Page 267 - Theory and Problems of BEGINNING CHEMISTRY

256 ACID-BASE THEORY [CHAP. 17

17.18. Calculate the pH of 0.100 M NH 3 . K b = 1.8 × 10 −5

−→
Ans. NH 3 + H 2 O ←− NH 4 + OH −
+
Initial 0.100 0 0
Change −x x x
Equilibrium 0.100 − x x x

−
[NH 4 ][OH ] (x)(x) −5
+
K b = = = 1.8 × 10
[NH 3 ] 0.100
2
x = 1.8 × 10 −6
x = 1.3 × 10 −3 = [OH ]
−
K w 1.0 × 10 −14
+ −12
[H 3 O ] = = = 7.5 × 10
−
[OH ] 1.3 × 10 −3
pH = 11.13
17.19. Calculate the pH, where applicable, of every solution mentioned in Problem 17.8 through 17.14.
Ans. Merely take the logarithm of the hydronium ion concentration and change the sign: Problem 17.8, 2.511;
Problem 17.9, 2.807; Problem 17.10, 5.37; Problem 17.11, 0.05; Problem 17.13, 7.00; and Problem 17.14,
0.70.

BUFFER SOLUTIONS
17.20. Is it true that when one places a weak acid and its conjugate base in the same solution, they react with
each other? That is, do they both appear on the same side of the equation?
Ans. They do not react with each other in that way. The conjugate acts as a stress as defined by Le Chˆatelier’s
principle, and does not allow the weak acid to ionize as much as it would if the conjugate were not
present.

17.21. According to Le Chˆatelier’s principle, what is the effect of NH 4 Cl on a solution of NH 3 ?
+
Ans. The added NH 4 will cause the ionization of NH 3 to be repressed. The base will not ionize as much as if the
ammonium ion were not present.

17.22. One way to get acetic acid and sodium acetate into the same solution is to add them both to water. State
another way to get both these reagents in a solution.
Ans. One can add acetic acid and neutralize some of it with sodium hydroxide (Sec. 10.3). The excess acetic acid
remains in the solution, and the quantity that reacts with the sodium hydroxide will form sodium acetate.
Thus, both acetic acid and sodium acetate will be present in the solution without any sodium acetate having
been added directly. See the next problem.

17.23. State another way that a solution can be made containing acetic acid and sodium acetate, besides the way
indicated in Problem 17.22.
Ans. One can add sodium acetate and hydrochloric acid. The reverse of the ionization reaction occurs, yielding
acetic acid:
+ −→
−
C 2 H 3 O 2 + H 3 O ←− HC 2 H 3 O 2 + H 2 O
If a smaller number of moles of hydrochloric acid is added than moles of sodium acetate present, some of the
sodium acetate will be in excess (Sec. 10.3), and will be present in the solution with the acetic acid formed
in this reaction.