CHM 235 Quantitative Analysis
Dr. S.A. Skrabal
Exam II
21 March 2006
NAME:
SOLUTIONS
Instructions: Read each question carefully before answering. Show all work on questions requiring calculations
to receive credit. Points will be taken off for incorrect significant figures. Circle or box in your final numerical
answer. The value of each question is given in parentheses after the question. Useful information is located on
the last page of the exam. Good luck!
1. You are hired by the Chargealot Wastewater Co. as a consultant. The company has a 520 L tank full of
wastewater that contains 12 mM dissolved Hg+ and 35 mM Pb2+ and have asked you for advice on how to
precipitate these ions as solids which can be removed easily from the tank. You determine that it is possible to
add dissolved sulfate ion (SO42-) to cause the following reactions to occur:
2 Hg+ (aq) + SO42- (aq)  Hg2SO4 (s)
Pb2+ (aq) + SO42- (aq)  PbSO4 (s).
The values of Ksp for Hg2SO4 and PbSO4 are 7.4 x 10-7 and 6.3 x 10-7, respectively. Calculate the minimum
concentration of dissolved SO42- needed in the tank to just begin precipitation of (a) Hg2SO4 and (b) PbSO4 in the
tank. Each of your answers must be supported by an appropriate calculation. (12)
Calculate the reaction quotient, Q, required so that Q = Ksp. Remember, when Q = Ksp, the solution is
saturated—the initial situation required for precipitation of the solid to occur.
For Hg2SO4: Q = Ksp = 7.4 x 10-7 = [Hg+ (aq)] 2 [SO42- (aq)]
7.4 x 10-7 = (12 x 10-3)2 [SO42- (aq)]
[SO42- (aq)] = 7.4 x 10-7 / (12 x 10-3)2 = 5.1 x 10-3 M
For PbSO4:
Q = Ksp = 6.3 x10-7 = [Pb2+ (aq)] [SO42- (aq)]
6.3 x 10-7 = (35 x 10-3) [SO42- (aq)]
[SO42- (aq)] = 6.3 x 10-7/35 x 10-3 = 1.8 x 10-5 M
2. Explain the significance of ex and em in a quantitative fluorescence or phosphorescence analysis. What do they
mean and why do you need to know them for such an analysis? (10)
ex = excitation wavelength at which absorbance of light occurs by some species
em = emission wavelength at which fluorescence or phosphorescence is being measured in an experiment
Both wavelengths are required in order to perform a quantitative analysis using these emission methods.
3. Calculate the solubility of silver phosphate (Ag3PO4; FW = 418.58; Ksp = 2.8 x 10-18 ) in a solution that already
contains 0.050 M PO43-. The dissolution reaction is:
Ag3PO4 (s)  3 Ag+ (aq) + PO43- (aq).
Express your answer in units of mol L-1. (12)
Ksp = [Ag+ (aq)3 [PO43- (aq)] = 2.8 x 10-18
Initial
Equilibrium
Ag3PO4 (s)
-----
Ag+ (aq)
0
3x
PO43- (aq)
0.050
x + 0.050
Ksp = (3x)3 (x + 0.050) = 2.8 x 10-18
Assume x << 0.050, then
Ksp = (3x)3 (0.050) = 2.8 x 10-18
(27x3)(0.050) = 2.8 x 10-18 = 1.35x3
x3 = 2.8 x 10-18/1.35; x = 1.3 x 10-6 M = [PO43- (aq)] = solubility since 1 mol PO43- per mol Ag3PO4
The assumption was good since 1.3 x 10-6 << 0.050.
4. The concentration of Fe3+ in aqueous samples can be determined by reacting it with hexacyanoruthenate(II)
ion, Ru(CN)64-, to form a FeRu(CN)6- complex that absorbs strongly at 550 nm. The molar absorptivity of the
FeRu(CN)6- complex is 9.970 x 103 M-1 cm-1.
