CHM 235 Quantitative Analysis
Dr. S.A. Skrabal
Final Exam (Exam 4)
28 April 2004
NAME:___________________________________
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. Express all pH values to two decimal places for
consistency. Justify any assumptions you make when solving problems. Note that the temperatures are assumed to
be 25 C unless otherwise noted. Note also that all solutions are aqueous. Useful information is located on the last
pages of the exam, including the structures and Ka values for the weak acids referred to in the exam (Table 1) and
activity coefficients (Table 3).
1. Aspartic acid (H3A+) is a triprotic amino acid whose structure is shown below. The pKa values for each
of the three exchangeable protons (in bold) are also shown.
NH3+
|
CHCH2CO2H
|
CO2H
pKa3 =10.002
pKa2 = 3.900
pKa1 = 1.990
(a) Show the structure of the form HA-. (3)
NH3+
|
CHCH2CO2|
CO2-
(b) Calculate the pH of a 0.200 M solution of HA-. (6)
2nd intermediate form of triprotic acid:
[ H ] 
K a 2 K a3 F  K a 2 K w
K a2  F

(10 3.900 )(10 10.002 )(0.200)  (10 3.900 )(10 14.000 )
10 3.900  0.200
pH = -log(1.12 x 10-7) = 6.95
or pH  (pKa2 + pKa3) / 2 = (3.900 + 10.002) / 2 = 6.95
 1.12 x10 7 M
1. (c) Calculate the pH of a 0.250 M solution of the fully acidic form, H3A+ (6).
Ka1 = x2/(F – x) = 10-1.990 = 1.02 x 10-2 = x2/ (0.250 – x)
x2 + 1.02 x 10-2x – 2.55 x 10-3 = 0
x = [H3O+] = 4.57 x 10-2 M
pH = -log(4.57 x 10-2) = 1.34
or assume x << F
Ka1 = x2/F = 1.02 x 10-2 = x2/0.250
x = 5.05 x 10-2 M; pH = 1.30
2. A solution contains 0.095 M disodium oxalate (Na2C2O4) and 0.120 M potassium hydrogen oxalate
(KHC2O4). What is the pH of 100.0 mL of this solution? (6)
Buffer in diprotic system: A2- (ox2-) and HA- (Hox-)
pH = pKa2 + log([ox2-]/[Hox-])
pKa2 = -log(5.42 x 10-5) = 4.27
pH = 4.27 + log(0.095/0.120) = 4.17
2
3. What is the pH of 100.0 mL of a 0.150 M solution of the fully basic form (i.e., A2-) of a diprotic weak
acid whose pKa1 = 2.451 and pKa2 = 6.883? (8)
Kb1 = Kw/Ka2 = 1.00 x 10-14/1.31 x 10-7 = 7.63 x 10-8 = x2/(F – x) = x2/(0.150 – x)
x2 + 7.63 x 10-8x – 1.15 x 10-8 = 0
x = [OH-] = 1.07 x 10-4 M
pOH = -log(1.07 x 10-4) = 3.97
pH = 14.00 – 3.97 = 10.03
Or assume Kb1 = x2/F, x << F, to get same answer.
4. A solution is prepared by adding 2.00 mL of 7.00 M HCl to 150.0 mL of 0.140 M NaHSO3 (sodium
hydrogen sulfite). What is the pH of the solution after mixing? (8)
SA reacts with WB completely:
H3O+ + HA-  H2A + H2O
H3O+ + HSO3-  H2SO3 + H2O
Initial mols HSO3- = (0.1500 L)(0.140 mol/L) = 2.10 x 10-2 mol HSO3Initial mols H3O+ = (2.00 x 10-3 L)(7.00 mol/L) = 1.40 x 10-2 mol H3O+
Final mols HSO3- = 2.10 x 10-2 – 1.40 x 10-2 = 7.0 x 10-3 mol HSO3Final mols H2SO3 = 1.40 x 10-2 mol H2SO3
Buffer formed:
pH = pKa1 + log(mols HSO3-/mols H2SO3)
pKa1 = -log(1.23 x 10-2) = 1.91
pH = 1.91 + log(7.0 x 10-3/1.40 x 10-2) = 1.61
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5. An aqueous solution is prepared for bacterial culture research. It contains 0.10 M NaCl, 0.0050 M
NaHCO3 (sodium bicarbonate), and 0.0050 M K2SO4 (potassium sulfate); all of these salts are fully
dissociated. Calculate the ionic strength (in M) of the solution. Ignore the insignificant association and
dissociation of the HCO3- and the insignificant association of the SO42-. (10)
μ = 0.5[0.10 M)(+1)2 + (0.10 M)(-1)2 + (0.0050 M)(+1)2 + (0.0050 M)(-1)2 + (2)(0.0050 M)(+1)2 +
(0.0050 M)(-2)2] = 0.12 M
6. Using activities, calculate the solubility (in M) of lanthanum fluoride (LaF3; Ksp = 2.0. x 10-19) in
0.040 M NaNO3. Interpolate activity coefficients as necessary. You may neglect the contribution of La3+
and F- to the ionic strength. The dissolution reaction of this solid is:
(10)
LaF3 (s)  La3+(aq) + 3 F-(aq) .
