CHM 235 Quantitative Analysis
Dr. S.A. Skrabal
Exam III
19 April 2007
NAME:
SOLUTIONS
Instructions: Read each question carefully before answering. Show all work on questions requiring calculations
to receive credit. Express pH values to 2 decimal places for consistency. Circle or box in your final numerical
answer. The value of each question is given in parentheses after the question. Note that the temperatures are
assumed to be normal room temperature (24-25o C) unless otherwise noted. Useful information is located on the
last page of the exam. State and justify any assumptions you use to solve a problem. Good luck!
1. Sodium hypochlorite (NaOCl) is used as a disinfectant and bleaching agent for many different applications such as
drinking water and swimming pool chlorination, hospital disinfection, and household cleaning products. A typical
concentration of commercial NaOCl would be 0.40 M. Calculate the following quantities for 100.0 mL of 0.40 NaOCl:
(a) [H3O+]; (b) [OH-]; (c) pH; and (d) percent association, %α. (16)
Weak base; Kb = Kw/Ka = 1.00 x 10-14/3.0 x 10-8 = 3.3 x 10-7 = x2 / (F – x) = x2 / (0.40 – x)
x2 + 3.3 x 10-7x – 1.32 x 10-7 = 0
Use quadratic eqn. to obtain x = [OH-]
(b) x = 3.63 x 10-4 M = [OH-]
(c) pOH = -log(3.63 x 10-4) = 3.44; pH = 14.00 – pOH = 14.00 – 3.44 = 10.56 = pH
(a) [H3O+] = 10-10.56 M = 2.75 x 10-11 M
(d) %α = 100(x/F) = 100(3.63 x 10-4 M/0.40 M) = 0.091%
Or assume x << F to obtain same answers.
2. A buffer used in a manufacturing process is prepared by the following process. A 50.0 L volume of 1.20 M NH4Cl
(ammonium chloride) is reacted in a Teflon-lined tank with 3.75 L of 5.00 M NaOH which is slowly pumped into the
tank and mixed carefully. What is the pH of the solution in the tank after complete mixing? (12)
Strong base reacts with weak acid completely:
OH- + NH4+  NH3 + H2O
Initial mols OH- = (3.75 L)(5.00 mol OH-/L) = 18.75 mols OHInitial mols NH4+ (WA) = (50.0 L)(1.20 mol NH4+/L) = 60.0 mol NH4+ (neglecting dissociation)
Final mols NH3 (WB) = 18.75 mols NH3
Final mols NH4+ (WA) = 60.0 mols – 18.75 mols = 41.2 mol NH4+
Buffer formed:
pH = pKa + log(mols NH3/mols NH4+)
pKa = -log(5.70 x 10-10) = 9.24
pH = 9.24 + log(18.8 mol/41.2 mol) = 8.90
3. Consider two weak bases, B1 with a pKb of 7.65 and B2 with a pKb of 9.41. Which is the stronger acid, the conjugate
of B1 or the conjugate of B2? Be sure to show the calculations that support your answer. (12)
Calculate Ka of each conjugate:
B1: Kb = 10-7.65 = 2.24 x 10-8
Ka = Kw/Kb = 1.00 x 10-14 / 2.24 x x10-8 = 4.46 x 10-7; pKa = 6.35
B2: Kb = 10-9.41 = 3.89 x 10-10
Ka = Kw/Kb = 1.00 x 10-14 / 3.89 x 10-10 = 2.57 x 10-5; pKa = 4.59
The conjugate acid of B2 is the stronger acid because it has the larger Ka value (and the smaller pKa).
Remember, the weaker the conjugate base, the stronger the conjugate acid and vice versa.
4. An artist uses an acidic solution during a metal-working process to clean surfaces before a finish is applied. The
solution is prepared by mixing 500 mL of 0.150 M hydrochloric acid with 1.00 L of deionized water. What is the pH of
this solution after mixing? (12)
HCl is a strong acid. Calculate final concentration of HCl (= concentration of H 3O+) after dilution:
CconcVconc = CdilVdil
Cdil = CconcVconc/Vdil = (0.150 M)(0.500 L)/(1.50 L) = 0.0500 M
Completely dissociated, so pH = -log(0.0500) = 1.30
5. A buffer is prepared for a biochemical experiment. A mass of 27.5 g of TES (N-Tris(hydroxymethyl)- methyl-2amino-ethanesulfonic acid; FW = 229.25) is dissolved in 500.0 mL of deionized water, then 10.00 mL of 8.00 M KOH
2
is added. What is the pH of the solution after mixing? (12)
Strong base (KOH) reacts with weak acid (TES) completely. TES = HA
OH- + HA  A- + H2O
Initial mols TES (HA) = (27.5 g TES)(1 mol TES/229.25 g TES) = 0.120 mol HA (neglecting dissociation)
Initial mols OH- = (10.00 x 10-3 L)(8.00 mol OH-/L) = 8.00 x 10-2 mol OHFinal mols HA = 0.120 mol – 8.00 x 10-2 mol = 4.00 x 10-2 mol HA
Final mols A- = 8.00 x 10-2 mol ABuffer formed:
pH = pKa + log(mols A-/mols HA)
pKa = -log(2.82 x 10-8) = 7.55
pH = 7.55 + log(8.00x 10-2/4.00 x 10-2) = 7.85
6. What would be the pH of the solution formed after mixing 25.00 mL of 2.50 M HCl with 250.0 mL of a buffer
containing 0.25 M benzoic acid and 0.35 M potassium benzoate? (12)
SA (HCl) reacts with weak base (benzoate) to form more weak acid (benzoic acid).
