Titration is a method to find out how much acid would neutralize an alkali Essays

Titration is a method to find out how much acid would neutralize an alkali   Essays Titration is a method to find out how much acid would neutralize an alkali   Essay Titration is a method to find out how much acid would neutralize an alkali   Essay The following equation shows the neutralisation reaction between iron and potassium manganate i will be carrying out this reaction in my experiment 5Fe2+(aq) + MnO4- (aq) + 8H+ 5Fe+3(aq)+Mn+2(aq) + 4H2O(l) We made the ammonium sulphate solution with the concentration of 0.05moldm3 with the volume 250cm3 To calculate the molecular mass i added up all the elements, (Fe(NH4)2 (SO4)2 6H2O), the molecular mass is 391.90 We can calculate the number of moles by using the following formula Number of moles = concentration x volume So 0.05 X 0.025(this has to be converted dm3) = 0.0125 APPARATUS: * Volumetric flask * 10 cm pipette * 100 cm conical flask * burette * distilled water * clamp stand * standard solution * Lab coats should be worn during the practical * Eye protection should also be worn Titration Rough 1 2 3 4 Initial burette reading 28.00 26.60 26.90 26.80 26.60 Titre 28.00 26.60 26.90 26.80 26.60 METHOD 1) set up apparatus 2) fill burette with potassium manganate solution 3) get 250ml of the standard solution transfer it to a flask using a pipette 4) while swirling add potassium manganate to the solution 5) use a white tile to see a colour change (pink) 6) record the end point the first will be rough repeat method to get concordant results Results Table Titration 1 2 3 4 5 Final reading 28.00 26.60 26.90 26.80 26.60 26.60 Average = 26.60cm3 5Fe2+(aq) + MnO4- (aq) + 8H+ 5Fe+3(aq)+Mn+2(aq) + 4H2O(l) The reaction shows that 5 moles of iron II neutralises 1 mole of potassium mangnate 0.00125/5 = 0.00025 From this we can work out the no of moles of potassium mangnate 0.000251000/26.60 = 0.0093 To find the percentage of iron i used this equation Moles of iron in 25cm3/moles in 25cm3XMr = 0.069/0.466=0.14263X100 = 14.8% i checked this value and compared my result to the original value of 14.3 and noticed i wasnt far off