Experiment Background: Standardization of 0.01 M Potassium Permanganate Solution

Question

Experiment Background:

Standardization of 0.01 M Potassium Permanganate Solution Obtain approximately 75 mL of un-standardized 0.01 M KMnO4 solution.

Reactions: 2MnO4- + 5C2O4 2- + 16H+ >>> 2Mn 2+ + 10CO2 + 8H2O

MnO4- + 8H+ +5Fe2+ >>>>> Mn2+ + 5 Fe 3+ + 4H2O

The optimal amount of titrant is around 20.00 mL. In this experiment the titrant has an approximate molarity of 0.01 M MnO4- and you use very pure K2C2O4• H2O to determine the molarity more precisely. Based on the optimal amount of titrant and its approximate molarity, how much pure K2C2O4• H2O should be your target mass?

Please Explain how to set up this calculation and Show any work.

Explanation / Answer

Given data

Molarity of MnO4- = 0.01 M

Volume of MnO4- = 20.0 ml

Mass of K2C2O4.H2O = ?

Lets first calculate the moles of the MnO4- using its volume and molarity

Moles = molarity * volume in liter

Moles of MnO- = mol per L * 0.020 L

                          = 0.0002 mol MnO4-

Now using the mole ratio of the balanced reaction lets calculate the moles of C2O42-

2MnO4- + 5C2O4 2- + 16H+ -------- > 2Mn 2+ + 10CO2 + 8H2O

Mole ratio of the MnO4- to C2O42- is 2 : 5

Therefore moles of C2O42- are calculated as follows

(0.0002 mol MnO4- * 5 mol C2O42- ) / 2 mol MnO4- = 0.0005 mol C2O42-

So moles of K2C2O4 needed are 0.0005 moles

Now lets convert this moles to the mass of K2C2O4.H2O

Mass = moles * molar mass

Molar mass of K2C2O4.H2O = 184.23 g per mol

Lets put the values in the formula and calculate mass of K2C2O4.H2O

mass of K2C2O4.H2O = 0.0005 mol * 184.23 g per mol

                                        = 0.0921 g K2C2O4.H2O

So the target mass of the K2C2O4.H2O = 0.0921 g

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