μ = 0.5[(0.040)(+1)2 + (0.040)(-1)2] = 0.040 M
Ksp = 2.0 x 10-19 = γLa3+[La3+] γF-3 [F-]3 = γLa3+[x] γF-3 [3x]3 , where x = [La3+]
Interpolate activity coefficients:
γLa3+: (0.05 – 0.04)/(0.05 – 0.01) = 0.25 = (0.245 – y)/(0.245 – 0.445) = (0.245 – y)/-0.20
-0.050 = 0.245 – y; y = 0.295
γF-: (0.05 – 0.04)/(0.05 – 0.01) = 0.25 = (0.81 – y)/(0.81 – 0.900) = (0.81 – y)/-0.090
-0.0225 = 0.81 – y; y = 0.832
2.0 x 10-19 = (0.295)(x) (0.832)3 (3x)3 = 4.59 x4
x4 = 2.0 x 10-19/4.59 = 4.36 x 10-20; x = [La3+] = solubility since 1 mol La3+ per mol LaF3
solubility = 1.4 x 10-5 M
4
7. A buffer was prepared by dissolving some disodium hydrogen phosphate (Na2HPO4) in deionized H2O
and adding some strong acid to adjust the solution to a pH of 7.50.
(a) What is the dominant conjugate acid-base pair in the solution? (5)
pH = pKax + log(CB/CA) in H3PO4 system
pKa1 = -log(7.11 x 10-3) = 2.15
pKa2 = -log(6.32 x 10-8) = 7.20
pKa3 = -log(7.1 x 10-13) = 12.15
pKa2 is closest to pH:
7.50 = 7.20 + log([HPO42-]/[H2PO4-])
So HPO42-/H2PO4- is dominant pair.
(b) What is the ratio of the base/acid concentrations in this solution. You do not need to calculate the
concentrations, just the ratio. (4)
7.50 = 7.20 + log([HPO42-]/[H2PO4-])
0.30 = log([HPO42-]/[H2PO4-])
100.30 = [HPO42-]/[H2PO4-] = 2.0
8. A solution contains 0.400 M NaH2AsO4( sodium dihydrogen arsenate) and 0.300 M Na2HAsO4 (disodium
hydrogen arsenate).
(a) What is the pH of this solution? (5)
Buffer in triprotic system: H2A- and HA2pH = pKa2 + log([HAsO42-]/[H2AsO4-])
pKa2 = -log(1.10 x 10-2) = 6.96
pH = 6.96 + log(0.300/0.400) = 6.84
5
8. (b) To 250.0 mL of the buffer in 8(a) is added 5.00 mL of 10.0 M NaOH. What is the pH of the
solution after mixing? (7)
SB reacts with WA completely:
OH- + H2AsO4-  HAsO4- + H2O
Initial mols HAsO42- = (0.2500 L)(0.300 mol/L) = 7.50 x 10-2 mol HAsO42Initial mols H2AsO4- = (0.2500 L)(0.400 mol/L) = 1.00 x 10-1 mol H2AsO4Initial mols OH- = (5.00 x 10-3 L)(10.0 mol/L) = 5.00 x 10-2 mol OHFinal mols HAsO4- = 7.50 x 10-2 + 5.00 x 10-2 = 1.25 x 10-1 mols HAsO42Final mols H2AsO4- = 1.00 x 10-1 – 5.00 x 10-2 = 5.00 x 10-2 mol H2AsO4pH = pKa2 + log(mol HAsO42-/mol H2AsO4-)
pH = 6.96 + log(1.25 x 10-1/5.00 x 10-2) = 7.36
9. How many grams of potassium hydrogen phthalate (FW = 204.23) must be added to 500.0 mL of
0.200 M phthalic acid to obtain a buffer with pH = 3.00? (9)
The buffer system is HPhth- and H2Phth.
