CHEMISTRY 31 EXAM 1 KEY
Some Useful Constants:
c = speed of light = 3.00 x 108 m s-1 h = Planck’s constant = 6.63 x 10-34 J s
A. Multiple Choice/Fill in the Blank Section. Only one correct answer for multiple choice
questions. (4 points for each question)
1. Write one complex ion forming reaction used in a quantitative analysis lab.
Reaction: _Mg2+ + HIn2- ó MgIn- + H+ , Ca2+ + Y4- ó CaY2- (Y = EDTA), AgCl(s) + 2NH3 ó
Ag(NH3)2+
2. In the reaction (forward direction): HPO42- + H2O ↔ H2PO4- + OH-, _H2O_ is the BronstedLowry acid.
3. Given the weak acids listed below and their pKa values, which base (below) is the strongest?
Acid
HSCN
HF
HONH3+
HBO3
pKa
0.89
3.17
5.96
9.24
a) NaSCN
b) KF
c) HONH2
d) NaBO3
Largest pKa means weakest acid means will have strongest conjugate base.
4. How would you expect the aqueous equilibrium reaction, Mg2+ + HIn2- ó MgIn- + H+ to shift
following the addition of some NaCl, assuming the main change is the ionic strength? (In =
indicator)
a) toward reactants b) toward products c) no change from addition of NaCl
reactant ions have a greater charge so will be stabilized more
5. Zinc carbonate ZnCO3 is a sparingly soluble solid. A student wants to calculate its solubility
in the presence of 0.010 M NH3 which complexes with Zn. In looking at using the systematic
method, there should be _________ mass balance equation(s) associated with solving this
problem (excluding mass balance equations associated with splitting of water).
a) 1
b) 2
c) 3
d) 4
(one mass balance relating Zn species to CO3 species – since the mass of ZnCO3 dissolved is
initially unknown – and one for the NH3 source)
6. If a photon is produced as an excited atom returns to its ground state, the process is called:
a) emission b) absorption c) infrared light
d) Beer’s law
7. In gas chromatography, the two factors affecting a compound's retention factor are:
a) the compound's volatility and polarity
b) the compound's polarity and the mobile phase's polarity
c) the compound's volatility and the flow rate
d) the carrier gas used and its flow rate
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B. Problem Section. Show all needed calculations to receive full credit. The number of points
are shown in parentheses. Use the back side of the page if needed.
1. Mercury(II) ion (Hg2+) reacts with CH3CO2- in the reaction, Hg2+ +2CH3CO2- ↔
Hg(CH3CO2)2(aq) forming a complex ion (K = 2.82 x 108). Calculate the concentration of
CH3CO2- at equilibrium if equilibrium concentrations of Hg2+ and Hg(CH3CO2)2(aq) are found
to be 1.0 x 10-5 M and 4.2 x 10-3 M, respectively. Ignore activity for this problem. (8 pts)
K=
[ Hg (CH 3CO2 ) (aq )]
[ Hg 2+ ][CH 3CO2− ]2
or [CH 3CO2− ]2 =
[ Hg (CH 3CO2 ) (aq )]
[ Hg 2+ ]K
4.2 x10 −3
=
1.0 x10 −5 2.82 x10 8
(
)(
)
[CH3CO2-] = 1.2 x 10-3 M
2. An oxalate buffer is made to have 0.0120 M NaHC2O4 and 0.0080 M Na2C2O4. Oxalic acid,
H2C2O4, is a weak diprotic acid. (16 pts)
a) Assuming each buffer species dissociates only to form Na+ as the cation with the remainder as
an anion (which do not react significantly further), calculate the ionic strength of the solution.
Ion concentrations: from NaHC2O4, [Na+] = 0.0120 M and [HC2O4-] = 0.0120 M
from Na2C2O4, [Na+] = 0.0160 M (2 Na+ per Na2C2O4) and [C2O42-] = 0.0080 M
or [Na+] total = 0.0280 M
µ = 0.5[0.0280(+1)2 + 0.0120(-1)2 + 0.0080(-2)2] = 0.5(0.72 M) = 0.036 M
b) If the hydrated radius of the H+ ion is 900 pm, calculate the activity coefficient of H+ in the
buffer.
