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Titration and Molarity Questions

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Part A. Standardization of the 0.02 M KMnO4 solution with Fe(NH4)2(SO4)2 ? 6H2O.
(Molecular weight of the ferrous sulfate hexahydrate is 392.16 g/mol).

If you weigh out 1.01 g of the Fe(NH4)2(SO4)2 ? 6H2O, how many moles of iron (Fe+2) are in the sample?

Part B
Assume you fill a buret with your KMnO4 solution and the volume initally reads 0.15 mL (don't waste your time trying to fill it drip by drip to exactly 0.00 mL, just record the starting volume). After you use the KMnO4 solution to titrate your iron sample you notice the buret then reads 25.68 mL.

What is the volume of KMnO4 used?

Part C
How will you know when the titration reaction is done and to stop titrating KMnO4 into the sample solution?

Part D
Looking at the stoichiometry of the reaction

8H+ + MnO4- + 5Fe2+ = Mn2+ + Fe3+ + 4H2O

how many moles KMnO4 must have been delivered by the buret to react with that much Fe(NH4)2(SO4)2 ? 6H2O?

4.38×10-04 moles

1.52×10-03 moles

2.58×10-03 moles

5.15×10-04 moles

Part E
Considering the moles and volume of KMnO4 above, what must the true molarity of your KMnO4 solution be?

1.33×10-02 M

2.02×10-02 M

0.020 M

6.72×10-03 M

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https://brainmass.com/chemistry/physical-chemistry/228320

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Part A: Answer --> 0.00258 moles
Part B: Answer --> 25.53 mL
Part C: The first drop of excess KMnO4 will leave a pink color

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Part D:

Looking at the stoichiometry of the reaction

8H+ + MnO4- + 5Fe2+ = Mn2+ + Fe3+ + 4H2O

how many moles KMnO4 must have been delivered by the buret to react with that much Fe(NH4)2(SO4)2 ? ...

Solution Summary

A range of questions related to titrations and using the collected information to extract the concentrations and molarity of the solutions involved.

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below are the answers so far that I have to get the final answer -please help. I am stumped on question 2 The concentration of the NaOH is known to be 0.1M

1. For your most exact titration, record the following:

(a) Volume of NaOH solution in the burette at the start (mL): 50ml

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(c) Volume of NaOH solution delivered to the flask (mL): 18 ml

(d) Volume of HCl solution in the flask (mL): 20 ml

2. Calculate the molarity of the HCl concentration from the equation:

C(acid) = C(base) * V(base) / V(acid)

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