Radioactive Decay And Acid Dissociation Constant Chap. 9 And 10 Worksheet With Answers - Middle Tennessee State University Page 11
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202 -
2 -
-
-
c)
Basic. CO
is a weak base. CO
+ H
O
HCO
+ OH
3
3
2
3
+
+
+
d)
Acidic. CH
NH
is a weak acid. CH
NH
+ H
O
H
O
+ CH
NH
3
3
3
3
2
3
3
2
9.109 A chemist has 20.00 mL samples of 0.100 M acid A and 0.100 M acid B in
separate flasks. Both acids are monoprotic. Unfortunately, the flasks are not
labeled, so the scientist does not know which sample is in which flask. But
fortunately, it is known that acid A is strong and acid B is weak. Before thinking
about the problem, the chemist adds 20.00 mL of 0.100 M NaOH to each flask.
Explain how the chemist could use a pH meter (or pH paper) to determine which
flask originally contained which acid.
Solution
The 20.00 mL of NaOH would react with all of the acid in both cases, giving
essentially a solution of the sodium salt of the acid. The salt of the strong
acid would not hydrolyze and would have a pH = 7. The salt of the weak acid
would hydrolyze, giving a basic solution (pH greater than 7).
9.111 How would the pH of equal molar solutions of the following salts compare
(highest, lowest, etc)? NaH
PO
, Na
HPO
, and Na
PO
2
4
2
4
3
4
Solution
3 -
The Na
PO
would have the highest pH, definitely basic, because the PO
3
4
4
is the strongest base of the three anions. The NaH
PO
would be the lowest
2
4
-
pH, definitely acidic, because H
PO
is the strongest acid of the three
2
4
anions. The pH of the Na
HPO
would be higher than NaH
PO
and lower
2
4
2
4
than Na
PO
.
3
4
Buffers (Section 9.13)
9.113 Could a mixture of ammonia (NH
), a weak base, and ammonium chloride
3
(NH
Cl) behave as a buffer when dissolved in water? Use equations to justify
4
your answer.
Solution
+
Yes. The NH
could react with added acid, and the NH
could react with
3
4
added base. The equations are:
+
+
+
-
NH
+ H
" NH
and NH
+ OH
" NH
+ H
O
3
4
4
3
2
9.115 Calculate the pH of a buffer made by dissolving 1 mol formic acid (HCOOH) and
1 mol sodium formate (HCOONa) in 1 L of solution (see Table 9.9).
Solution
-
[B
]
pH = pK
+ log
a
From Table 9.9, pK
= 3.74
Using Equation 9.54,
[HB]
a
-
[B
] = [HB] = 1.0
pH = 3.74 + log 1 = 3.74
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