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Sample Lab Questions

Investigations #20 and #21



These questions have been taken from lab quizzes given in previous years.
Some of the answers are listed at the end.

Note: The subscripts and superscripts may not appear as such. It depends upon your web browser.



1.
(1 pt) Define end point.

2.
(1 pt) Define heterogeneous mixture.

3.
(1 pt) What is the color of a H2SO4 solution?

4.
(1 pt) Give one safety rule for the titrations performed in Investigations #20 or #21.

5.
(2 pts) Balance the folowing hypothetical half-reactions in acid solution:
(a) MnO4- Mn4+
(b) C2O42- CO2

6.
Given the following unbalanced equation for a redox reaction:
KCN(aq) + KMnO4(aq) + H2O(l) MnO2(s) + KCNO(aq) + KOH(aq)
a. (1pt) Which compound is the oxidizing agent? (2pts) Which element is reduced? What are its oxidation numbers before and after the reaction?
b. (1pt) Balance the reduction half-reaction:
MnO4- + H2O + e- MnO2 + OH-
c. (2pts) Give the balanced equation for the total reaction.
d. (2pts) Give the balanced net ionic equation for this reaction.

7.
To determine the percent of sodium oxalate (Na2C2O4) in an unknown sample, a student took 0.5946 g of the unknown sample and dissolved it in 150 mL of water. After adding some H2SO4 and MnSO4, he titrated it with 0.01612 M KMnO4 solution.
a. (1pt) Balance the equation for the titration reaction:
KMnO4 + Na2C2O4 + H+ Mn2+ + Na+ + K+ + CO2 + H2O
b. (4pts) Determine the percent of Na2C2O4 in the unknown sample if the titration required 28.33 mL of the KMnO4 solution.

8.
(2pts) What is the mass of KMnO4 (in grams) needed to prepare 250mL of a 0.025M solution of KMnO4?

9.
(3pts) Given the following ionic equation:
2MnO4- + 5C2O42- + 16H+ 2Mn2+ + 10 CO2 + 8 H2O
A freshly prepared KMnO4 solution was standardized by titrating 0.5638 g of sodium oxalate (Na2C2O4) with 40.25 mL of the KMnO4 solution Calculate the molarity of the KMnO4 solution.


Answers

6. (a) KMnO4 (b) Mn +7 +4 7. (b) 25.76% 8. 0.99 g 9. 0.04181 M


To report any corrections, please e-mail Dr. Wendy Keeney-Kennicutt.