Question

2.5 grams of a metallic sample containing an unknown amount of zinc is dissolved in 500 ml of acidic water, releasing the zinc into the water as Zn²⁺ ions. A 10.00 ml aliquot of this solution is extracted with 10 ml of CCl₄ containing an excess of 8-hydroxy quinoline. The CCl₄ and aqueous phases separate and 8-hydroxy quinoline forms a fluorescent complex with Zn²⁺ ions that partitions entirely into the CCl₄ phase. The CCl₄ phase is separated and then diluted to 25 ml. This solution gives a fluorescence intensity of 155 (arbitrary units). A similar procedure is done with 10.00 ml of the unknown zinc solution plus 8.00 ml of an aqueous 2.50 ppm Zn²⁺ solution, and the final fluorescence intensity is 247. Calculate the Zn²⁺ concentration in the original 500 ml solution and the amount of zinc in the original solid.

Answer 1

Ans. Given,

Fluorescence intensity of unknown [Zn²⁺] solution = 155 units

Fluorescence intensity of (unknown [Zn²⁺] + Standard) solution = 247 units

Increase in fluorescence intensity = (247 - 155) unit = 92 units

Conclusion: Increase in Fluorescence intensity of 92 units is due to addition of 8.0 mL of standard Zn²⁺ solution of concentration 2.50 ppm to the unknown [Zn²⁺] solution.

It’s assumed that fluorescence intensity increases linearly with concentration of Zn²⁺ for the given range of concentrations.

Thus,

            92 units fluorescence intensity is equivalent to 2.50 ppm [Zn²⁺]

Hence, 1 unit fluorescence intensity is equivalent to (2.50 ppm/ 92 units ) = 0.0271739 ppm/ unit.

Given: Fluorescence intensity of unknown [Zn²⁺] solution = 155 units

Therefore,

[Zn²⁺] in the solution of unknown concentration = 155 units x [Zn²⁺] equivalent to 1 unit

                                                            = 155 units x (0.0271739 ppm/ unit)

                                                            = 4.212 ppm

Now, [Zn²⁺] = 4.212 ppm belongs to the concentration of the 10 mL aliquot taken from 500 mL stock solution. Thus, the concentration of the 500 mL solution is also 4.212 ppm because it also represents the same solution taken for further analysis.

That is, [Zn²⁺] in 500 mL solution = 4.212 ppm

Total amount of Zn²⁺ in 500 mL solution = Concentration of Zn²⁺ x volume of solution in Liters

                                                            = 4.212 ppm x 0.5 L                   ; [1 ppm = 1 mg/ L]

                                                            = (4.212 mg / L) x 0.5 L

                                                            = 2.106 mg

Since, 500 mL solution is prepared from 2.5 g solid sample, it represents the amount of Zn present in that given solid sample.

Therefore, the amount of Zn²⁺ in 2.5 g solid sample = 2.106 mg = 0.002106 g

[Zn] in terms of % (w/w) in solid sample = (mass of Zn / Mass of total sample taken) x 100

                                                            = (0.002106 g / 2.5 g) x 100

                                                            = 0.08424 %

Thus, [Zn] in the solid sample = 0.08424 % (w/w)