Stainless steel is an alloy of iron ,nickel and chromium. 0.500g sample of the s

Question

Stainless steel is an alloy of iron ,nickel and chromium. 0.500g sample of the steel is dissolved and diluted to 250ml. a 50 ml aliquot required 36.16 ml of EDTA in order to completely titrate all the ions. A second 50ml aliquot had chromium masked through addition of hexamethylenetetriamine, and required 30ml of the same EDTA solution to complete the titration. finally, a third aliquot had both iron and chromium masked through the addition of pyrophosphate aliquot required 0.68ml of EDTA solution to completely titrate. Determine iron, nickel, and chromium percentages of original sample (15 points)

Explanation / Answer

1)titration of stainless steel(Cr+Ni+Fe)

molarity of steel solution=M1=?

volume of steel solution=50ml=V1

volume of EDTA=V(EDTA)=36.16ml

molarity of EDTA=M(EDTA)

using eqn,M1*V1=M(EDTA)*V(EDTA)

M1=M(EDTA)*V(EDTA)/V1=M(EDTA)*36.16ml/50ml=0.723 *M(EDTA).

2)titration of stainless steel(Ni+Fe)

molarity of steel solution=M2=?

volume of steel solution=50ml=V2

volume of EDTA=V(EDTA)=30ml

molarity of EDTA=M(EDTA)

using eqn,M2*V2=M(EDTA)*V(EDTA)

M2=M(EDTA)*V(EDTA)/V2=M(EDTA)*30ml/50ml=0.6 *M(EDTA).

3)titration of stainless steel(Ni)

molarity of steel solution=M3=?

volume of steel solution=50ml=V3

volume of EDTA=V(EDTA)=0.68ml

molarity of EDTA=M(EDTA)

using eqn,M3*V3=M(EDTA)*V(EDTA)

M3=M(EDTA)*V(EDTA)/V3=M(EDTA)*0.68ml/50ml=0.0136 *M(EDTA).

M(Ni)=0.0136 *M(EDTA).

M(Ni+Fe)=0.6 *M(EDTA).

M(Cr+Ni+Fe)=0.723 M(EDTA)

%Ni in the sample=(0.0136/0.723)*100=1.9%

%(Ni+Fe)=(0.6/0.723 )*100=83%

So % Cr=100-83=17%

and % Fe=100-(17+1.9)=81.1%

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