A 0.684g sample of an unknown ore containing manganese was dissolved in acid and

Question

A 0.684g sample of an unknown ore containing manganese was dissolved in acid and diluted to 100.00 mL. A 10.00mL aliquot was taken, sodium persulfate was added to oxidize the manganese to permanganate and the solution diluted to 50 mL. The absorbance of this solution was measured in a 1.00cm cell to be 0.426. A standard sample was prepared by dissolving 0.100 g of potassium permanganate in water and diluting to 1.00 L. A 10.00mL aliquot was diluted to 500 mL and the absorbance was measured in a 1.00cm cell to be 0.483. What is the percentage manganese in the ore sample?

Explanation / Answer

Absorption of the ore solution, A1 = 0.426   

absorbacne o stanard solution, A2 = 0.483

calculation of concentration of standard solution:

1000 ml solution contains 0.1 g KMnO4;

No of moles of KMnO4 = 0.1/158 = 6.33 x 10-4 mole

Concentration of the solution = 6.33 x 10-4 mole/ 1L = 6.33 x 10-4 M

10 ml solution of 6.33 x 10-4 M was dilute to 500 mL

so concentration = (6.33 x 10-4 x 10)/500 = 1.27 x 10-5 M

concentraion of KMnO4, c2 = 1.27 x 10-5 M

A1/A2 = c1/c2

c1 = (A1 x c2) /A2 = (1.27 x 10-5 x 0.426)/0.483 = 1.12 x 10-5 M

The concentration of permangate in the ore solution = 1.12 x 10-5 M

Since amount of permanganate solution is 50 mL so

no of moles of permangate = 1.12 x 10-5 x 50 /1000 = 5.6 x 10-7 mole

10 ml solution contain 5.6 x 10-7 mole

100 ml solution contain = 5.6 x 10-6 mole

1 equiv of Mn will produce 1 equiv of KMnO4 so

amount of Mn = 5.6 x 10-6 mole = 5.6 x 10-6 x 55 = 308 x 10-6 g

Perrcentage of Mn in the ore = (308 x 10-6 g / 0.684 ) x 100% = 0.05 %

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