In: Chemistry

# A 2.67 g sample of plant tissue was analyzed for its zinccontent. 23.25 ml of...

A 2.67 g sample of plant tissue was analyzed for its zinc content. 23.25 ml of a standardized 0.010 Molar solution of EDTA was used to titrate the sample (by complexing with the zinc ions), determine the mass of zinc in the sample and express the value as;

a. ppm zinc?

b. ppb zinc?

c. What mass of zinc product would form if the zinc (2+) ions were instead allowed to precipitate as zinc phosphate?

3Zn2+(aq) + 2PO43-(aq)Zn3(PO4)2(s)

## Solutions

##### Expert Solution

The mass of zinc present is calculated:

m Zn = M * V * MM = 0.01 M * 0.02325 L * 65.38 g / mol = 0.015 g

a) The ppm is calculated:

ppm = mg / Kg = 0.015 g Zn / 2.67 g * (1000 mg / 1 g) * (1000 g / 1 Kg) = 5618 ppm

b) The ppb is calculated:

ppb = ppm * 1000 = 5618 * 1000 = 5618000

c) The mass of Zn3 (PO4) 2 formed is calculated:

m Zn3 (PO4) 2 = 0.015 g Zn * (1 mol / 65.38 g) * (1 mol Zn3 (PO4) 2/3 mol Zn) * (386.1 g Zn3 (PO4) 2/1 mol) = 0.030 g

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