CHM 235 Quantitative Analysis
Dr. S.A. Skrabal
Exam III
25 November 2003
NAME: SOLUTIONS
Instructions: Read each question carefully before answering. Show all work on questions requiring
calculations to receive credit. For consistency, express all pH and pOH values to two decimal places.
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. Good luck!
1. Sodium benzoate (C6H5CO2Na) is a common ingredient in many shampoos. What are the pH and
pOH of 250 mL of a 0.350 M aqueous solution of sodium benzoate? (13)
Weak base:
Kb = Kw/Ka = 1.00 x 10-14/6.28 x 10-5 = 1.59 x 10-10 = x2 / (F – x) = x2 / (0.350 – x)
Assume x << 0.350, then Kb = 1.59 x 10-10 = x2 / F = x2 / 0.350
x = [OH-] = 7.46 x 10-6 M
pOH = -log(7.46 x 10-6) = 5.13
pH = 14.00 – 5.13 = 8.87
Since x is << 0.350, the assumption was good.
2. If the pH of a 0.245 M solution of a weak acid is 3.22, what are (a) the fraction of dissociation () and
(b) the Ka of this acid? (13)
 = x/F
x = [H3O+] = 10-3.22 = 6.03 x 10-4 M
Ka = x2/(F – x)
 = 6.03 x 10-4/0.245 = 2.46 x 10-3 or 0.246%
Ka = (6.03 x 10-4)2 / (0.245 – 6.03 x 10-4) = 1.49 x 10-6
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3. Calculate the pH of a 8.97 x 10-2 M solution of phenylacetic acid. (12)
Weak acid:
Ka = x2 / (F – x) = 4.90 x 10-5 = x2 / (8.97 x 10-2 – x)
4.395 x 10-6 – 4.90 x 10-5x = x2
x2 + 4.90 x 10-5x – 4.395 x 10-6 = 0
Use quadratic eqn. to obtain x = 2.07 x 10 -3 M = [H3O+]
pH = -log(2.07 x 10-3) = 2.68
Or assume Ka = x2/F to get x = 2.10 x 10-3 M; pH = 2.68
4. Calculate the pH and pOH of a 0.098 M solution of lithium hydroxide (LiOH). (13)
Strong base:
LiOH  Li+ + OH[OH-] = 0.098 M
pOH = -log(0.098) = 1.01
pH = 14.00 – 1.01 = 12.99
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5. An aqueous solution is prepared by dissolving 35.0 g of potassium formate (HCO 2K) and 11.0 mL of
5.00 M HCl in a total volume of 500.0 mL. What is the pH of this solution? (15)
Strong acid reacts with weak base completely: H 3O+ + HCO2-  HCO2H + H2O
Initial mols of WB = (35.0 g HCO2K)(1mol HCO2K/84.12 g HCO2K) = 0.416 mol HCO2K or HCO2Initial mols of H3O+ = (11.0 x 10-3 L)(5.00 mol H3O+/L) = 5.50 x 10-2 mol H3O+
Final mols of WB = 0.416 – 5.50 x 10-2 = 0.361 mol HCO2Final mols of WA = 5.50 x 10-2 mol HCO2H
Buffer: pKa = pH + log(mol HCO2-/mol HCO2H)
pKa = -log(1.80 x 10-4) = 3.74
pH = 3.74 + log(0.361/5.50 x 10-2) = 4.56
6. How many grams of sodium acetate (CH3CO2Na; FW = 82.03) must be added to 300.0 mL of
0.75 M acetic acid (CH3CO2H) to give a buffer of pH 4.90? (15)
Buffer:
pH = pKa + log (mol CH3CO2-/mol CH3CO2H)
pKa = -log (1.75 x 10-5) = 4.76
mol CH3CO2H = (300.0 x 10-3 L)(0.75 mol CH3CO2H/L) = 0.225 mol CH3CO2H
4.90 = 4.76 + log(mol CH3CO2-/0.225); let x = mol CH3CO20.14 = log (x / 0.225)
100.14 = x / 0.225 = 1.38
x = 0.310 mol CH3CO2- needed
(0.310 mol CH3CO2-)(1 mol CH3CO2Na/1 mol CH3CO2-) (82.03 g CH3CO2Na/1 mol CH3CO2Na)
= 25.4 g of CH3CO2Na
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7. (a) A buffer is prepared by mixing 50.0 mL of 0.98 M NH 3 with 60.0 mL of 0.55 M NH4Cl. What would
be the pH of the solution after mixing? (12)
Buffer:
mol NH3 = (50.0 x 10-3 L)(0.98 mol/L) = 0.049 mol NH3
mol NH4+ = (60.0 x 10-3 L)(0.55 mol/L) = 0.033 mol NH4+
pKa = -log(5.70 x 10-10) = 9.24
pH = 9.24 + log(0.049/0.033) = 9.41
7. (b) What would be the pH of the buffer in 7(a) after it is mixed with 3.0 mL of 1.5 M NaOH? (7)
Strong base reacts with weak acid completely:
OH- + NH4+  NH3 + H2O
Initial mols NH3 = 0.049 mol
Initial mols NH4+ = 0.033 mol
Initial mols OH- = (3.0 x 10-3 L)(1.5 mol/L) = 4.5 x 10-3 mol
Final mols NH3 = 0.049 + 4.5 x 10-3 = 0.0535 mol NH3
Final mols NH4+ = 0.033 – 4.5 x 10-3 = 0.0285 mol NH4+
]
Still a buffer:
pH = 9.24 + log(0.0535/0.0285) = 9.51
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Useful information
Quadratic formula: x 
 b  b 2  4ac
2a
Monoprotic Acids and Bases
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-
-
Weak base: F = [BH+
] / ([A-] + [HA]) = x / F
+]
/ ([BH+] + [B]) = x / F
Table 1. Dissociation constants of acids at 25º C (Exchangeable protons in bold)
Name
Acetic acid
Ammonium ion
Benzoic acid
Formula
CH3CO2H
NH4+
Ka
1.75 x 10-5
5.70 x 10-10
6.28 x 10-5
--CO2H
Formic acid
HCO2H
Phenylacetic acid
1.80 x 10-4
4.90 x 10-5
--CH2CO2H
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
5
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