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

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1L
0.250molNaOH
1molHI = 6.79 x10
- 3
a) molesHI= 27.15 mL x
x
x
molHI
1000 mL
1L
1molNaOH
- 3
1000 mL
6.79 x10
molHI x
MHI=
=0.272 MHI
25mL
1L
1L
0.109molKOH
1molH
SO
- 4
2
4
b) molesH
SO
=11.12mL x
x
x
molH
SO
=6.06x10
2
4
2
4
1000mL
1L
2molKOH
- 4
1000mL
6.06x10
molH
SO
2
4
MH
SO
x
SO
=
=0.303MH
2
4
2
4
20mL
1L
1L
0.250 molNaOH
1molHCl = 4.60 x10
- 3
c) molesHCl=18.40mL x
x
x
molHCl
1000 mL
1L
1molNaOH
1000 mL
- 3
4.60 x10
molHCl x
MHCl=
=0.184MHCl
25mL
1L
9.103 A 20.00 mL sample of diprotic oxalic acid (H
C
O
) solution is titrated with 0.250
2
2
4
M NaOH solution. A total of 27.86 mL of NaOH is required. Calculate:
a) the number of moles of oxalic acid in the 20.00 mL sample
b) the molarity of the oxalic acid solution
c) the number of grams of oxalic acid in the 20.00 mL sample
Solution
1L
0.250molNaOH
1molH
C
O
2
2
4
- 3
molH
C
O
x
x
molH
C
O
=27.86mL
=3.48x10
2
2
4
2
2
4
a)
1000mL
1L
2molNaOH
- 3
3.48 x10
molH
C
O
1000 mL
2
2
4
M=
x
= 0.174 MH
C
O
2
2
4
1L
b)
20 mL
90.0 g H
C
O
- 3
2
2
4
3.48 x10
mol H
C
O
x
= 0.313 g H
C
O
2
2
4
2
2
4
1mol H
C
O
c)
2
2
4
Hydrolysis Reactions of Salts (9.12)
9.105 A solution of solid NH
Cl in pure water is acidic (the pH is less than 7). Explain.
4
Solution
+
The NH
is a Br
nsted acid, establishing an equilibrium with H
O:
Ø
2
4
+
+
NH
+ H
O
H
O
+ NH
4
2
3
3
+
The H
O
formed makes the solution acidic (pH is less than 7).
3
9.107 Predict the relative pH (greater than 7, less than 7, etc) for water solutions of the
following salts. Table 9.9 may be useful. For each solution in which the pH is
greater or less than 7, explain why and write a net ionic equation to justify your
answer. a) potassium sulfite, K
SO
b) lithium nitrite, LiNO
c) sodium
2
3
2
carbonate, Na
CO
; d) methylammonium chloride, CH
NH
Cl (CH
NH
is a
2
3
3
3
3
2
weak base)
Solution
2 -
2 -
-
-
a)
Basic. SO
is a weak base. SO
+ H
O
HSO
+ O H
2
3
3
3
-
-
-
b)
Basic. NO
is a weak base. NO
+ H
O
HNO
+ OH
2
2
2
2

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