Starch gives a dark blue colouration in the presence of iodine.8n= f( n P  | C (Acuthiog   | VAParchmentS" ?0d   | 0J:d 8K  qA B C D 0 2 f    | 0lO:   OD 2f  B | s *޽h ? ̙33y___PPT10Y+D=' = @B +H  0L0 WO (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0 W:q~ r$COPPER(II) / THIOSULPHATE TITRATIONS2%(2$ff  0\:"`0 ]Practical details!    0`:"` { eContinue 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.@/      P   C (Acuthiog    0h:d 8K  A B C D ET 2ff     0$n:   OE 2f  B  s *޽h ? ̙33y___PPT10Y+D=' = @B +Q  0L0 ` X (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0u:q~ bTYPICAL CALCULATIONS2(2ff  0D:"`0*  ^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.!Z-Z%Z!!h   4    '    @`B  s *޽h ? ̙33y___PPT10Y+D=' = @B +' 0L0 6 .  (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0\{:q~ bTYPICAL CALCULATIONS2(2ff  0:"`0TLD___PPT9&  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.Z-ZZCZ.!h   ) !!!!!!!' )  !!!!   "!!!!!!K7     @`B  s *޽h ? ̙33y___PPT10Y+D=' := @B +  0L0   (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0:q~ ~0CALCULATIONS  Example 12(2ff\  0:"` B25cm3 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.@  !  A  Z=3.!HfZ   Z   .   S B  s *޽h ? ̙33y___PPT10Y+D=' := @B +  0L0   p4 (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  00rq~ ~0CALCULATIONS  Example 12(2ffJ  0xr"` B25cm3 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@  !  A  Z=3vZ   Z   .   S   0X/r"`]| .tFrom 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)G6@> #           2      B  s *޽h ? ̙33y___PPT10Y+D=' := @B + 0L0 (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0 Prq~ ~0CALCULATIONS  Example 12(2ffJ  0(Vr"` B25cm3 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@  !  A  Z=3vZ   Z   .   S   08lr"`]J  zhFrom 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 10-3G6@ #           2     33333)33 33 333333333 5 B  s *޽h ? ̙33y___PPT10Y+D=' := @B + 0L0 t(  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0rq~ ~0CALCULATIONS  Example 12(2ffJ  0xr"` B25cm3 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@  !  A  Z=3vZ   Z   .   S   0r"`]  nFrom 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 = 0.107fG6@ #           2     33333)33 33 333333333 3333333 m B  s *޽h ? ̙33y___PPT10Y+D=' := @B + 0L0 `*(  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0@rq~ ~0CALCULATIONS  Example 12(2ffJ  0r"` B25cm3 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@  !  A  Z=3vZ   Z   .   S   0t"`]  $BFrom 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 G6@% #           2     33333)33 33 333333333 33333333 333  B  s *޽h ? ̙33y___PPT10Y+D=' := @B +! 0L0 0(P(  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0/tq~ ~0CALCULATIONS  Example 12(2ffJ  0 6t"` B25cm3 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@  !  A  Z=3vZ   Z   .   S .  0Mt"`]  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 = 6.79g % of Cu2+ in compound = 6.79 / 26.50 x 100 = 25.64%G6@` #           2     33333)33 33 333333333 33333333 33%3333  B  s *޽h ? ̙33y___PPT10Y+D=' := @B + 0L0 p`(  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  08tq~ ~0CALCULATIONS  Example 12(2ff\  0t"` B25cm3 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.@  !  A  Z=.!HZ   Z   .   S   0t"`]  HBFrom 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 = 6.79g G6@% #           2     )   !!!!!!!! !!!!  B  s *޽h ? ̙33y___PPT10Y+D=' := @B +E 0L0 TL(  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0tq~ ~0CALCULATIONS  Example 12(2ff\  0t"` B25cm3 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.@  !  A  Z=.!HZ   Z   .   S @  0H u"`]L  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 = 6.79g molar mass of compound = mass/moles = 26.50 / 0.107 = 247.66 G6@j #           2     )   !!!!!!!! !!!F!  B  s *޽h ? ̙33y___PPT10Y+D=' := @B + 0L0 .&`(  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  00uq~ ~0CALCULATIONS  Example 12(2ffJ  06u"` B25cm3 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@  !  A  ZkHfZ   Z   .   