Page 269 - Theory and Problems of BEGINNING CHEMISTRY

258 ACID-BASE THEORY [CHAP. 17

Ans. (a) The hydronium ion is H 3 O , which is a proton added to a water molecule. (b) The Brønsted-Lowry
+
theory defines acids as proton donors and bases as proton acceptors. (c) A proton in the Brønsted sense
is a hydrogen nucleus—a hydrogen ion. (d) An acid is a proton donor. (e) A base in the Brønsted sense
is a proton acceptor. (f ) A conjugate is the product of a proton loss or gain from a Brønsted acid or base.
−
+
(g) An acid or a base that reacts completely with water to form H 3 O or OH ions, respectively. The common
strong acids are HCl, HClO 3 , HClO 4 , HBr, HI, HNO 3 , and H 2 SO 4 (first proton). All soluble hydroxides are
strong bases. (h) The equilibrium constant for the reaction of a weak acid with water. (i) The equilibrium
constant for the reaction of a weak base with water. ( j) An acid-base reaction of a substance with itself,
−→
for example, 2 H 2 O ←− H 3 O + OH .(k)pH =−log[H 3 O ]. (l) K w = [H 3 O ][OH ].
+
+
−
−
+
17.29. 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 NH 3 will not ionize as much as if the
+
ammonium ion were not present. This is the same as Problem 17.21, whether or not Le Chˆatelier’s principle
is mentioned.
17.30. What chemicals are left in solution after 0.100 mol of NaOH is added to 0.200 mol of NH 4 Cl?
Ans. The balanced equation is
NaOH + NH 4 Cl −→ NH 3 + NaCl + H 2 O
The limiting quantity is NaOH, so that 0.100 mol NaOH reacts with 0.100 mol NH 4 Cl to produce 0.100 mol
NH 3 + 0.100 mol NaCl + 0.100 mol H 2 O. There is also 0.100 mol excess NH 4 Cl left in the solution. This
is a buffer solution of NH 3 plus NH 4 in which the Na and Cl are inert.
+
+
−
−3 +
17.31. Fluoroacetic acid, HC 2 H 2 FO 2 , has a K a of 2.6 × 10 . What concentration of the acid is required to get [H 3 O ] =
−3
1.5 × 10 ?
Ans. HC 2 H 2 FO 2 + H 2 O ←− C 2 H 2 FO 2 + H 3 O +
−→
−
[C 2 H 2 FO 2 ][H 3 O ] −3
−
+
K a = = 2.6 × 10
[HC 2 H 2 FO 2 ]
−→
HC 2 H 2 FO 2 + H 2 O ←− C 2 H 2 FO 2 + H 3 O +
−
Initial x 0 0
Change −0.0015 0.0015 0.0015
Equilibrium x − 0.0015 0.0015 0.0015
If you try to neglect 0.0015 with respect to x, you get a foolish answer. This problem must be solved exactly.
(0.0015) 2 −3
K a = = 2.6 × 10
x − 0.0015
−3
2.25 × 10 −6 = (2.6 × 10 )x − 3.9 × 10 −6
x = 2.4 × 10 −3 = [HC 2 H 2 FO 2 ]

−→
17.32. For the reaction W + X ←− Y + Z
(a) Calculate the value of the equilibrium constant if 0.100 mol of W and 55.40 mol of X were placed in a 1.00-L
vessel and allowed to come to equilibrium, at which point 0.010 mol of Z was present. (b) Using the value of
the equilibrium constant calculated in (a), determine the concentration of Y at equilibrium if 0.200 mol of W and
55.40 mol of X were placed in a 1.00-L vessel and allowed to come to equilibrium. (c) Calculate the ratio of the
concentration of W at equilibrium to that initially present in part b. Calculate the same ratio for X. (d) Calculate the
value of
[Y][Z]

K =
[W]