3. The amount of iron in a sample can be determined by titration with potassium

Question

3. The amount of iron in a sample can be determined by titration with potassium dichromate, K2Cr2O7, in acid solution. The two half-reactions occurring are: A 1.885 g solid sample containing some FeSO4 is dissolved in water. The resulting solution is acidified and titrated to the endpoint with 20.48 mL of 0.02540 M K2Cr2O7. a. Write the balanced overall net-ionic equation for the titration reaction. b. How many moles of iron(II) ion are in the solid sample? c. How many grams of iron(II) sulfate are in the solid sample? d. What is the percent iron(II) sulfate in the solid sample?

Explanation / Answer

The amount of iron in a sample can be determined by titration with potassium dichromate, K2Cr2O7, in acid solution. The two half-reactions occurring are:

Fe2+(aq) -> Fe3+(aq) + e-
Cr2O2-(aq) + 14H+(aq) + 6e- -> 2Cr3+(aq) + 7H2O(l)

A 1.885g solid sample containing some FeSO4 is dissolved in water. The resulting solution is acidified and titrated to the endpoint with 20.48mL of 0.02540M K2Cr2O7.

a. Write the balanced overall net-ionic equation for the titration reaction:

6Fe2+(aq) + Cr2O72-(aq) + 14H+(aq) -> Fe3+(aq) + 2Cr3+(aq) + 7H2O(l)

b. How many moles of iron(ii) ion are in the solid sample?

5.2019*10^-4 mol K2CrO7 x 6 mol Fe2+/1mol K2CrO7 = 3.121*10^-3 mol Fe2+

c. How many grams of iron(II) sulfate are in the solid sample? (I solved using only Fe2+, assuming they come out with same value as sulfate, but I may be wrong. Please explain if I did this part incorrectly.)

3.121*10^-3 mol Fe2+ x 55.85g Fe/1mol Fe = 0.1743g FeSO4

d. What is the percent iron(II) sulfate in the solid sample?

0.17432g Fe2+/1.885g Sample x 100% = 9.248%

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