S ,  0lNu"`] |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 = 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 = 247.66 - 159.50 = 88.16 moles of water (x) = mass / molar mass = 88.16 / 18 = 4.9 (5)G6@B #           2     )    %4cfffffff5ff3f3+ff3f3f3  B  s *޽h ? ̙33y___PPT10Y+D=' := @B +  0L0 P(  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0uq~ ~0CALCULATIONS  Example 12(2ff\  0`u"` B25cm3 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.@  !  A  Z=3.!HfZ   Z   .   S   0x"`] FFrom 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 = 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 = 88.16 moles of water (x) = mass / molar mass = 88.16 / 18 = 4.9 (5)$G6@' #           2     33333)33 33 333333333 33333333 33%3343F!fffffffff3f3+ff3f3f3  B  s *޽h ? ̙33y___PPT10Y+D=' := @B +  0L0 @(  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0`xq~ ~0CALCULATIONS  Example 22(2ff\  0x"` B25cm3 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.@  !  A  Z=3.!HfZ   Z   .   S   0X%x"`] FFrom 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 = 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 = 88.16 moles of water (x) = mass / molar mass = 88.16 / 18 = 4.9 (5)$G6@' #           2     33333)33 33 333333333 33333333 33%3343F!fffffffff3f3+ff3f3f3  B  s *޽h ? ̙33y___PPT10Y+D=' := @B + 0L0 0(  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0`jxq~ ~0CALCULATIONS  Example 22(2ff^  0tnx"` 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.  O  T Z3@b  W   < B  s *޽h ? ̙33y___PPT10Y+D=' := @B +  0L0   J (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0~xq~ ~0CALCULATIONS  Example 22(2ff^  0 t"` 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.  O  T Z3@b  W   <   0x"` 0vFrom 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) G6@? #           2      B  s *޽h ? ̙33y___PPT10Y+D=' := @B +  0L0   ^ (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0xq~ ~0CALCULATIONS  Example 22(2ff^  0Px"` 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.  O  T Z3@b  W   <   0x"`  DFrom 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 "G6@} #           2     33333*333  B  s *޽h ? ̙33y___PPT10Y+D=' := @B +  0L0    (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  04xq~ ~0CALCULATIONS  Example 22(2ff^  0px"` 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.  O  T Z3@b  W   <   0Hy"`  nFrom 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 G6@ #           2     * 33 333333333 8 B  s *޽h ? ̙33y___PPT10Y+D=' := @B +  0L0    (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0)yq~ ~0CALCULATIONS  Example 22(2ff^  0-y"` 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.  O  T Z3@b  W   <   0Cy"`B  zFrom 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 fG6@ #           2     *   33 33333 s B  s *޽h ? ̙33y___PPT10Y+D=' := @B + 0L0    ` (   dB   <D)P0Q   0Pp@ 0 0dB  @ <D)PXP   0X0@ 0 0   0;yq~ ~0CALCULATIONS  Example 22(2ff^   0ry"` 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.  O  T Z3@b  W   <    0(y"`v  FRFrom 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 G6@- #           2     *     33 333  B   s *޽h ? ̙33y___PPT10Y+D=' := @B +1 0L0 @8 (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0tyq~ ~0CALCULATIONS  Example 22(2ff^  0y"` 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.  O  T Z3@b  W   < *  0{"`  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%G6@h #           2     *     <3  B  s *޽h ? ̙33y___PPT10Y+D=' := @B + 0L0 .& (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0hyq~ ~0CALCULATIONS  Example 22(2ff^  0{"` 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.  O  T Z3@b  W   <   0{"`  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%G6@h #           2     33333*33 33 333333333 33 333333 33Z3  B  s *޽h ? ̙33y___PPT10Y+D=' := @B +  0L0 !@(   X @ @ <̙? x @D 0 0 @ 04x m i/ 2010 JONATHAN HOPTON & KNOCKHARDY PUBLISHING0(20 f @ 0dV M v  ATHE END(2fJ  @ C "A KTRE  !@ 0T 3AN INTRODUCTION TO COPPER / THIOSULPHATE TITRATIONSD4(2 f!0f<fH @ 0޽h ? ̙33y___PPT10Y+D=' := @B +r@d*+ &1GGv0nRZ_l}4;u*N1W, B&^htKOh+'0T hp  No Slide TitleHOPTONJONATHAN HOPTON1021Microsoft PowerPoint@H=@@hNx@X! 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