CHM 235 Quantitative Analysis
Dr. S.A. Skrabal
Exam III
16 April 2002
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. Note
that the temperatures are assumed to be normal room temperature unless otherwise noted. Useful
information, including dissociation constants and formulae, is located on the last page of the exam.
All solutions are aqueous. Good luck!
1. Dr. Wattnott pipets 1.00 mL of 1.5 M HCl and 2.00 mL of 2.5 M HNO3 into 997 mL of
deionized H2O. What will be the pH of the solution? (10)
Both HCl and HNO3 are strong acids:
{(1.00 x 10-3 L)(1.5 mol H+/L) + (2.00 x 10-3 L)(2.5 mol H+/L)} / 1.00 L = 6.5 x 10-3 mol H+/L
pH = -log [H+] = -log(6.5 x 10-3) = 2.19
2) What is the pH of 1.00 mL of pure H2O at 50C? (10)
Kw = [H+] [OH-] = 5.47 x 10-14 (at 50C)
[H+] = (5.47 x 10-14)0.5 = 2.34 x 10-7 M
pH = -log(2.34 x 10-7) = 6.63
3) A 0.0175 M solution of a weak base is 0.251% associated (i.e., = 0.00251). What is the pH
of this solution? (10)
 = x/F = [OH-]/F = 0.00251 = [OH-]/0.0175 M
[OH-] = (0.0175)(0.00251) = 4.39 x 10-5
pOH = -log(4.39 x 10-5) = 4.36
pH = 14.00 - 4.36 = 9.64
4. There is trouble at the chemical factory. The boss, Dr. Heyrovsky, has prepared a 20.0 L
batch of very pure 0.00200 M aqueous potassium hydroxide, KOH. He suspects that his clumsy
lab technician, Guy Laporche, has spilled 1.0 L of 0.050 M CH3CO2H (acetic acid) into the
batch. If in fact the acetic acid was spilled into the KOH, what pH would you expect the solution
to have? (12)
Strong base (OH-) reacts with weak acid (CH3CO2H).
Initial mol OH- = (20.0 L) (0.00200 mol OH-/L) = 0.0400 mol OH- (from KOH)
Initial mol CH3CO2H = (1.0 L) (0.050 mol/L) = 0.050 mol CH3CO2H
We are neglecting the small amount of dissociation and association of the weak acid and weak
base, respectively (a good assumption for most buffers).
Initial
Reaction
Equilibrium
OH+
0.0400 mol
-0.0400
neglect
CH3CO2H
-->
0.050 mol
(0.050 - 0.0400)
0.010
CH3CO2neglect
+0.0400
0.0400
+
H2O
Buffer( use Henderson-Hasselbalch eqn.):
I used moles of WA and WB since it’s the same volume for each:
pH = pKa + log(mol CH3CO2-/mol CH3CO2H) = 4.76 + log(0.0400/0.010) = 5.36
5. Dr. Heyrovsky then prepares 5.0 L of a 0.0950 M aqueous solution of ammonium chloride,
NH4Cl. What is the pH of this solution? (12)
NH4Cl  NH4+ + ClNH4+ is a weak acid:
NH4+ + H2O  NH3 + H3O+
Ka = x2/(F – x) = 5.70 x 10-10 = x2/(0.0950 –x)
(0.0950 – x) (5.70 x 10-10) = x2
5.41 x 10-11 – 5.70 x 10-10x = x2
x2 + 5.70 x 10-10x – 5.41 x 10-11 = 0
Use quadratic formula to get x = [H3O+] = 7.35 x 10-6 M
pH = -log(7.35 x 10-6) = 5.13
Or use approximation Ka = x2/F = 5.70 x 10-10 = x2/0.0950
To obtain x = 7.36 x 10-6 M; pH = 5.13
6. Suppose a student prepares a buffer consisting of valeric acid (CH3CH2CH2CH2CO2H) and its
conjugate base, valerate ion (CH3CH2CH2CH2CO2-). If the pH of the student’s buffer is 5.01,
what must be the ratio between the concentrations of valerate ion and valeric acid in this
solution? (12)
pH = pKa + log([valerate]/[valeric acid])
pKa = -log(1.44 x 10-5) = 4.84
5.01 = 4.84 + log([valerate]/[valeric acid])
0.17 = log([valerate]/[valeric acid])
100.17 = [valerate]/[valeric acid] = 1.48
7. A volume of 5.0 mL of 2.0 M HCl is added to 1.0 L of a buffer solution containing 0.050 M
formic acid (HCO2H) and 0.080 M sodium formate (HCO2Na) . What is the pH of the resulting
solution? (12)
SA reacts completely with WB in buffer.
Initial mol H+ = (5.0 x 10-3 L)(2.0 mol H+/L) = 0.010 mol H+ (from HCl)
Initial mol HCO2H = (1.0 L)(0.050 mol HCO2H/L) = 0.050 mol HCO2H
Initial mol HCO2- = (1.0 L)(0.080 mol HCO2-/L) = 0.080 mol HCO2We are neglecting the small amount of dissociation and association of the weak acid and weak
base, respectively (a good assumption for most buffers).
Initial
Reaction
Equilibrium
H+
+
0.010 mol
-0.010
?
HCO2-->
0.080 mol
(0.080 - 0.010)
0.070
HCO2H
0.050 mol
(0.050 + 0.010)
0.060
Still a buffer!
I used moles of WA and WB since it’s the same volume for both:
pH = pKa + log(mol HCO2-/mol HCO2H) = 3.74 + log(0.070/0.060) = 3.81
8. A volume of 28 mL of 0.045 M hydrofluoric acid solution (HF) is mixed with 35 mL of 0.033
M sodium fluoride solution (NaF). What is the pH of this solution? (12)
HF and F- are a conjugate acid-base pair.
Find concentrations or moles of each, then use H-H eqn. (I used moles).
pKa of HF = -log(6.8 x 10-4) = 3.17
mol of F- (from NaF  Na+ + F-) = (35 x 10-3 L)(0.033 mol F-/L) = 1.16 x 10-3 mol Fmol of HF = (28 x 10-3 L)(0.045 mol/L) = 1.26 x 10-3 mol HF
pH = 3.17 + log(1.16 x 10-3/1.26 x 10-3) = 3.13
9. Given the choices of weak acids in the table at the end of the exam, what conjugate
acid/conjugate base pair would you use to prepare a buffer of pH 8.9? (10)
We are looking for pKa of the weak acid in the buffer to be  to the desired pH of the buffer.
Calculate pKa’s for the weak acids in table at end of test: HF (3.17); benzoic acid (4.20); formic
acid (3.74); NH4+ (9.24); valeric acid (4.84); acetic acid (4.76).
NH4+/NH3 would the conjugate acid-base pair to use since pKa of NH4+ is closest to desired
buffer pH.
Useful information
x
 b  b 2  4ac
2a
Weak acids (Exchangeable protons in bold)
Name
Hydrofluoric acid
Benzoic acid
Formic acid
Ammonium ion
Valeric acid
Acetic acid
Formula
HF
C6H5CO2H
HCO2H
NH4+
CH3CH2CH2CH2CO2H
CH3CO2H
Ka
6.8 x 10-4
6.28 x 10-5
1.80 x 10-4
5.70 x 10-10
1.44 x 10-5
1.75 x 10-5
Dissociation constant of H2O
T ( C)
0
5
25
50
100
Kw
1.14 x 10-15
1.85 x 10-15
1.01 x 10-14
5.47 x 10-14
5.45 x 10-13