(^98) › STEP 4. Review the Knowledge You Need to Score High
❯ Answers and Explanations
It is possible to simplify the calculations by replacing the definition of molarity
mole
L
with the equivalent expression
mole
1,000 mL.
- D—The reaction is H 2 SO 4 (aq) + 2 KOH(aq) → K 2 SO 4 (aq) + 2 H 2 O(l).
The calculation is
(^) (50.0 mL base) 0.200 mol base
1,000 mL base
1 mol acid
2 mol base
1,000 mL acid
0.100 mol acid
= 50.0 mLThis calculation simplifies to 50.0 mL base
0.200 mol base
1,000 mL base1,000 mL acid
0.200 mol acid
()
- C—The reaction is H 2 C 2 O 4 (aq) + 2 NaOH(aq) → Na 2 C 2 O 4 (aq) + 2 H 2 O(l).
The calculation is
(45.20 mL base)
0.1200 mol base
1,000 mL base1 mol acid
2 mol base1
20.00 mL1,000 mL
L
= 0.1356 M acid
As always, round the values to get an estimate and pick the closest answer.- C—Moles acid = ()
50.0 mL
0.20 mol acid
1,000 mL = 0.0100 moleMoles base = 50.0 mL
0.20 mol base
()1, 000 mL
= 0.0100 moleThere is sufficient base to react completely with only one of the ionizable hydrogen
ions from the acid. This leaves H 2 AsO 4 -. Answer D cannot be correct because it is a
cation.- A—
(45.20 mL Cr O)0.1000 mol CrO
1, 000 ml Cr O6 mol Fe
1 mol Cr O1
50.00 mL1, 000 mL(^27) L
2 27
2
27
2
2
27
2
−−
−+
−^ = 0.5425 M
This is a perfect example of where simplification is important. Change the above
calculation to(45.20 mL Cr O)0.1000 mol CrO
1, 000 ml Cr O6 mol Fe
1mol CrO1
50.00 mL1,000 mL(^27) L
2 27
2
27
2
2
27
2
−−
−+
−^ = 0.5425 M
This becomes(45.20)0.1000
1, 000
6 mol Fe
11
50.00
1, 000
L
^2
+
= 0.5425 MNext, round and simplify to(50)0.1000
1
6 mol Fe
11
50.00
1
L
^2
+
= 0.6 M