Radioactive Decay And Acid Dissociation Constant Chap. 9 And 10 Worksheet With Answers - Middle Tennessee State University Page 9

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c) 0.100 M HCl
d) 10.00 g of H
PO
in 250 mL solution
3
4
e) 0.150 mol H
MoO
in 500 mL solution f) 0.215 mol H
MoO
in 700 mL
2
4
2
4
solution
Solution
+
-
At the equivalence point, the mol H
= mol OH
or meq acid = meq base.
V
M
= V
M
, where the molarity of the acid and base must be the molarity of
a
a
b
b
+
-
H
or OH
.
-
For the 0.120 M NaOH, the solution is 0.120 M in OH
.
+
a) For 0.180 M HCI, the solution is 0.180 M in H
.
0.180 M
M
a
V
= V
x
= 20.00 mL x
= 30.0 mL NaOH
b
a
M
0.120 M
b
+
+
b) For 0.180 M H
SO
, there are 2 H
per H
SO
; so it is 0.360 M in H
.
2
4
2
4
0.360 M
M
a
V
= V
x
= 20.00 mL x
= 60.0 mL NaOH
b
a
M
0.120 M
b
+
c) For 0.100 M HCI, the solution is 0.100 M in H
.
0.100 M
M
a
V
= V
x
= 20.00 mL x
= 16.7 mL NaOH
b
a
M
0.120 M
b
1molH
PO
3
4
molH
PO
= 10.00 gH
PO
x
= 0.1021mol H
PO
3
4
3
4
3
4
97.99 g H
PO
d)
3
4
0.1021molH
PO
3
4
M=
= 0.408 MH
PO
3
4
0.250 L soln
+
+
For 0.408 M H
PO
, there are 3 H
per H
PO
; so it is 1.23 M H
.
3
4
3
4
1.23 M
M
a
V
= V
x
= 20.00 mL x
= 205 mL NaOH
b
a
M
0.120 M
b
0.150 molH
MoO
2
4
e) M ofH
MoO
= 0.300 M H
MoO
=
2
4
2
4
0.500 L
+
+
Since there are 2H
per H
MoO
, the solution is 0.600 M H
2
4
0.600 M
M
a
V
= V
x
= 20.00 mL x
= 100 mL NaOH
b
a
M
0.120 M
b
0.215 mol H
MoO
2
4
f) M of H
MoO
=
= 0.307 MH
MoO
2
4
2
4
0.700 L
+
+
Since there are 2H
per H
MoO
, the solution is 0.614 M H
2
4
0.614 M
M
a
V
= V
x
= 20.00 mL x
= 102 mL NaOH
b
a
M
0.120 M
b
9.101 The following acid solutions were titrated to the equivalence point with the base
listed. Use the titration data to calculate the molarity of each acid solution.
a) 25.00 mL of HI solution required 27.15 mL of 0.250 M NaOH solution
b) 20.00 mL of H
SO
solution required 11.12 mL of 0.109 M KOH solution
2
4
c) 25.00 mL of gastric juice (HCl) required 18.40 mL of 0.0250 M NaOH
solution
Solution

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