Notes
a) 3 points
HCOOH + H2O <===> H3O+ + HCOO¯
Ka = ([H3O+] [HCOO¯]) / [HCOOH]
| [HCOOH] | [H3O+] | [HCOO¯] | |
| Initial: | 0.40 | 0.60 | |
| Equil: | 0.40 - y | y | 0.60 + y |
Ka = 1.8 x 10¯4 = (y (0.60 + y)) ÷ (0.40 - y)
neglect the +y and the -y
[H3O+] = y = 1.2 x 10¯4
pH = - log (1.2 x 10¯4) = 3.92
Alternative Approach:
pH = pKa + log [A¯] / [HA]
= - log (1.8 x 10¯4) + log (0.60 / 0.40)
= 3.92
b) 1 point
The pH remains unchanged because the ratio of the formate and formic acid stays the same
c) 3 points
Initial amounts of concentrations:
1.00 M HCl x 0.00500 L = 0.00500 mole HCl
0.40 M HCOOH x 0.100 L = 0.040 mole HCOOH
0.60 M HCOO¯ x 0.100 = 0.060 mole HCOO¯
or
1.00 M HCl x (5.00 ml / 105 ml) = 0.0476 M HCl
0.40 M HCOOH x (100 ml / 105 ml) = 0.38 M HCOOH
0.60 M HCOO¯ x (100 ml / 105 ml) = 0.57 M HCOO¯
Amounts or concentrations after H+ reacts with HCOO¯ :
0.040 + 0.005 = 0.045 mole HCOOH
0.060 - 0.005 = 0.055 mole HCOO¯
or
0.43 M HCOOH and 0.52 M HCOO¯
[H3O+] = 1.8 x 10¯4 x ((0.045/vol.) / (0.055/vol.)) = 1.5 x 10¯4 M
d) 2 points
0.800 liter x 2.00 M HCOOH = 1.60 moles HCOOH
0.200 liter x 4.80 M NaOH = 0.96 mole OH¯
1.60 moles HCOOH + 0.96 mole OH¯ yields 0.64 mole HCOOH and 0.96 mole HCOO¯
[H3O+] = 1.8 x 10¯4 x (0.64 mole/liter) / (0.96 mole/liter) = 1.2 x 10¯4
The alternative approach in part (a) can also be used in part (d)
a) 3 points
Cathode:
2H2O + 2e¯ ---> H2 + 2 OH¯
or
2H+ + 2e¯ ---> H2
Anode:
2 H2O ---> O2 + 4H+ + 4e¯
or
2 O2¯ ---> O2 + 4e¯
Overall:
2 H2O ---> 2 H2 + O2
or
4 H+ + 2 O2¯ ---> 2 H2 + O2
b) 1 point
Coulombs = 10.0 amp x 100 min x 60 sec/min = 6.00 x 104 coulombs
c) 2 points - the 2 pts were distributed by assigning 1 pt. for each factor
moles of O2 = (60,000 coulombs / 96,500 coulombs/mole of e¯) x (1 mole O2 / 4 moles of e¯)
= 0.155 mole of O2
d) 3 points
moles of H2 = (2 moles H2 / 1 mole O2) x 0.155 mole O2 = 0.311 mole H2
one mole H2 yields 1 mole H2O
Therefore:
[delta]H = 0.311 mole H2O x (-242 kJ / mole H2O) = - 75.2 kJ
or
heat liberated = (242 kJ / mole) x 0.311 mole = 75.2 kJ
Credit in later parts is given if an error is made in balancing the reactions in part (a) and if the factors used are consistent with the equation.
a) 2 points
95 g F x (1 mole F / 19.0 g F) = 5 mole F atoms
5 g H x (1 mole H / 1.00 g H) = 5 mole H atoms
Therefore HF
One point was awarded if H5F5 or other 1:1 ratios were given.
b) 2 points
Amount of F in UF6 = (4.267 g) ((6 x 19.0) / (238 + 6 x 19.0))
= 1.38 g F
Fraction of F in HF = (0.970 g x 0.95) / 1.38 g
Fraction of F in solid = 1 - 0.67 = 0.33
c) 4 points
Mass of U = 4.267 g - 1.38 g = 2.89 g U
Mass of F = 1.38 g x 0.333 = 0.46 g F
Mass of O = 3.730 g - (2.89 + 0.46) = 0.38 g O
2.89 g U x (1 mole U / 238 g U) = 0.012 mole U atoms
0.460 g F x (1 mole F / 19.0 gF) = 0.024 mole F atoms
0.38 g O x (1 mole O / 16.0 g O) = 0.024 mole O atoms
Therefore solid product is UO2F2
d) 1 point
UF6 + 2 H2O ---> UO2F2 + 4 HF
4) The 5 choices were scored at 3 points each: In general 1 point for the reactions and 2 points for the products except for part (b) where the weighting was reversed. Partial credit was given for equations with multiple products. No credit was given to species with incorrect formulas or charges.
