Title: Review my chemistry titration lab calculations PLEASE!? Post by: rinaann on May 8, 2013 Aim: To determine the amount of Fe2+ in a dissolved wire solution.
Method: 25cm^3 of a sol. Of iron wire in dilute sulphuric acid(some in the form of Fe3+ and some in the form of Fe2+) was pipetted and placed into a conical flask to which 10cm^3 of dil. H2SO4 was added. The contents were titrated using Potassium Manganate (3.16 gdm^-3 KmnO4) until a pale pink colour was observed. (1M H2SO4 was used) Calculations: Title value =22.73 Calculate the concentration in mol/dm^3 Potassium Manganate, KMnO4 Mass ok KmnO4 = 3.16 gdm^-3 Molecular mass of KMnO4 = 158g Concentration = Mass/Mol. Mass = 3.16/158 = 0.02 mol dm^-3 What is the concentration in mol/dm^3 of MnO4- ions in the sol. Mass of MnO4/mass of KMnO4 x conc. Of KMnO4 = 119/158 x 0.02 = 0.015 mol dm^-3 Calculate the no. of moles of MnO4- ions run from the burette into the titration flask If in 1000cm^3 of KMnO4 sol, there is 0.015 mol dm^-3 of MnO4- Then in 22.7 cm^3 (title value), there will be: 1000/22.73 x 0.015 mol dm^-3 = 0.0003 mol I have to write half equations for the reaction: MnO4-(aq) + 8H+(aq) + 5e- ------> Mn2+(aq) + 4H2O(l) and Fe2+(aq) -----> Fe3+(aq) + e- To combine the no. of electrons used must be equal to the no. formed so the second equation needs to be multiplied by five: MnO4-(aq) + 8H+(aq) + 5e- ------> Mn2+(aq) + 4H2O(l) 5Fe2+(aq) -----> 5Fe3+(aq) + 5e- Then I add them together to give: MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) ------> Mn2+(aq) + 4H2O(l) + 5Fe3+(aq) Find the no. of moles Fe2+ ions present in the titration flask: Ratio MnO4- : 5Fe2+ So no. of moles Fe2+ present = 0.0003 x 5 = 0.0015 mol Calculate the mass of Fe2+ ions in 1 dm^3: If in 25 cm^3, there was 0.0015 mol Then in 1000cm^3, there will be: 1000/25 x 0.0015 = 0.06 mol dm^-3 PLEASE CORRECT ME! Title: Review my chemistry titration lab calculations PLEASE!? Post by: leenamu on May 8, 2013 I have added comments after the ####
Method: 25cm^3 of a sol. Of iron wire in dilute sulphuric acid(some in the form of Fe3+ and some in the form of Fe2+) was pipetted and placed into a conical flask to which 10cm^3 of dil. H2SO4 was added. The contents were titrated using Potassium Manganate (3.16 gdm^-3 KmnO4) until a pale pink colour was observed. (1M H2SO4 was used) Calculations: Title value =22.73 #### is this the volume of KMnO4 used? Calculate the concentration in mol/dm^3 Potassium Manganate, KMnO4 Mass ok KmnO4 = 3.16 gdm^-3 Molecular mass of KMnO4 = 158g Concentration = Mass/Mol. Mass = 3.16/158 = 0.02 mol dm^-3 ##### OK but keep 3 sig figs 0.0200 M What is the concentration in mol/dm^3 of MnO4- ions in the sol. Mass of MnO4/mass of KMnO4 x conc. Of KMnO4 = 119/158 x 0.02 = 0.015 mol dm^-3 #### this is wrong [MnO4-] = 0.0200 M Calculate the no. of moles of MnO4- ions run from the burette into the titration flask If in 1000cm^3 of KMnO4 sol, there is 0.015 mol dm^-3 of MnO4- Then in 22.7 cm^3 (title value), there will be: 1000/22.73 x 0.015 mol dm^-3 = 0.0003 mol #### should be 0.0200 x 0.02273 = 0.0004546 mol I have to write half equations for the reaction: MnO4-(aq) + 8H+(aq) + 5e- ------> Mn2+(aq) + 4H2O(l) and Fe2+(aq) -----> Fe3+(aq) + e- To combine the no. of electrons used must be equal to the no. formed so the second equation needs to be multiplied by five: MnO4-(aq) + 8H+(aq) + 5e- ------> Mn2+(aq) + 4H2O(l) 5Fe2+(aq) -----> 5Fe3+(aq) + 5e- Then I add them together to give: MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) ------> Mn2+(aq) + 4H2O(l) + 5Fe3+(aq) Find the no. of moles Fe2+ ions present in the titration flask: Ratio MnO4- : 5Fe2+ So no. of moles Fe2+ present = 0.0003 x 5 = 0.0015 mol #### 0.0004546 x 5 = 0.002273 mol Calculate the mass of Fe2+ ions in 1 dm^3: If in 25 cm^3, there was 0.0015 mol Then in 1000cm^3, there will be: 1000/25 x 0.0015 = 0.06 mol dm^-3 #### 1000/25 x 0.002273 = 0.09092 mol #### so mass Fe2+ = 0.09092 x 55.84 = 5.08 g to 3 sig figs |