Question

a 0.1965g sample containing copper is analysed iodometrically. copper(ii) is reduced to copper(i) by iodide:

2Cu2+ + 4I- --> 2Cul + I2

calculate the percentage of copper in the sample if 25.0mL of 0.1005M Na2S2O3 is required for the titration of the liberated iodine.

Answer 1

               2Cu2+ + 4I- --> 2Cul + I2

2Na2S2O3 + I2 ----------> 2NaI + Na2S4O6

no of moles of Na2S2O3 = molarity * volume in L

                                         = 0.1005*0.025   = 0.0025125moles

2moles of Na2S2O3 react with 1 moles of I2

0.0025125 moles of Na2S2O3 react with = 1*0.0025125/2   = 0.00125625 moles of I2

2Cu2+ + 4I- --> 2Cul + I2

1 moles of I2 obtained from 2 moles of Cu⁺²

0.00125625 moles of I2 obtained from = 2*0.00125625 = 0.0025125 moles of Cu⁺²

mass of Cu⁺²   = no of moles * atomic mass of Cu

                        = 0.0025125*63.5   = 0.15954g of Cu

percentage of Cu   = 0.15954*100/0.1965   = 81.19%