15. 0.2058 g of a steel sample was dissolved in acid. The manganese was oxidized

Question

15. 0.2058 g of a steel sample was dissolved in acid. The manganese was oxidized to permanganate, MnO4 (FW 118.936), using potassium persulfate and was diluted to 250 mL. 2Mn® + 5lOr + 9H2O 2MnOa. +510s +6H3O+ 20.00 mL portions of the sample were transferred and treated according to the following table and diluted to 100 mL. The concentration of the standard manganese solution was 0.1219 M. The average of 3 absorbance readings of each sample was measured and is reported in the table below. What is the percentage of Mn (AW 54.938) in the original sample? (14 points) H3PO4 Aliquot Aliquot 2 Aliquot 3 Std Mn (mL) KIO (9) 2 10 0.3997Abs 0.4000 0.07061 0.10773 0.20087 0.24315 5 0.3999 Aliquot 4 5 0.3999 Absorbance vs Volume of Standard Solution 0.3 y 17.734x+0.0802 0.2 0.15 a 0.1 0.05 o 2 46 8 10 1 Volume of Standard Solution Added (mL)

Explanation / Answer

Y-intercept, b = 0.0802

Slope, m = 17.734

Standard manganese solution concentration, Cs = 0.1219 M

Volume of manganese sample added, Vx = 20 mL

Concentration of manganese (MnO4-) in treated sample Cx = b Cs / (m Vx)

= 0.0802 * 0.1219 / (17.734 * 20) M

= 2.76 x 10-5 M

Moles of manganese (MnO4-) in treated sample = 2.76 x 10-5 M * 100 mL / 1000 (mL/L)

= 2.76 x 10-6 moles

Moles of manganese (MnO4-) in original 250 mL = 250 mL / 20mL * 2.76 x 10-6 moles

= 3.45 x 10-5 moles

Moles of Manganese oxidized = Moles of manganese (MnO4-) in original 250 mL

= 3.45 x 10-5 moles

Mass of Manganese present in original sample = 3.45 x 10-5 moles * AW (54.938 g/mol)

= 0.0019 g

% of Mn in original sample = 0.0019 g / 0.2058 g * 100 = 0.92 %

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