− 0.51(+1) 2 (0.036) 0.5 − 0.0968
=
= −0.0620
1 + 900(0.036) 0.5 / 305 1.5599
γ (H+) = 10-0.062 = 0.867
log γ H + =
c) The pH of the buffer is measured to be 3.89, determine [H+] in M to 2 significant figures.
pH = -log{γ(H+)[H+]} or 10-pH = γ(H+)[H+] or [H+] = 10-3.89/0.867 = 1.5 x 10-4 M
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3. A solution is prepared in which 0.0020 moles of FeCl3 is dissolved completely in water
making a 1.000 L solution. The following reactions are known to occur:
1) FeCl3 → Fe3+ + 3ClNote: FeCl3 does not exist as a complex.
2) Fe3+ + OH- ↔ FeOH2+
3) FeOH2+ + OH- ↔ Fe(OH)2+
Assume no other complexation reactions occur. (18 pts)
a) Add one other reaction to the list above. Hint: This reaction would be needed if you wanted
to calculate the pH of the solution.
H2O(l) ↔ H+ + OH-
b) Give a charge balance equation based on the above reactions (including the one in a)).
3[Fe3+] + 2[FeOH2+] + [Fe(OH)2+] + [H+] = [Cl-] + [OH-]
c) Give two mass balance equations based on the listed reactions and the initial concentration of
the dissolved ionic compound.
[FeCl3]o = initial ionic compound concentration = 0.0020 mol/ 1.000 L = 0.0020 M
As the FeCl3 splits, we expect: [Fe]any form/[Cl]any form = 1/3 (based on stoichiometry) and also
[Fe]any form = [FeCl3]o = 0.0020 M
Rewriting the first of these two equations, we get 3[Fe]any form = [Cl]any form = 3(0.0020 M)
Since Cl doesn't react further, we know [Cl]any form = [Cl-] = 0.0060 M (the first MB
equation)
Since Fe does react further, we need to put the Fe equation in terms of existing species:
[Fe3+] + [FeOH2+] + [Fe(OH)2+] = 0.0020 M
d) Write the equilibrium equations for all equilibrium reactions.
Reaction #1 is not an equilibrium reaction so is not included.
Krxn2 = [FeOH2+]/[Fe3+][OH-]
Krxn3 = [Fe(OH)2+]/[FeOH2+][OH-]
Kw = [H+][OH-]
4. Some X-ray light detectors give signals proportional to the energy of the light. Calculate the
energy of a photon (in J) that has a wavelength of 0.840 nm. 1 nm = 10-9 m. (7 pts)
E = hc/λ = (6.63 x 10-34 Js)(3 x 108 m s-1)/[(0.840 nm)(10-9 m/nm)] = 2.37 x 10-16 J
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5. A chemist measures Fe2+ in samples by reacting Fe2+ with excess ferrocene (a complexing
agent) to produce a complex that absorbs at 562 nm. If the molar absorptivity of the Fe2+ferrocene complex is 16200 M-1 cm-1, and the absorbance is measured as 0.676 using a cell with
a width of 0.50 cm, calculate the Fe2+ concentration in M. (7 pts)
A = εbC or C = A/(εb) = 0.676/[(16200 M-1 cm-1)(0.50 cm)] = 8.3 x 10-5 M
6. Given the following reverse-phase HPLC chromatogram using a solvent of 20% methanol
and 80% water and data table below, determine:
a) the retention factor for adenosine (4 pts)
k = (tr - tm)/tm = (3.564 – 1.056)/1.056
k = 2.38
b) the resolution between
uracil and sulfanilamide (4 pts)
resolution = Δt/wave
resolution = (2.489 – 2.182)/[0.5(0.136 + 0.155)]
resolution = 2.11
c) which compound is the least polar (explain). (4 pts)
Adenosine is the least polar because it is most retained on a non-polar stationary phase.
d) The plate number (N) based on one of the retained peaks. The peak width given is the
baseline width. (4 pts)
N = 16(tr/w)2 see table for values
Data Table
Peak # Ret Time
Name
(min.)
1
1.056 unretained
2
2.182 uracil
3
2.489 sulfanilamide
4
3.564 adenosine
Area (mV*s) Peak Width
(min.)
1074.343
1617.349
0.136
1645.061
0.155
1525.604
0.219
N = 4120
N = 4130
N = 4240
Bonus) If samples were found not to contain sulfanilamide, what solvent change should be made
to speed up the separation (3 pts). Use a higher % methanol to elute compounds faster.
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