H3O+ + HA  A- + H2O
Initial mols HA = (250.0 x 10-3 L)(0.25 mol/L) = 6.25 x 10-2 mol HA
Initial mols A- = (250.0 x 10-3 L)(0.35 mol/L) = 8.75 x 10-2 mol AInitial mols H3O+ = (25.00 x 10-3 L)(2.50 mol/L) = 6.25 x 10-2 mol H3O+
Final mols HA = 6.25 x 10-2 mol + 6.25 x 10-2 mol = 0.12 mol HA
Final mols A- = 8.75 x 10-2 mol – 6.25 x 10-2 mol = 2.5 x 10-2 mol AStill a buffer:
pH = pKa + log (mols benzoate/mols benzoic acid)
pKa = -log(6.28 x 10-5) = 4.20
pH = 4.20 + log(2.5 x 10-2/0.12) = 3.52
7. Chloroacetic acid (ClCH2CO2H) is used in the manufacture of detergents, drilling mud, herbicides, pharmaceuticals,
and other chemical reagents. Over 1 million lbs. are produced in the U.S. annually. What are the (a) pH and
(b) percent dissociation ( %α) of 400.0 mL of a 0.125 M solution of chloroacetic acid? (12)
3
Weak acid:
Ka = x2/(F – x) = 1.36 x 10-3 = x2/(0.125 – x) where x = [H3O+]
x2 + 1.36 x 10-3x – 1.70 x 10-4 = 0
Use quadratic formula to obtain x = [H3O+] = 1.31 x 10-2 M
pH = -log(1.31 x 10-2) = 1.88
%α = 100([H3O+]/F) = 100(x/F) = 100(1.31 x 10-2 M/0.125 M) = 10.5%
Assuming Ka = x2/F gives x = 1.30 x 10-2 M; pH = 1.89
%α = 10.4%
Good assumption since x (1.30 x 10-2 M) << F (0.125 M)
8. What mass of sodium glycolate (HOCH2CO2Na FW = 98.03) must be added to 400.0 mL of 1.50 M glycolic acid
(HOCH2CO2H) to produce a buffer that has a pH of 4.20? (12)
pH = pKa + log(mol glycolate/mol glycolic acid)
pKa = -log(1.48 x 10-4) = 3.83
4.20 = 3.83 + log(mols glycolate/mols glycolic acid)
0.37 = log (mols glycolate/mols glycolic acid)
100.37 = 2.34 = mols glycolate/mols glycolic acid
mols glycolic acid = (400.0 x 10-3 L)(1.50 mol/L) = 0.600 mol glycolic acid (neglecting dissociation)
mols glycolate needed = (2.34)(0.600 mol) = 1.40 mol glycolate
(1.40 mol glycolate)(1 mol sodium glycolate/1 mol glycolate)(98.03 g sodium glycolate/1 mol sodium glycolate)
= 137 g sodium glycolate
Useful information
Quadratic formula: x

 b  b 2  4ac
2a
Monoprotic Acids and Bases
4
Ka = [H3O+] [A-] / [HA]
Kb = [BH+] [OH-] / [B]
Ka = x2 / (F-x); x = [H3O+] = [A-]
Kb = x2 / (F-x); x = [OH-] = [BH+]
pH = pKa + log ([A-] / [HA])
Ka Kb = K w
Weak acid: F = [HA] + [A-]
For WA:  = [A-] / ([A-] + [HA]) = x / F
Weak base: F = [BH+] + [B]
For WB:  = [BH+] / ([BH+] + [B]) = x / F
Table 1. Weak acids (Exchangeable protons in bold)
Name
Formula
Ka
-5
Acetic acid
CH3CO2H
1.75 x 10
Ammonium ion
NH4+
5.70 x 10–10
Benzoic acid
C6H5CO2H
6.28 x 10-5
Chloroacetic acid
ClCH2CO2H
1.36 x 10-3
Glycolic acid
HOCH2CO2H
1.48 x 10-4
Hypochlorous acid
HOCl
3.0 x 10-8
N-Tris(hydroxymethyl))methyl2-amino-ethanesulfonic acid
(TES)
(HOCH2)3CN+H2CH2CH2SO3-
2.82 x 10-8
Table 2. Dissociation constant of H2O
T (C)
Kw
5
1.85 x 10-15
24
1.00 x 10-14
100
5.45 x 10-13
5