Notes
(a) (i) Cu(OH)2(s) <===> Cu2+(aq) + 2 OH¯(aq) (one point)
Correct stoichiometry and charges (but not phases) necessary
No credit earned if water as a reactant or product
(ii) 1.72 x 10¯6 g / 97.57 g/mol = 1.763 x 10¯8 mol Cu(OH)2 (one point)
1.763 x 10¯8 mol Cu (OH)2 / 0.100 L = 1.76 x 10¯7 moles per liter (one point)
One point earned for conversion of mass to moles (need not to be computed explicity)
One point earned for calculation of moles per liter
(iii) [ Cu2+] = 1.76 x 10¯7 M
[OH¯] = 2 x (1.76 x 10¯7 M ) = 3.52 x 10¯7 (one point)
Ksp = [Cu2+] [OH¯]2 = (1.76 x 10¯7) (3.52 x 10¯7)2 = 2.18 x 10¯20 (one point)
One point earned for correct [Cu2+] and [OH¯]
One point for correct substitution into Ksp expression and answer
Response need not include explicit statement of [OH¯] if Ksp expression is written with correct values of [Cu2+] and [OH¯]
(b) (i) pH =9.35 ----> pOH = 4.65 ----> [OH¯] = 2.24 x 10¯5 M (one point)
[Zn2+] = Ksp / [OH¯]2 = 7.7 x 10¯17 / (2.24 x 10¯5)2 = 1.5 x 10¯7 M (one point)
One point earned for correct determintion of [OH¯]
One point for correct answer (assume [Zn2+] equals solubility in moles per liter)
No points earned if [OH¯] is assumed equal to twice [Zn2+]
(b) (ii)
| Zn2+ | + | 2 OH¯ | ----> | Zn(OH)2 | |
| initial amount | 0.0050 mol | 0.0150 mol | 0 mol | ||
| final amount | 0 mol | 0.0050 mol | 0.0050 mol |
OR
[OH¯] = 0.0050 mol / 0.100 L = 0.0050 M (one point)
One point earned if precipitation reaction is clearly indicated and moles or concentration of OH¯ is calculated correctly
| Zn(OH)2(s) ---> | Zn2+ | + | OH¯ |
| x | (0.050 + 2x) |
Ksp = 7.7 x 10¯17 = [Zn2+] [OH¯]2 = (x)(0.050 + 2x)2 = (x)(0.050)2 ----> [Zn2+] = 3.1 x 10¯14 M (one point)
OR
| Zn(OH)2(s) ---> | Zn2+ | + | 2 OH¯ |
| (0.050-x | (0.150-2x) |
Ksp = 7.7 x 10¯17 = [Zn2+] [OH¯]2 = (0.050 - x)(0.150 - 2x)2 (one point)
Solve for x, then subtract x from 0.050 M to obtain [Zn2+] (one point)
2)
(a) Assume a 100 gram sample ( not necessary for credit ):
65.60g C x (1 mol C / 12.01 g C) = 5.462 mol C
9.44g H x (1 mol H / 1.0079 g H) = 9.366 mol H
mass O = [100 - (65.60 + 9.44)] = 24.96 g O
24.96 g O x (1 mol O / 15.9994 g O) = 1.560 mol O
C5.462H9.366O1.560 ---> C3.5H6.0O1.0 ---> C7H12O2
One point earned for determining moles of C and moles of H
One point earned for determining moles of O
One point earned for correct empirical formula
(b) m = DT / Kf = 15.2 °C /40.0 K kg mol¯1 = 0.380 mol / kg
0.01608 kg x (0.380 mo / 1 kg) = 0.00611 mol
molar mass = 1.570 g/ 0.00611 mol = 257 g / mol
One point earned for determination of molarity
One point earned for conversion of molarity to molar mass
OR,
moles solute = (DT x kg solvent) / Kf = 0.00611 mol (one point)
molar mass = 1.570 g / 0.00611 mol = 257 g / mol (one point)
OR,
molar mass = (mass x Kf) / (DT x kg solvent) = 257 g / mol (two points)
empirical mass of C7H12O2 = 7(12) + 12(1) + 2(16) = 128 g/mol
128 g/mol = 1/2 molar mass ---> molecular formula = 2x ( empirical formula) -----> molecular formula = C14H24O4 (one point)
One point earned if molecular formula is wrong but is consistent with empirical formula and molar mass
No penalty for simply ignoring the van't Hoff factor
Only one point earned for part (b) if response indicates that DT= (15.2 + 273) = 288 K and molar mass = 13.6 g / mol
(c) n = (pV) / (RT) = [(1 atm) (0.577 L)] / [(0.0821 L atm mol°1 K°1) (573 K)] = 0.0123 mol (one point)
molar mass = mass of sample / moles in sample = 1.570 g / 0.0123 mol = 128 g/mol (one point)
Only one point can be earned for part (c) if wrong value for R is used and/or T is not converted from C to K
(d) The compound must form a dimer in solution, because the molar mass in solution is twice that it is in the gas phase,
OR,
the compound must dissociate in the gas phase ( A (g) --> 2B (g)) because the molar mass in the gas phase is half that it is in solution.
