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Redox Titration Help

Just began A2 work and struggling with these questions:

1. 9.80g of pure FeSO4.(NH4)2SO4. 6H2O was completely dissolved in dilute (1 mole dm-3) sulphuric acid and the solution made up to 250cm3. 25.0cm3 of the solution reacted with 24.6cm3 of KMnO4 solution. Calculate the concentration of the KmnO4.

2. 1.15g of impure iron wire was dissolved in 30cm3 of 1 mol dm-3 sulphuric acid (a large excess), and the solution made up to 250cm3. A 25.0cm3 sample of the resulting solution of Iron(II) sulphate was titrated with 0.0200mol dm-3 KMnO4; it took 20cm3 to reach and end-point. Calculate:

a) The mols of Fe2+ in 25.0cm3 solution
b) The moles of iron in the sample of iron wire.
c) The % Purity of the wire.

Any help would be great, thanks in advance.

Reply 1

GDN

1. using the mass of the salt you can calculate the number of moles of Fe2+

= 9.8/392

= 0.025 moles

now use n=cv to calculate the conc of Fe2+

c = n/v

= 0.025/0.25

= 0.1 mol dm -3

when Fe2+ and MnO4 - react 1 mol of MnO4 - reacts with 5 mol of Fe2+

the number of mol of Fe2+ reacting = 0.1 x 0.025

= 0.0025

so the no of mol of MnO4 - = 0.0025/5

= 0.0005

note this is in 24.6 cm3 so use n=cv again

c = n/v

= 0.0005/0.0246

= 0.0203 mol dm-3 (3 s.f.)