(a) What is the molarity of Fe3+ in a solution having an absorbance of 0.216 at 550 nm in a 1.00 cm cell? (10)
Aλ = ελ b c
c = A550 nm / (ε550 nm)(b)
c = (0.216) / (9.970 x 103 M-1 cm-1) (1.00 cm) = 2.17 x 10-5 M
(b) What is the percent transmittance of this solution? (6)
A = -log T
T = 10-0.216 = 0.608
%T = 60.8%
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5. A spectrophotometric method for phosphorus was used in which the P reacts with molybdenum(VI) to form a
molybdophosphate complex, which is then reduced to give an intensely colored blue complex known as
molybdenum blue. This complex has a maximum absorbance at 650 nm. Using a 1.00 cm cell, a series of P
standards ranging in concentrations between 0 and 75 µM was analyzed, and the following calibration equation
was obtained:
absorbance = 0.00671 (conc.) + 0.00241,
where the concentration is in µM.
(a) A sample of rinse water from a detergent manufacturing process was analyzed for dissolved P using the above
method. The sample absorbance was 0.433 and the blank gave an absorbance of 0.002. What is the
concentration of P in the sample in units of µM? (10)
Corrected absorbance = 0.433 – 0.002 = 0.431
Conc. = (0.431 – 0.00241) / 0.00671 = 63.9 µM
(b) Using the slope of the calibration curve, determine the molar absorptivity (in M-1 cm-1) of the molybdenum
blue complex. (7)
A = εb c
Abs = m (conc.) + b
Ignoring the y-intercept (b), m = εb, where b is the pathlength in Beer’s Law
m = (0.00671 µM-1)(106 µM/M) = 6710 M-1
ε = m/b = 6710 M-1/1.00 cm = 6.71 x 103 M-1 cm-1
(c) Suppose a second unknown sample gave an absorbance of 0.717. Should you trust the concentration
you obtain for this sample? Why or why not? (6)
Corrected absorbance = 0.717 – 0.002 = 0.715
Conc. = (0.715 – 0.002410 / 0.00671 = 106 µM
This concentration exceeds the range of the standard curve. There is no assurance that the standard curve is
linear beyond this range, so the value can not be trusted.
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6. A Volhard titration employs both precipitation and a back-titration to determine anions that precipitate with
Ag+ (e.g., Cl-, Br-, SCN-, I-). Suppose a 10.00 mL sample of liquid medical waste is analyzed for I- by initially
treating it with 10.00 mL of 0.1095 M AgNO3. The excess Ag+ was then back-titrated with 2.652 mL of 0.09884
M KSCN (potassium thiocyanate). What is the molarity of I- in the waste solution? The analysis reactions are:
Forward titration:
Back titration:
Ag+ (aq) + I- (aq)  AgI (s)
Ag+ (aq) + SCN- (aq)  AgSCN (s)
(15)
Initial mols of Ag+ = (10.00 x 10-3 L)(0.1095 mol Ag+/L) = 1.095 x 10-3 mol Ag+
Excess mols of Ag+ = (2.652 x 10-3 L)(0.09884 mol SCN-/L)(1 mol Ag+/1 mol SCN-) = 2.621 x 10-4 mol Ag+
Mols of Ag+ reacted with I- = 1.095 x 10-3 – 2.621 x 10-4 =
= 1.095 x 10-3 – 0.2621 x 10-3 = 0.833 x 10-3 mol Ag+
0.833 x 10-3 mol Ag+/(1mol I-/1 mol Ag+)(1/10.00 x 10-3 L) = 8.33 x 10-2 mol I-/L
7. A 0.4336-g solid sample containing malonic acid (CH2(CO2H)2); FW = 104.06) was dissolved completely in
deionized water, indicator was added, and the solution was titrated to the endpoint with 19.54 mL of 0.1094 M
NaOH, according to the following reaction:
CH2(CO2H)2 + 2 NaOH  CH2(CO2-)2 + 2 Na+ + 2 H2O.
Calculate the weight percent malonic acid in the solid sample. (12)
(19.54 x 10-3 L)(0.1094 mol NaOH/L)(1 mol m.a./2 mol NaOH)(104.06 g m.a./1 mol m.a.)(100/0.4336 g) =
25.56%
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Useful information
A = bc
T = P/Po
A = log (Po/P) = - log T
A = 2 - log %T
E = hc/
E = h
c = 
y = mx + b
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