SAMPLE EXAMINATION 2 263 PART II A second standardized solution of 0.0214 M KMnO
ID: 1041305 • Letter: S
Question
SAMPLE EXAMINATION 2 263 PART II A second standardized solution of 0.0214 M KMnO, is used to analyze an opened bottle of drugstore H2Og solution to determine the % by mass of H2O.. The net ionic equation for this reaction is 6 H(ag)+ 2 MnO,(aq)+ 5 H202(a) 2 Mn2 (ay)+ 502(8)+ 8 H20(a?) A 3.00 mL sample of the H202 solution is placed in a flask, acidifed with 2 mL of 6M H2SO, and titrated to the endpoint with the KMno, delivered from the buret. The volume of 0.0214 M KMnO required to reach the endpoint is 34.70 mL. Assume the density of the H202 solution is 1.00g/ mL. Volume of? % H,02 Volume of 0.0214 M KMnO434.70 ml 3.00 mL (e) Calculate the % by mass of the H2O2 solution. Assume the density of the H2O2 solution is 1.00g/ mL. (f) If the endpoint of the titration reaction is overshot, will the student's calculated value for the % H,0, increase, decrease, or remain the same? Explain. (g) i. Define the following terms Analyte: Titrant: ii. In this lab, there was a su substance and explain how it acted as both.Explanation / Answer
Ans. #e. Moles of KMnO4 consumed = Molarity x Vol. of soln. in liters
= 0.214 mol x 0.03470 L
= 0.0074258 mol
# Following stoichiometry of balanced reaction, 2 mol KMnO4 (= MnO4-) neutralized 5 mol H2O2.
So,
Moles of H2O2 in sample = (5 / 2) x Moles of KMnO4 = (5/2) x 0.0074258 mol
= 0.0185645 mol
Now,
Mass of H2O2 in sample = moles x MW = 0.0185645 mol x (34.01468 g/ mol)
= 0.631 g
# At density of 1.00 g/ mL, mass of H2O2 sample = 3.00 g
Now,
% H2O2 = (Mass of H2O2 / Mass of sample) x 100
= (0.631 g / 3.0 g) x 100
= 21.03 %
#f. In case of overshooting endpoint, the apparent moles of KMnO4 would becomes greater than its actual value. A higher value of moles of KMnO4 gives higher moles of H2O2 in sample, and in tur, higher (INCREASED) % H2O2 in the sample.
#g. #I. Titrant: The solution filled in burette during titration is called a titrant. In this case, KMnO3 is the titrant.
Analyte: The chemical species being analyzed/measured is called an analyte. In this case, H2O2 is an analyte.
#II. Not enough information for part of question provided here. See previous part of the question NOT provided in the image.
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