One point earned for a reference to either or both the ideas of dimerization and dissociation,br> -No point earned for a " non - ideal behavior " argument
(a) 2.000g x (1 mol / 94.113 g) = 0.02125 mol phenol (one point)
Heat released per mole = 64.98kJ / 0.02125mol = 3,058 kJ/mol (one point)
or, DHcomb = - 3,058 kJ/mol
Units not necessary
(b) DHcomb = - 3,058 kJ/mol (one point)
- 3,058 kJ = [6 (-395.5) + 3 (-285.85)] - [DH°f phenol] (one point)
DH°f phenol = - 161 kJ (one point)
One point earned for correct sign of heat of combustion, one point for correct use of moles/coefficients, and one point for correct substitution
(c)DS° = [3 (69.91) + 6 (213.6)] - [7 (205.0) + 144.0 = - 87.67 J/K (one point)
DG° = DH° - TDS° = 3,058 kJ - (298 K) (-0.08767 kJ/K ) = -3,032 kJ (one point)
Units not necessary; no penalty if correct except for wrong DHcomb from part (a)
(d) moles gas = 9 x [moles from part (a)] = 9 (0.02125 mol0 = 0.1913 moles gas (one point)
P = (nRT) / V = [(0.193 mol) (0.0821 L atm mol¯1K¯1) (383 K)] / 10.0 L = 0.601 atm (one point)
Units necessary; no penalty for using Celsius temperature if also lost point in part (c) for same error
4)
(a) Sn2+ +Fe3+ ---> Sn4+ +Fe2+
Two points earned if only error is wrong symbol for tin (e.g., Ti)
(b) Co2+ + OH¯ ---> Co(OH)2
(c) C2H4 + O2 ---> CO2 + H2O
No penalty for other oxidized forms of carbon as products (e.g.,C, CO)
(d) H3PO4 + OH¯ ---> H2PO4¯ + H2O
One point earned for H+ + OH¯ ---> H2O
Two points earned for removal of H+ from any HxPyOz species and H2O as product
(e) CaSO3 ----> CaO + SO2
Two points earned for CaSO4 CaO + SO3
(f) H+ + Cl¯ + [Ag(NH3)2]+ ---> AgCl + NH4+
Cl¯ + [Ag(NH3)2]+ ---> AgCl + NH3 ( or NH4+ ) earns two points
H+ + [Ag(NH3)2]+ ---> Ag+ + NH4+ earns two points
(g) Na2O + H2O ---> Na+ + OH¯
Two points earned if reactants correct but only product is NaOH
(h) Zn + H+ ---> Zn2+ + H2
Two points earned for Zn + H+ + Br¯ ---> ZnBr2 + H2
Two points earned for Zn+ HBr ---> Zn2+ + Br¯ + H2
5)
(a) 4 essential steps (2 points)
1) weigh KHP
2) fill buret with NaOH solution
3) add indicator (phenolphthalein)
4) titrate to endpoint (color change)
Two points earned for all 4 steps; one point earned for 2 or 3 steps
Titration of acid into base accepted if described correctly
(b) moles KHP = Mass KHP / molar mass KHP (one point)
moles KHP = moles OH¯ at equivalence and (moles OH¯/ liters NaOH) = [ OH¯] (one point)
Acceptable if some parts of part (b) appear in (a)
(c) Curve should have 3 important features: (2 points)
1) curve begins above pH 1, but below pH 7
2) equivalence point at 25 mL
3) equivalence point above pH 7
Both points earned for all 3 features
One point earned for any 2 of the 3 features
(d) At the halfway point in the titration, pH = pKa. (one point) (e) A point A in the titration, the anion in highest concentration is Y2¯.
Also acepted: Y¯2, Y¯¯, Y=, and specific anions such as SO42¯,
SO32¯
HY¯, Y¯, and "Y ion" not accepted
6)
(a) Response must clearly indicate ( and distinguish between ) Eact and DHrxn on graph
Each earns one point
(b) i. Response shows a softly curving line that approaches the time axis and whose slope changes continually.
No penalty if curve crosses time axis or levels out above time axis.
Curve must drop initially and continually. No credit earned if [N2O5]
increases
ii. Reaction rate is the slope of the line tangent to any point on the curve. (one point)
Rate must be tied somehow to slope of the graph
Answer may be indicated directly on the graph
Instantaneous rate ust be indicated rather than the average rate
iii. Since "rate = slope = k[N2O5]", the value of k can be determined algebraicallly from the slope at a known value of [N2O5]. (one point)
No penalty for "rate = 2k [N2O5] as a reaction stoichiometry
could suggest this answer.
