Question

Fundamentals Of Analytical Chemistry 9th Edition - Skoog

A 1.51g sample of steel wire was dissolved in excess of dilute sulphuric acid and the solution made up to 250mL in a standard graduated flask. A 25.0mL aliquot of this solution was pipetted into an Erlenmeyer flask and titrated with 25.45mL of 0.0200M KMnO₄ for complete oxidation. Calculate the percentage of iron in the steel wire sample?

The unbalanced ionic equation is

Fe²⁺ + Mn0₄^- --> Fe³⁺ + Mn²⁺

Answer 1

First we need to balance the equation

Fe²⁺ + Mn0₄^- --> Fe³⁺ + Mn²⁺ (acidic solution)

write down the two half reactions

Fe2+(aq) Fe3+(aq) + e-

MnO4-(aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H2O

Mn is changed from +7 to +2 state so it needs 5 electrons for reduction.1 Fe2+ giving only 1 electron so for each mole of MnO4- we need 5 moles of Fe2+. since it is an acedic solution we are ading H+ in the equation in reactants side.combining two equations and balancing

5Fe2+(aq) + MnO4-(aq) + H+(aq) 5Fe3+(aq) + Mn2+(aq) + H2O

4 oxygens on reactant side, so we need to balance oxygen

5Fe2+(aq) + MnO4-(aq) + H+(aq) 5Fe3+(aq) + Mn2+(aq) + 4H2O

8 oxygens on product side, so balance H on bothsides

5Fe2+(aq) + MnO4-(aq) + 8H+(aq) 5Fe3+(aq) + Mn2+(aq) + 4H2O

No of moles of KMnO4 consumed=molarityxvolume in L=0.0200M x0.02545L=0.00509

1mole of KMnO4 is reacted with 5 moles of Fe2+.so No Of Moles of Fe2+ in this reaction=0.00509molx5mol/1mol=0.002545mol

amount of Iron in the sample=No of molesxatomic mass=0.002545molx55.845g/mol=0.1421g

% of Iron in the steel wire=

write down the two half reactions

Fe2+(aq) Fe3+(aq) + e-

MnO4-(aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H2O

Mn is changed from +7 to +2 state so it needs 5 electrons for reduction.1 Fe2+ giving only 1 electron so for each mole of MnO4- we need 5 moles of Fe2+. since it is an acedic solution we are ading H+ in the equation in reactants side.combining two equations and balancing

5Fe2+(aq) + MnO4-(aq) + H+(aq) 5Fe3+(aq) + Mn2+(aq) + H2O

4 oxygens on reactant side, so we need to balance oxygen

5Fe2+(aq) + MnO4-(aq) + H+(aq) 5Fe3+(aq) + Mn2+(aq) + 4H2O

8 oxygens on product side, so balance H on bothsides

5Fe2+(aq) + MnO4-(aq) + 8H+(aq) 5Fe3+(aq) + Mn2+(aq) + 4H2O

No of moles of KMnO4 consumed=molarityxvolume in L=0.0200M x0.02545L=0.00509

1mole of KMnO4 is reacted with 5 moles of Fe2+.so No Of Moles of Fe2+ in this reaction=0.00509molx5mol/1mol=0.002545mol

amount of Iron in the sample=No of molesxatomic mass=0.002545molx55.845g/mol=0.1421g

% of Iron in the steel wire=0.1421g/1.51gx100=9.41