A sample contains an unknown amount of Fe^2+ which is titrated with a KMno_4 sol

Question

A sample contains an unknown amount of Fe^2+ which is titrated with a KMno_4 solution in acidic conditions to determine the iron composition by mass. A standardized KMno_4 solution is needed for the analysis of this selected is ferrous ammonium sulfate hexahydrate (MW = 392.21 g/mol) to find the exact molarity of the KMnO_4 solution. The balanced net ionic equation for this redox reaction is as follows: A. The student weighs 2.094 grams of FAS and finds that 12.34 mL of KMnO_4 solution is needed to titrate the FAS sample, what is the molarity of this KMno_4 solution? [KMnO_4] = ____ M The student then weighs 1.211 grams of the unknown solid and finds it requires 1 mL of the standardized KMnO_4 solution (from part A) to titrate. What is the mass of iron 55.85 g/mol) this sample? Mass of Fe^2+ in unknown sample = _____ g c. What is the iron by mass in the unknown sample for Part B? % mass of Fe^2+ in unknown sample = ____ % Suppose a student spilled some FAS after weighing it but before performing the titration to standardize the KMno, solution. What effect, if any, would this have on the standard molarity of the KMnO_4 solution? Explain your answer clearly following the steps in the calculation.

Explanation / Answer

5moles of FAS will react with 1 mol of KMnO4.

moles of FAS = 2.094 gm/392.21 g/mol = 5.34*10^-3 moles

Moles of KMnO4 required = 5.34*10^-3 moles/5 = 1.07*10^-3 moles

Molarity of KMnO4 = 1.07*10^-3 moles/ 0.01234 L = 0.087 M

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moles of KMnO4 required = 0.01345 L * 0.087 M = 1.16*10^-3 moles

moles of Fe present in thesample = 5* 1.16*10^-3 moles = 5.82*10^-3 moles

Mass of Fe^2+ present in the sample = 5.82*10^-3 moles * 55.85g/mol = 0.325 gm

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% Fe = (0.325 gm/1.211 gm) *100% = 26.8 %

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If the student spills some FAS before titration, actual concentration of Fe will be less in the solution. So, less volume of KMnO4 will be required. Calculated molarity will be higher.

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