Unit 5: Acids & Bases Page 2
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2Name: ANSWER KEY
2
4. What is the molar concentration of a 50-mL solution of Ba(OH)
that is titrated to an end point by
2
15 mL of a 0.00300 M solution of HCl?
(aq) → BaCl
2 HCl(aq) + Ba(OH)
(aq) + 2 H
O (l)
2
2
2
M
= 0.00300
M
= M
A
B
B
B
B
V
= 15
V
= 50
A
B
M
V
= 2 M
V
A
A
B
B
(0.00300) (15) = 2 (M
) (50)
answer
B
B
×
= 4.5 × 10
-4
-4
M
M
[Ba(OH)
] = 4.5
10
M
B
2
2
5. What is the molarity of a 21 mL nitric acid solution that completely neutralizes 25.0 mL of a
0.300 M solution of NaOH?
(aq) + NaOH(aq) → NaNO
HNO
(aq) + H
O(l)
3
3
2
M
= M
M
= 0.300
A
A
B
B
V
= 21
V
= 25.0
A
B
M
V
= M
V
A
A
B
B
(M
) (21) = (0.300) (25.0)
A
M
= 0.357 M
Answer: [HNO
] = 0.357 M
A
3
2
6. What is the molar concentration of a 45.0 mL solution of KOH that is completely neutralized by 15.0
mL of a 0.500 M H
SO
solution?
2
4
+ 2 KOH → K
H
SO
SO
+ 2 H
O
2
4
2
4
2
M
= 0.500
M
= M
A
B
B
B
B
V
= 15
V
= 45
A
B
M
V
= 2 M
V
A
A
B
B
(0.500) (15) = 2 (M
) (45)
B
B
M
= 0.333 M
Answer: [KOH] = 0.333 M
B
2
.
7
A neutral solution is produced when 42.00 mL of a 0.150 M NaOH solution is used to titrate 50.00 mL
of a sulfuric acid (H
SO
) solution. What is the concentration of the sulfuric acid solution before
2
4
titration?
+ 2 NaOH → Na
H
SO
SO
+ 2 H
O
2
4
2
4
2
M
= M
M
= 0.150
A
A
B
B
V
= 50.00
V
= 42.00
A
B
2 M
V
= M
V
A
A
B
B
2 (M
) (50) = 2 (0.150) (42.00)
A
M
= 0.063 M
Answer: [H
SO
] = 0.063 M
A
2
4
Unit 5: Acids & Bases
Assignment 3
Page 2 of 2
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