356 ❯ STEP 5. Build Your Test-Taking Confidence- C—The reaction is taking place in aqueous
solution. For this reason, there is no significant
change in the H 2 O concentration. It is not pos-
sible to determine the effect of a reactant if the
concentration does not change. - B—The first step is to calculate the true con-
centration of the sample to see how the student
results compare. The balanced chemical equation
is H 2 C 2 O 4 (aq) + 2 NaOH(aq) → Na 2 C 2 O 4 (aq)- 2 H 2 O(l)
All the volumes are similar; therefore, it is pos-
sible to use any one of them and calculate an
approximate molarity using rounded numbers
for simplicity.
- 2 H 2 O(l)
M H 2 C 2 O 4 ≈ (40mLNaOH) ×
×
0.050molNaOH
1000 mL1mol HCO
2mol NaOH1
10 mL224
1000 mL
L≈ 0.1 M
The calculated value should be about half what
the student reported. This indicates that the stu-
dent did not include the 1:2 mole ratio relating
the acid to the base or that the student incorrectly
used a relationship such as M 1 V 1 = M 2 V 2.- B—The relationship is: ΔG = - nFE ° (given on
the exam)
The number of electrons transferred = 2 = n
F = 96,500 coulombs mol-^1 and volt = 1 joule
coulomb-^1 (both given on the exam), these two
relationships lead to F = 96,500 J V-^1 mol-^1.
E ° = (0.76 − 0.28) V = 0.48 V
Entering this information in the equation gives
ΔG = - (2)(96,500 J V-^1 mol-^1 )(0.48 V) ≈
- (2)(100,000)(0.50) ≈ - 100,000 J mol-^1 (actual
value = - 9.26 × 104 J mol-^1 )
- A—At point G, all the CO 32 - has been con-
verted to HCO 3 - and the moles of HCO 3 - will
equal the moles of CO 32 - originally present. It
will require an equal volume of acid to titrate
an equal number of moles of HCO 3 - as required
for the CO 32 -. For pure sodium carbonate, F will
always be 2G.
16. A—At point G, all the CO 32 - has been converted
to HCO 3 - and the moles of HCO 3 - will equal the
moles of CO 32 - originally present plus the quan-
tity of HCO 3 - originally present. It will require
a greater volume of acid to titrate a greater
number of moles of HCO 3 - as required for the
CO 32 -.
17. C—It would be necessary to titrate the strong
base and the CO 32 - to reach G. However, it is
only necessary to titrate the HCO 3 - to reach H,
which means less acid is necessary.
18. B—At G the CO 32 - is now HCO 3 - , so no CO 32 -
remains. The Na+ did not react, so it is still
present as ions. The Cl- is from the HCl and
remains as separate ions in solution. After G, the
H+ from the acid begins to convert HCO 3 - to
form H 2 CO 3 , which is complete at point F leav-
ing no HCO 3 - in the solution. Other than water,
all species are strong electrolytes and exist as ions
in solution. The H 2 CO 3 will be decomposing to
H 2 O and CO 2 (g).
19. D—The pH will equal the pKa2 when the con-
centration of HCO 3 - equals the concentration
of H 2 CO 3. This occurs when one-half of the
HCO 3 - has been converted to H 2 CO 3.
20. D—The average kinetic energy, not the aver-
age speed, is the same if the temperatures are
the same. Each container has one mole of gas,
which means that at the same volume and tem-
perature they will have the same pressure. The
greater the molar mass, divided by a constant
volume, the greater the density. One mole of gas
will have Avogadro’s number of molecules.
21. B—The decrease in temperature indicates that
the system absorbed heat, meaning that this is
an endothermic process. For an endothermic
process to be spontaneous, the entropy must
increase.
22. D—Hydrogen bonding may occur when hydro-
gen is attached directly to N, O, or F.
23. C—The two molecules are hydrogen bonded
together. Hydrogen bonding is a relatively strong
intermolecular force. Acetyl chloride cannot
exhibit anything stronger than dipole-dipole
forces, which are, in general, weaker than hydrogen
bonds.
21-Moore_PE02_p341-370.indd 356 31/05/18 1:54 pm