CHM 235 Quantitative Analysis
Dr. S.A. Skrabal
Exam II
4 November 2003
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. Calculate the volume of 0.04547 M KMnO4 that would be needed to reach the equivalence point in a
titration of 50.00 mL of 0.07545 M Fe2+ solution. The titration analysis reaction is:
(10)
MnO4- (aq) + 5Fe2+ (aq) + 8H+ (aq)  Mn2+ (aq) + 5Fe3+ (aq) + 4H2O (l).
(50.00 x 10-3 L)(0.07545 mol Fe2+/L)(1 mol MnO4-/5 mol Fe2+)(1 L/0.04547 mol MnO4-)(1000 mL/L) =
16.59 mL
2. Calculate the solubility (in mole/L) of Y(IO3)3 (yttrium iodate; Ksp = 7.1 x 10-11) in deionized H2O. The
dissolution reaction is:
(12)
Y(IO3)3 (s)  Y3+ (aq) + 3IO3- (aq).
Initial
Equilibrium
Y(IO3)3 (s)
-----
Y3+ (aq)
0
x
IO3- (aq)
0
3x
Ksp = [Y3+ (aq)] [IO3-]3 = 7.1 x 10-11
x (3x)3 = 7.1 x 10-11
27x4 = 7.1 x 10-11
x = 1.3 x 10-3 M = [Y3+] = solubility in mol solid per L since 1 mol Y3+ per mol Y(IO3)3
3. The calcium carbonate (CaCO3) in a commercial antacid tablet was determined by dissolving the tablet in
an excess quantity of HCl (reaction 1) then back-titrating the excess HCl with standardized NaOH (reaction
2).
CaCO3 (s) + 2HCl (aq)  CaCl2 (aq) + H2O (l) + CO2 (g)
(1)
HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)
(2)
Suppose one tablet of the antacid is completely dissolved in 20 mL of H2O and 5.00 mL of 6.00 M HCl. The
excess acid requires 33.73 mL of 0.5895 M NaOH to reach the phenolphthalein endpoint. How many
milligrams of CaCO3 are in the tablet? (15)
Initial mol HCl = (5.00 x 10-3 L HCl)(6.00 mol HCl/L) = 3.00 x 10-2 mol HCl
Excess mol HCl = (33.73 x 10-3 L NaOH)(0.5895 mol NaOH/L)(1 mol HCl/1 mol NaOH) = 1.99 x 10-2 mol
HCl
Mol HCl reacted with CaCO3 = 3.00 x 10-2 mol – 1.99 x 10-2 mol = 1.01 x 10-2 mol HCl
(1.01 x 10-2 mol HCl)(1 mol CaCO3/2 mol HCl)(100.09 g CaCO3/1 mol CaCO3)(1000 mg/g) =
505 mg CaCO3
4. Calculate the solubility (in mole/L) of Ag2CO3 (silver carbonate; Ksp = 8.2 x 10-12) in a solution of
0.11 M Na2CO3 (which dissociates completely into 2 Na+ + CO32-). The dissolution reaction is: (15)
Ag2CO3 (s)  2Ag+ (aq) + CO32- (aq).
Ksp = [Ag+ (aq)]2 [CO32- (aq)] = 8.2 x 10-12
Initial
Equilibrium
Ag2CO3 (s)
-----
Ag+ (aq)
0
2x
CO32- (aq)
0.11
0.11 + x
Ksp = (2x)2 (0.11 + x) = 8.2 x 10-12
Assume x << 0.11, then (2x)2 (0.11) = 8.2 x 10-12
0.44x2 = 8.2 x 10-12
x = 4.3 x 10-6 M = [CO32-] = solubility of Ag2CO3 since 1 mol of CO32- per mol Ag2CO3
Assumption was good since x = 4.3 x 10-6 << 0.11
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5. A spectrophotometric analysis for dissolved Fe2+ is carried out by reacting the Fe-containing
standards and samples in acid and 1,10-phenanthroline, forming a complex which absorbs strongly at
510 nm. A series of standards is made with Fe2+ concentrations from 0 to 4.00 mg L-1. The resulting
calibration equation (from measurements in a 1.00 cm cell) is:
Absorbance = 0.2152 (conc. in mg L-1) - 2.16 x 10-3.
The blank absorbance was 0.002. A 100.0-mL sample prepared the same way as the standards gave
an absorbance of 0.393.
(A) What is the concentration (in mg L-1) of Fe2+ in the sample? (10)
Corrected absorbance = 0.393 – 0.002 = 0.391
Rearrange calibration equation to solve for conc.:
conc. 
0.391  (2.16 x 10 3 )
0.2152
 1.83 mg / L
(B) How many milligrams of Fe2+ are in the 100.0 mL sample? (Note: There is a 1:1 molar relationship
between Fe2+ and the Fe2+-1,10-phenanthroline complex.) (6)
(1.83 mg/L)(0.1000 L) = 0.183 mg Fe2+
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6. Suppose that you are performing a spectrophotometric analysis for some chromophoric analyte
which has a formula weight of 203.54 and a molar absorptivity of 1.45 x 103 M-1 cm-1 at an analysis
wavelength of 671 nm.
(A) If a solution of this analyte had a transmissivity of 25.8% in a 5.00 cm cell at 671 nm, what would be
the absorbance of the solution? (10)
A = -log T = -log (0.258) = 0.588 or
A = 2 - log(%T) = 2 – log(25.8) = 0.588
(B) Suppose another solution of this analyte is made up to a concentration of 10.0 mg L-1. Calculate the
absorbance of this solution if it is measured at 671 nm in a 5.00 cm cell? (10)
A671 nm = 671 nmbc
Convert concentration from mg/L to M:
(10.0 mg/L)(g/1000 mg)(1 mol/203.54 g) = 4.91 x 10-5 mol/L
A671 nm = (1.45 x 103 M-1 cm-1)(5.00 cm)(4.91 x 10-5 M) = 0.356
7. What concentration of Cu+ is required to start the precipitation of CuI (cuprous iodide;
Ksp = 5.1 x 10-12) in a solution of 0.0012 M potassium iodide (KI)? The reaction is:
(12)
Cu+ (aq) + I- (aq)  CuI (s).
Ksp = [Cu+ (aq)] [I- (aq)] = 5.1 x 10-12
[Cu+ (aq)] (0.0012) = 5.1 x 10-12
[Cu+ (aq)] = 4.2 x 10-9 M
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Useful Information
A = bc
T = P/P0
A = log10 (P0/P) = - log T
A = 2 - log %T
E = hc/
E = h
c = 
y = mx + b
E = mc2
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