How do you solve this titration calculation?

Question

35.0 g of hydrated iron(II) ammonium sulfate #sf(Fe(NH_4)_2(SO_4)_(2).6H_2O)# is dissolved in 1 litre of water. 25.00 ml is added to a flask and acidified. When titrated against #sf(KMnO_4)# solution an average titre of 21.00 ml was recorded. What is the concentration of the Mn(VII) solution in g/l ?

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Answer 1 · 2017-06-10T13:38:06

You can do it like this:

Start with the two 1/2 equations:

#sf(MnO_(4(aq))^(-)+8H_((aq))^++5erarrMn_((aq))^(2+)+4H_2O_((l))" "color(red)((1)))#

#sf(Fe_((aq))^(2+)rarrFe_((aq))^(3+)+e" "color(red)((2)))#

To get the electrons to balance we multiply #sf(color(red)((2)))# by 5 and add to #sf(color(red)((1))rArr)#

#sf(MnO_(4(aq))^(-)+8H_((aq))^++cancel(5e)+5Fe_((aq))^(+)rarrMn_((aq))^(2+)+4H_2O_((l))+5Fe_((aq))^(2+)+cancel(5e))#

This tells us that 1 mole Mn(VII) reacts with 5 moles Fe(II).

The #sf(M_r)# of the iron(II) salt is 392.13

The no. moles of iron(II) in 1000 ml is therefore #sf(35.0/392.13=0.0892)#

#:.##sf([Fe^(2+)]=0.0892color(white)(x)"mol/l")#

#sf(c=n/v)#

#:.##sf(n=cxxv)#

#:.##sf(n_(Fe^(2+))=0.0892xx25.00/1000=2.231xx10^(-3))#

From the mole ratio we can say that:

#sf(n_(MnO_4^-)=(2.231xx10^-3)/5=0.4463xx10^(-3))#

#sf(c=n/v)#

#:.##sf([MnO_4^-]=(0.4463xx10^(-3))/(0.021)=0.0212color(white)(x)"mol/l")#

#sf(M_r[KMnO_4]=158.034)#

In grams/litre the concentration is given by:

#sf(c=0.0212xx158.034=3.35color(white)(x)"g/l")#

This is equal to 0.335 g/100 ml or 0.33% (w/v) to 2 sig fig.