Point can be earned for rate constant = slope of graph if ln[N2O5]
vs. time since reaction is first order.
Use half-life or integrated rate law to solve for k can be accepted.
iv. The value of the rate constant is independent of the reactant concentrations, so adding more reactant will not affect the value of k. (one point)
no point earned for simply stating that value of k will not change.
Response must distinguish between rate and rate constant.
(c) i. Rate = k[A] or ln([A]/[A]o) = kt. Since graph of ln[A] vs. time is linear, it must be a first-order reaction. (one point)
Either from of the rate law is acceptable, and both the equation and the brief
justification are required to earn the point.
No point earned if response indicates first order because the first graph is not linear.
ii. Determine the slope of the second graph and set "k = -slope." (one point)
Response must indicate both the negative sign and the slope.
7)
(a) The number of moles of CO will decrease (one point)
because
adding H2, will make the reaction shift to the left, (one point)
OR
adding H2s will make the reaction quotient larger than K, thus the reaction shifts to the left.
(b) The number of moles of CO will increase (one point)
because
since the reaction is endothermic, adding of the heat (as a reactant) will drive the reaction to the right. (one point)
(c) The number of moles of CO will decrease. (one point)
because
`there are more moles of gas (2) on the right than on the left (1), thus decreasing the volume which increases the pressure causes the reaction to shift to the left. (one point)
(d) The number of moles of CO will stay the same (one point)
because
Solids are not involved in the equilibrium expression (one point)
OR
solids have no effect on the equilibrium.
8)
(a) 2 Ag+(aq) + Cd(s) --> 2 Ag(s) + Cd2+(aq) (one point)
equation must be balanced and net ionic, phases not necessary
reaction direction and ion charges must be correct
0.08 - (-0.40) = 1.20 V (one point)
evidence of where numbers came from should be present; if equation is exactly reversed, -1.20 V earns the point
(b) Anions (or NO3¯ ions) will follow to the Cd2+ solution or from the Ag+ solution to balance the charges (one point)
OR
Anions will flow to the left to balance the positive charge of the new Cd2+
ions
both the correct direction and justification needed to earn this point
direction may be indicated by arrow marked on diagram
(c) The cell voltage will increase. (one point)
Ag+ is a reactant, so increasing [Ag+] will increase the driving force (stress) for the forward (spontaneous) reaction and the potential will increase (one point)
OR
Since Q = [Cd2+] / [Ag+]2, increasing [Ag+] will decrease Q. According to the Nernst equation, E = E - ( 0.0592 log Q ) / n, if Q decreases, then the voltage increases.
(d) The cell voltage will decrease. (one point)
Adding the NaCl will have no effect on the Cd cell, but will cause AgCl to precipitate
in the Ag cell (Ag+ + Cl¯ --> AgCl ). Thus [Ag+] causes a decrease in
voltage. (one point)
One credit earned for " decreasing [Ag+] results in decreased voltage
" or "opposite part of (c) "
(e) Since Q = [Cd2+] / [Ag+]2, diluting both solution
by the same amount will increase the value of Q. According to the Nernst equation, E = E°
- (0.0592 log Q)/n, if Q increases, then voltage dcreases. (one point)
No credit earned for "since the solutions are diluted, the voltage will decrease.
"
9)
(a) At the higher altitude the ambient presssure is significantly less than 1.0 atm. Under reduced pressure, water boils at less than 100 °C. (2 points)
Two points earned for " At the higher altitude water boils at less than 100 °C, and at the lower temperature the chemical/physical process ("the cooking") take longer.
(b) Cu2+ (aq) + 2 OH¯ (aq) --> Cu(OH)2(s)
Response must indicate that an insoluble hydroxide forms, but equation is not neccessary.
(one point)
Cu(OH)2(s) + 4 NH3(aq) ---> [Cu(NH3)4]2+
(aq)
Response must indicate some cationic ammine complex with a reasonable coordination number.
(c) CH3CH2OH
Response must indicate that a clear ethanol structure ( Lewis diagram not neccessary )
(one point)
The hydroxyl group forms hydrogen bonds with water molecules. (one point)
Response must mention/indicate involvement of hydroxyl group
Point earned for " Ethanol is more polar than dimethyl ether ," but no point
earned for " dimethyl ether is linear ( or nonpolar )"
(d) Au3+, Co3+, or F2 These oxidants are below the Cl2/Cl¯ reduction half-reactions, so they would spontaneously oxidize Cl¯ to Cl2 (one point)
Any species to the right of the arrow and above the Cl2/ Cl¯ reduction half -reaction on the standard reduction potential table. (one point)
Identification and justification needed to earn each point - justification should minimally make some reference to relative positions in the reduction potential table.