Question

Your TA made a series of dilutions to a liquid, Mn-containing sample (an ore already dissolved with acid to make a solution) that you analyzed by AAS. Her dilutions involved taking 1.00 mL of the original ore solution containing an unknown concentration of manganese and boiling it with concentrated nitric and hydrofluoric acid for 15 minutes (the boiling nitric acid oxidizes all metals and boiling HF destroys any silicate substructure in the ore). The liquid left was diluted to 100.0 mL with Mn-free water with mixing. A 25 µL aliquot of that solution was diluted to 25.00 mL volume with more Mn-free water and mixed well. Finally 5 mL of that well-mixed solution was transferred to a 50-mL volumetric, 5 mL of concentrated HNO3 was added to the flash which was diluted to the line with Mn-free water and mixed. When analyzed by AAS, that solution was determined to have 23.45 ppb Mn. What was the percentage by mass of Mn in the original ore solution assuming that solution has a density of 1?

Answer 1

This is a dilution problem; we need to work backward to find out the concentration of Mn in the original sample. Let us number the solutions as 1 (original solution), 2 (100 mL solution), 3 (25 mL solution and 4 (50 mL solution).

The final diluted solution shows a Mn concentration of 23.45 ppb. This was the Mn concentration in solution 4. Find out the concentration of Mn in solution 3 by using the dilution factor.

The dilution factor was (50 mL/5 mL) = 10; the concentration of Mn in solution 3 = (concentration in solution 4)*(dilution factor) = (23.45 ppb)*(10) = 234.50 ppb.

Again, solution 3 was prepared by taking 25 µL of solution 2 and diluting to a final volume of 25 mL. Remember that 1 mL = 1000 µL; therefore, the dilution factor is (25 mL)/(25 µL) = (25 mL)/[(25 µL)*(1 mL/1000 µL)] = 1000.

The concentration of Mn in solution 2 is (concentration in solution 3)*(dilution factor) = (234.50 ppb)*(1000) = 234500 ppb = (234500 ppb)*(1 ppm/1000 ppb) [1 ppm = 1000 ppb] = 234.50 ppm = 234.50 mg/L (1 ppm = 1 mg/L).

Finally, find out the dilution of solution 2. The dilution factor is (100 mL/1 mL) = 100. The concentration of Mn in solution 1 = (concentration in solution 2)*(dilution factor) = (234.50 mg/L)*(100) = 23450 mg/L = (23450 mg/L)*(1 g/1000 mg) = 23.45 g/L.

We had 1.00 mL of the original solution so that the mass of Mn in the solution is (1.00 mL)*(1 L/1000 mL)*(23.45 g/L) = 0.02345 g.

The density of the solution is 1.00 g/mL; therefore, mass of the solution = (volume of solution)*(density of solution) = (1.00 mL)*(1 g/mL) = 1.00 g.

Percentage of Mn in the sample = (mass of Mn)/(mass of solution)*100 = (0.02345 g/1.00 g)*100 = 2.345% (ans).