(a) H2 + Fe2O3 ---> Fe + H2O (or FeO)
(b) I¯ + IO3¯ ---> I2 + H2O (H2O not required)
(c) H+ + SO42¯ + CaF ---> HF + CaSO4 (or HSO4¯)
If H+ + F¯ ---> HF or Ca + SO42¯ ---> CaSO4, a maximum of two points was awarded.
(d) (NH4)2CO3 ---> H2O + CO + NH3
One point for any correct product and 2 points for all three products.
(e) CH4 + Cl2 ---> CCl4 + HCl
(CH3Cl, CH2Cl2, CHCl3, also accepted)
(f) Zn(OH)2 + NH3 ---> Zn(NH3)42+ + OH¯
(2 points were awarded for Zn2+ + NH3 ---> Zn (NH3)42+)
(g) H2O2 + Br¯ + H+ ---> Br2 + H2O (H+ not required)
(h) Hg22+ + Cl¯ ---> Hg2Cl2
a) 2 points
Real molecules exhibit finite volumes, thus excluding some volume from compression.
Real molecules exhibit attractive forces, thus leading to fewer collisions with the walls and a lower pressure.
b) 3 points
SO2 is the least ideal gas.
It has the largest size or volume.
It has the stongest attractive forces ( van der Waals forces or dipole-dipole interactions).
c) 3 points
High temperature results in high kinetic energies.
This energy overcomes the attractive forces.
Low pressure increases the distance between molecules. (So molecules comprise a small part of volume or attractive forces are small)
a) 1 point
Mg 1s2 2s22p6 3s2
Ar 1s2 2s22p6 3s23p6 (both must be correct)
b) 3 points
Valence electrons for Mg and Ar are in the same principal energy level, but Ar is the smaller and it has a greater nuclear charge. Therefore, the first and second ionization energies of Mg are less than those of Ar. The removal of the third electron from Mg involves the n = 2 energy level, and this electron experiences a very large nuclear charge.
(A maximum of 2 points was awarded for explanations based only on the stability of an octet.)
c) 2 points
MgCl2 forms, since Mg readily loses its two valence electrons to form a stable configuration.
d) 2 points
The formula is QCl
A very high second ionization energy indicates that there is only one valence electron.
a) 2 points, using any of the last three equations received one point
Ba2+ + 2 OH¯ + 2 H+ + SO42¯ ---> BaSO4(s) + 2 H2O
(or H+ + HSO4¯)
or
Ba(OH)2 + H2SO4 ---> BaSO4(s) + 2 H2O
or
H+ + OH¯ ---> H2O
or
Ba2+ + SO42¯ ---> BaSO4
b) 3 points
The initial conductivity is high because of the presence of Ba2+ and OH¯ ions. The conductivity also decreases because Ba2+ forms insoluble BaSO4 with the addition of SO42¯ ions.
The conductivity also decreases because OH¯ combines with the addition of H+ ions by forming H2O.
Beyond the equivalence point, conductivity increases as H+ and SO42¯ (or HSO4¯) ions are added.
c) 1 point
# moles Ba(OH)2 = # moles H2SO4
= (0.1 mol / L) x 0.04 L
= 0.004 mol
d) 2 points
BaSO4(s) dissociates slightly to form Ba2+ + SO42¯ (or H2O ionizes slightly to form H+ + OH¯).
a) 3 points
Add water to the mixture.
CaCO3 does not dissolve in water whereas the CaCl2 does dissolve.
Filter the solution. The aqueous CaCl2 solution passes through the filter paper and the CaCO3 is collected on the paper.
b) 3 points
Pipet an aliquot of known volume into a flask.
Add excess AgNO3 solution to precipitate AgCl.
Filter, dry, and then weigh the AgCl.
wt. AgCl / M.W. AgCl = # moles AgCl = # moles NaCl
M = moles AgCl (or NaCl) / Volume Aliquot in L
or
Take a known volume of solution. Evaporate solution to dryness and weigh the NaCl residue.
wt. NaCl / M.W. NaCl = # moles NaCl
M = # moles NaCl / vol. NaCl in L
Full credit for this problem could also be achieved by using a weight measurement of the solution (and obtaining the the weight of the solvent), evaporating the solvent, weighing the residue, and calculating the molality. Through the proper application of a colligative property, it was also possible to obtain full credit.
c) 2 points
Fractional distillation
(An ordinary distillation or separation received 1 point.)
a) 4 points
One point if only one or two structures were given.
Full credit for CO2 was awarded for giving only the first structure above.
b) 1 point
CO has the shortest bond distance because it has the greatest number of electrons between carbon and oxygen.
c) 3 points
CO32¯ is a trigonal planar. There are 3 bonded pairs of electrons and no lone pairs around C. To minimize the repulsion of the bonded pairs there will be 120° bond angles with all the atoms in the same plane, or C is sp2 hybridized with a delocalized pi bond, therefore, CO32¯ must be planar.
CO2 is linear. It has 2 bonded pairs and no lone pairs of electrons. Or C uses two sp hybridized orbitals.
CO is linear. Two points or atoms make a straight line.
One point was given in part (c) for 3 correct geometrics with no explanations or incorrect or inadequete explanations.
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Educational Testing Service. All rights reserved.