pH = pKa1 + log(mol HPhth-/mol H2Phth)
pKa1 = -log(1.12 x 10-3) = 2.95
mols of H2Phth = (0.5000 L)(0.200 mol/L) = 1.00 x 10-1 mol
3.00 = 2.95 + log(x/1.00 x 10-1)
0.05 = log(x/1.00 x 10-1)
100.05 = x/1.00 x 10-1 = 1.12; x = mol HPhth- = 0.112 mol HPhth- needed.
(0.112 mol HPhth-)(1 mol KHPhth/1 mol HPhth-)(204.23 g KHPhth/1 mol KHPhth) = 22.9 g KHPhth
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10. A solution for a quantitative analysis lab is prepared by pipetting 25.00 mL of a 53.0 g L-1 solution of
potassium acetate (CH3CO2K; FW = 98.15) into a 200.0 mL volumetric flask and diluting exactly to volume.
(a) What is the molarity of CH3CO2K in the diluted solution? (4)
CconcVconc = CdilVdil
Cconc = (53.0 g CH3CO2K/L)(1 mol CH3CO2K/98.15 g CH3CO2K) = 0.540 mol/L
Cdil = (CconcVconc)/Vdil = (0.540 mol/L)(25.00 mL)/200.0 mL = 6.75 x 10-2 mol/L
(b) What is the pH of the diluted solution? (5)
WB
Kb = Kw/Ka = 1.00 x 10-14/1.75 x 10-5 = 5.71 x 10-10 = x2/(6.75 x 10-2 – x)
x2 + 5.71 x 10-10x – 3.85 x 10-11 = 0
x = [OH-] = 6.20 x 10-6 M; pOH = 5.21
pH = 14.00 – 5.21 = 8.79
Or assume x << F to obtain same answer.
11. Give two examples of an amphiprotic species. (4)
Can function as either acid or base, depending on the context.
Examples:
Hydrogen phthalate ion (HPhth-)
Hydrogen phosphate ion (HPO42-)
Dihydrogen phosphate ion (H2PO4-)
Water (H2O)
Numerous other possibilities
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Useful information
Ionic strength,  = 0.5  ci zi2
Quadratic formula:
x
 b  b 2  4ac
2a
Table 1. Dissociation constants of acids at 25 C (Exchangeable protons in bold)
Name
Acetic acid
Arsenic acid
Aspartic acid
Sulfurous acid
(hydrogen sulfite)
Formula
CH3CO2H
O
||
HO-As-OH
|
OH
NH3+
|
CHCH2CO2H
|
CO2H
O
||
HOSOH
Phthalic acid
Oxalic acid
Phosphoric acid
HO2CCO2H
H3PO4
Ka or Ka1
1.75 x 10-5
5.8 x 10-3
Ka2
----1.10 x 10-7
Ka3
----3.2 x 10-12
1.02 x 10-2
1.26 x 10-4
9.95 x 10-11
1.23 x 10-2
6.6 x 10-8
-----
1.12 x 10-3
3.90 x 10-6
-----
5.60 x 10-2
7.11 x 10-3
5.42 x 10-5
6.32 x 10-8
----7.1 x 10-13
Table 2. Dissociation constant of H2O
T ( C)
0
5
25
100
Kw
1.14 x 10-15
1.85 x 10-15
1.01 x 10-14
5.45 x 10-13
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Table 3. Activity coefficients for aqueous solutions at 25 C
Ion
H
OH-, F-, SCNCl-, Br-, I-, K+
Li+
CH3CO2Ag+, NH4+
Ca2+, Cu2+, Co2+
Mg2+, Be2+
SO42La3+
+
µ = 0.001
0.967
0.964
0.964
0.965
0.964
0.964
0.870
0.872
0.867
0.738
0.005
0.933
0.926
0.925
0.929
0.928
0.924
0.749
0.755
0.740
0.540
0.01
0.914
0.900
0.899
0.907
0.902
0.898
0.675
0.69
0.660
0.445
0.05
0.86
0.81
0.805
0.835
0.82
0.80
0.485
0.52
0.445
0.245
0.1 M
0.83
0.76
0.755
0.80
0.775
0.75
0.405
0.45
0.355
0.180
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