Chemistry
Name:
§04
Date:
KEY
Reactions in Aqueous Solutions – Molarity Practice
1. A reagent bottle is labeled 0.250 M K2CrO7.
a. How many moles of K2CrO7 are present in a 0.333 mL sample?
b. How many milliliters of this solution are required to provide 0.800 mol of potassium
ion?
c. Assuming no volume change, how many grams of K2CrO7 are do you need to add to
2.00 L of this solution to make 1.000 M of K2CrO7?
d. If 50.0 mL of this solution is added to enough water to make 125 mL of solution,
what is the molarity of the diluted solution?
ANSWER
Document1 (2/9/16)
§04: Practice
p. 2
ACID-BASE
2. What is the molarity of a solution of nitric acid if 0.216 g of barium hydroxide is needed to
neutralize 20.00 mL of concentrated (15.8 M) nitric acid?
ANSWER
REDOX
3. Write the balanced chemical equations for the following reactions:
CrO42– + SO32–  Cr3+ + SO42– (acid solution)
ANSWER
A. Balance Mass.
half-reactions
CrO42–  Cr3+
add waters
CrO42–  Cr3+ + 4H2O
add H+’s
8H+ + CrO42–  Cr3+ + 4H2O
B. Balance Charge.
half-reactions
check charges
8H+ + CrO42–  Cr3+ + 4H2O
+6  +3
SO32–  SO42–
1H2O + SO32–  SO42–
1H2O + SO32–  SO42– +2H+
1H2O + SO32–  SO42– +2H+
-2  0
add/lose e–‘s
write equations
(gains 3 e–; CrO42– is reduced)
(loses 2 e–; SO32– is oxidized)
3e– + 8H+ + CrO42–  Cr3+ + 4H2O
1H2O + SO32–  SO42– +2H+ + 2e–
make e–‘s equal
2(3e– + 8H+ + CrO42–  Cr3+ + 4H2O)
3(1H2O + SO32–  SO42– +2H+ + 2e–)
6e– + 16H+ + 2CrO42–  2Cr3+ + 8H2O
3H2O + 3SO32–  3SO42– +6H+ + 6e–
§04: Practice
C. Add the two half reactions together & simplify.
6e– + 16H+ + 2Cr2O42 – + 3H2O + 3SO32–  2Cr3+ + 8H2O + 3SO42– +6H+ + 6e–
10H+ + 2Cr2O42 – + 3SO32–  2Cr3+ + 5H2O + 3SO42–
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