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A series of dilutions were made to a liquid, Mn-containing sample (an ore alread

ID: 846218 • Letter: A

Question

A series of dilutions were made to a liquid, Mn-containing sample (an ore already dissolved with acid to make a solution) that was analyzed by AAS. The dilutions involved taking 1.00 mL of the original ore solution containing the 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 was diluted to 100.0 mL with Mn-free water with mixing. A 25 uL 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?

Explanation / Answer

Let's assume the mass of Mn in original ore solution as x g/ml

1 ml of this original solution contains = x*1 = x g of Mn

This 1 ml was diluted to 100 ml

So, this new solution cotains : x/100 g/ml of Mn

25 uL ( = 0.025 ml ) of this new solution contains = (x/100)*0.025 g of Mn

This 0.025 ml was diluted to 25 ml volume

Thus, this second new solution contains : (x/100)*0.025 / 25 = x*10-5 g/ml of Mn

5 ml of this second new solution contains : 5*x*10-5 g of Mn

This 5 ml was diluted to 50 ml

Thus, this final new solution contains : 5*x*10-5/50 = x*10-6 g/ml of Mn

Also, by AAS, we know the conc. of this final new solution is : 23.45 ppb = 23.45*10-6 g/L = 23.45*10-9 g/ml

Thus, x*10-6 = 23.45*10-9.

Solving we get : x = 0.02345

Thus, original ore solution contained 0.02345 g/ml of Mn

Density of solution = 1 g/ml

Thus, in 1 ml of solution, mass of Mn = 0.02345 , mass of solution = 1g

Thus, mass % = mass of Mn/mass of soln *100 = 0.02345/1*100 = 2.345%

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