HI EXPERT! B) A permanganate solution is prepared by dissolving 20.0123 g KMnO 4

Question

HI EXPERT!

B) A permanganate solution is prepared by dissolving 20.0123 g KMnO4 in 500 mL of distilled water and boiled for 1 hour to remove any organic material. Following sintered-glass filtration, the solution is quantitatively transferred to a 1.0 L volumetric flask and diluted to volume with distilled water. The permanganate solution was titrated against 0.1023 M oxalic acid prepared in sulfuric acid solution. A 50.00 mL aliquot of oxalic acid solution required 16.68 mL of permanganate solution. A titration blank required 0.04 mL of permanganate. What is the permanganate molarity?

5 H2C2O4 + 2 MnO4- + 6H+ <==> 10 CO2 + 2 Mn2+ + 8H2O

C) The percent purity of a malonic acid reagent must be calculated. A student weighs out 0.1511 grams of malonic acid, CH2(CO2H)2, and quantitatively transfers the solid to a 250 mL Erlenmeyer flask. To the flask 50.00 mL of distilled water is added to dissolve the malonic acid. After dissolution, the student uses a 50 mL volumetric pipet to add 50 mL of 0.200 M Ce4+, which is allowed to react with the malonic acid. After complete reaction, the excess Ce4+ is back titrated with 14.58 mL of 0.100 M Fe2+ solution. Calculate the percent purity malonic acid.

CH2(CO2H)2 + 2 H2O + 6 Ce4+ 2 CO2 + HCO2H + 6 Ce3+ + 6 H+

Ce4+ + Fe2+ Ce3+ + Fe3+

THANK YOU IN ADVANCE!

Explanation / Answer

Molar mass: 158.034 g/mol of Potassium permanganate

wt. of KMnO4= 20.0123 g dissolved in 1000 mL water.

(MV)KMnO4=(MV)Oxalic acid

MKMnO4=(MV)Oxalic acid/ VKMnO4= (0.1023 M x 50 mL)/ 16.64 mL=0.3074 M

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