quant analysis chemistry

Question

Calamine, which is used for relief of skin irritations, is a mixture of zinc and iron oxides. A 1.022g sample of dried calamine was dissolved in acid and diluted to 250 mL. Potassium fluoride was added to a 10.00-mL aliquot of the diluted solution to mask the iron; after adjustment of the pH the zinc required 38.78 mL of 0.01294 M Edta for titration. A second 10.00-mL aliquot required 47.66mL of 0.01294 M Edta for titration of both the zinc and iron. What is the percentage of zinc and iron in the calamine?

Answer 1

This is a difference titration. Neither buret reading gives you iron directly - you get zinc alone from the masked run, everything from the unmasked run, and iron by subtraction.

EDTA binds every metal 1:1, so millimoles of EDTA equal millimoles of metal in each case.

Step 1 - the two titrations, per 10.00 mL aliquot

Zinc only (fluoride masks the iron):
(38.78 mL)(0.01294 M) = 0.5018 mmol Zn

Zinc plus iron:
(47.66 mL)(0.01294 M) = 0.6167 mmol total

Iron by difference:
0.6167 - 0.5018 = 0.1149 mmol Fe

Doing the subtraction in millimoles, not in milliliters, is the safe habit. It happens to work in milliliters here because both titrations used the same EDTA, but it fails the moment the concentrations differ.

Step 2 - scale the aliquot back to the whole sample

The 10.00 mL aliquot is 10.00/250.0 = 1/25 of the dissolved sample, so multiply by 25:

Zn: (0.5018)(25) = 12.545 mmol
Fe: (0.1149)(25) = 2.873 mmol

Step 3 - convert to mass and divide by the sample

Zn: (12.545 mmol)(65.38 mg/mmol) = 820.2 mg
Fe: (2.873 mmol)(55.85 mg/mmol) = 160.4 mg

%Zn = 820.2 / 1022 x 100 = 80.26% Zn
%Fe = 160.4 / 1022 x 100 = 15.70% Fe

Why fluoride is in the procedure at all

Fe3+ forms a very strong EDTA complex, stronger than zinc's, so it would be titrated right along with the zinc and you could never separate the two. Fluoride ties the iron up as fluoride complexes that EDTA cannot compete with, which makes the first titration selective for zinc. That masking step is the entire experimental idea; the arithmetic is just bookkeeping around it.

An aside, if your course asks you to sanity-check results

The two metals together account for 96.0% of the sample mass, leaving 4.0% for oxygen. But if the zinc were present as ZnO and the iron as Fe2O3, the oxygen alone would need about 26% of the mass. So the numbers in this problem do not quite close as a real oxide mixture - the buret readings are idealized. That does not change what you should turn in, but noticing it is exactly the habit that catches genuine errors in the lab.