324 ❯ STEP 5. Build Your Test-Taking Confidence- D—This is due to hydrolysis of the conjugate
base of the acid (IO 3 – , IO 4 – , NO 2 – , and CN–).
The smaller the Ka of the acid, the greater the
Kb of the conjugate base. A larger Kb means a
stronger base. - A—In order to get good results when titrating
a weak base, it is important to use as strong an
acid as possible to get a sharp endpoint. HIO 3 is
the strongest acid in the table. - B—The position ½ H is halfway to the first
equivalence point. The pH at the halfway point
is equal to pKa. The value of Ka1 is 8.0 × 10 -^5 ,
which gives an approximate pKa of 5 (actual 4.1). - A—To reach H, it is necessary to convert all
the ascorbic acid to the hydrogen ascorbate ion,
which means the moles of hydrogen ascorbate
ion formed must equal the moles of ascorbic
acid originally present. The moles of hydrogen
ascorbate ion formed would require the same
number of moles of base as that required to
convert the ascorbic acid to hydrogen ascorbate.
Equal moles would mean equal volumes added. - D—In any acid-base titration, it is always
easier to use a strong base (and a strong acid).
Potassium hydroxide, KOH, is the only strong
base among the choices. - B—The titration of a weak acid with a strong
base always has an equivalence point above 7. - D—Since Ka2 is Ka2 = 1.6 × 10 -^12 , the pH at
the equivalence point would be greater than 12,
which is too high for a simple titration. - B—All can behave as Brønsted bases (accept a
hydrogen ion). Only B cannot behave as an acid
(donate a hydrogen ion). - C—Initially, doubling the volume will result in
halving the concentrations. Next, consider the
reaction. The balanced equation is: K 2 CrO 4 (aq) +
2 AgNO 3 (aq) → Ag 2 CrO 4 (s) + 2 KNO 3 (aq).
The silver ion is the limiting reagent, so very
little remains in solution (due to its Ksp). The
precipitation of silver chromate reduces the chro-
mate concentration from 0.050 M. The nitrate
does not change (soluble) remaining 0.050 M.
Since two potassium ions (soluble) are formed
per potassium chromate, after mixing the potas-
sium ion concentration was 0.10 M and does not
change (soluble).- D—To determine the molar mass of the gas it
is necessary to know the mass of the gas (mea-
sured) and the moles of the gas. The ideal gas
equation is necessary to determine the number
of moles of gas present. To use the equa-
tion, it is necessary to know the temperature
(measured), volume (measured = volume of
displaced water), and pressure of the gas. The
pressure of the gas is equal to the barometric
pressure minus the vapor pressure of water.
Water, whenever present, will contribute its
vapor pressure. - A—
[H 2 (g) + 1/2 O 2 (g) → H 2 O(l)] (-300 kJ)
2[C(s) + O 2 (g) → CO 2 (g)] 2(-400 kJ)
- [ H 2 O(l) + 2 CO 2 (g) →
C 2 H 2 (g) + 5/2 O 2 (g)] -(-1300 kJ)
2 C(s) + H 2 (g) → C 2 H 2 (g) 200 kJ
- A—Since the compound is less soluble in hot
water, the solution process must be exothermic.
Exothermic processes shift toward the start-
ing materials (solid cerium(III) sulfate) when
heated. - C—The gases are all at the same temperature;
therefore their average kinetic energies are the
same. Since kinetic energy is equal to ½ mv^2 ,
½ m 1 v 12 = ½ m 2 v 22 , the subscripts refer to two
different gases. Setting m 1 = 16 g mol-^1 and v 2 =
½ v 1 gives: ½ 16v 12 = ½ m 2 (½ v 1 )^2 ; this leads to:
16 v 12 = m 2 (1/4 v 12 ). Rearranging and cancelling
v 1 yields m 2 = 4(16) = 64 g mol-^1. - A—The smaller the molecule and the less polar
(more nonpolar) the gas is, the smaller the devia-
tion from ideal gas behavior. - A—Escaping gas would decrease the number
of moles. Less gas remaining in the container
would mean less pressure. The faster moving
gas molecules would escape faster, lower the
average velocity of those remaining in the con-
tainer. A lower average velocity means a lower
temperature.
20-Moore_PE01_p307-340.indd 324 31/05/18 1:58 pm