40 mL of 0.01% Mn2+ standard solution was boiled with 4 mL of 85% phosphoric aci

Question

40 mL of 0.01% Mn2+ standard solution was boiled with 4 mL of 85% phosphoric acid and 0.04 g of potassium periodate. After the solution was cooled., it was diluted to 100 mL in a volumetric flask. Calculate the concentration of the 0.01% Mn2+ standard solution in unit of g/L. assume that the density of water is 1.0 g mL. Using the answer for part (a), calculate the concentration (in g/L) of Mn2+ in the final 100 mL solution. To make one of the standard solution, 6 mL of the 100 mL Mn2+ solution was placed into a beaker and 2 mL of water was added. What is the concentration (g/L) of this standard?

Explanation / Answer

1) The reaction here would be,

2Mn2+ + 5KIO4 + 3H2O ---> 2MnO4- + 5KIO3 + 6H+

a) 0.01% Mn2+ solution will have, 0.01 g of Mn2+ in 100 g of solvent

density of solvent = 1 g/ml

So volume of solvent = 100 ml

concentration of solution = 0.01g/54.94 g/mol x 0.1 L = 1.82 x 10^-3 M

= 1.82 x 10^-3 x 54.94 g/mol = 0.1 g/L

b) 40 ml of 1.82 x 10^-3 M Mn2+ solution is diluted to 100 ml

concentration of Mn2+ in new solution = 1.82 x 10^-3 M x 40 ml/100 ml = 7.28 x 10^-4 M

= 7.28 x 10^-4 x 54.94 g/mol = 0.04 g/L

c) 6 ml of 7.28 x 10^-4 M Mn2+ solution is diluted to final volume of 4 ml

concentration of Mn2+ in solution = 7.28 x 10^-4 x 2/4 = 3.64 x 10^-4 M

= 3.64 x 10^-4 M x 54.94 g/mol = 0.02 g/L

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