Question

How to solve: A solution of K2Cr2O7 is standardized against Fe(NH4)2(SO4)2 6H20. Exactly 0.625g of the iron salt are equivalent to 22.5 mL of the dichromate solution. A sample of unknown iron (II) is then titrated with the dichromate. Exactly 25.0 mL of the iron are equivalent to 18 mL of the K2Cr2O7 solution. Calculate the number of millimoles of iron per mL of unknown solution. The answer is 2.98 mg/mL

Answer 1

6 Fe(NH4)2(SO4)2 + 7H2SO4 + K2Cr2O7 → 3Fe2(SO4)3 + Cr2(SO4)3 + 6(NH4)2SO4 + K2SO4 + 7H2O

Mohr’s salt (Fe(NH4)2(SO4)2 6H20),molar mass=392.13g/mol

Moles of the salt=mass/molar mass=0.625g/392.13g/mol=0.00159 moles

Given

0.625g of the iron salt = 22.5 mL of the dichromate solution

Or, 0.00159 moles of Fe(II) salt =0.0225L of K2Cr2O7

From the balanced eqn above,

6 moles of mohr’s salt reacts with 1 mole k2Cr2o7

So, 0.00159 moles of Fe(II) salt=1/6*0.00159 moles =0.000265 moles of K2Cr2O7

This suggests 0.000265 moles of K2Cr2O7 are present in 0.0225L of K2Cr2O7 solution

Concentration of K2Cr2O7=[k2Cr2O7]=0.000265 mol/0.0225 L=0.01178 mol/L or M

Titration with unknown

25.0 mL of the iron are equivalent to 18 mL of the K2Cr2O7 solution

Using eqn,M1*V1=M2*V2=constant for reactants

Where M=molarity ,V=volume

25.0 ml *M(Fe(II))= 0.01178 mol/L *18 ml

M(Fe(II))= 0.01178 mol/L *18 ml/25.0ml=0.00848 mol/L or M

M(Fe(II))=0.00848 mol/L =0.00848 mol*392.13 g/mol/L=3.325 g/L=3.325*1000 mg/1000 ml=3.325 mg/ml=3.3 mg/ml(approx)