CHM 235 Quantitative Analysis
Spring 2007
Dr. S.A. Skrabal
SOLUTIONS TO PROBLEM SET 10
Monoprotic buffers and acid-base equilbria
10 April 2007
Please refer to Appendix G in your textbook for required values of Ka.
Please refer to Appendix G in your textbook for required values of Ka.
Note: I have expressed values of pH and pOH to only two decimal places for convenience.
1. (A) What is the pH of a 0.075 M NH3 (ammonia) solution? (B) What is the pH of a 0.025 M NH4Cl
(ammonium chloride) solution? (C) What is the pH of 50 mL of a solution consisting of 0.075 M NH3
and 0.025 M NH4Cl?
(A) Weak base: Kb = Kw/Ka = 1.00 x 10-14 / 5.69 x 10-10 = 1.76 x 10-5 = x2 / (F – x)= x2 / (0.075 – x)
Multiply out to obtain x2 + 1.76 x 10-5x – 1.32 x 10-6 = 0
Use quadratic formula to obtain x = [OH-] = 1.14 x 10-3 M
pOH = -log(1.14 x 10-3) = 2.94
pH = 14.00 – 2.94 = 11.06.
Or assume x << F to obtain x = 1.15 x 10-3 M and pH = 11.06. Good assumption since 1.14 x 10-3 <
0.075.
(B) Weak acid: Ka = x2 / (F – x) = 5.69 x 10-10 = x2 / (F – x) = x2 / 0.025 – x)
Multiply out to obtain x2 + 5.69 x 10-10x – 1.42 x 10-11 = 0
Use quadratic formula to obtain x = [H3O+] = 3.77 x 10-6 M
pH = -log(3.77 x 10-6) = 5.42
Or assume x << F to obtain x = 3.77 x 10-6 M and pH = 5.42. Good assumption since 3.77 x 10-6 <<
0.025.
(C) Buffer: pH = pKa + log([NH3]/[NH4+] = 9.24 + log(0.075/0.025) = 9.72
2. (A) What is the pH of 250 mL of 0.040 M benzoic acid? (B) A solution is prepared by adding 2.50
mL of 2.0 M KOH to 250 mL of 0.040 M benzoic acid. What is the pH of the solution after mixing?
(A) Weak acid: Ka = 6.28 x 10-5 = x2 / (F – x) = x2 / (0.040 – x)
Multiply out to obtain x2 + 6.28 x 10-5x – 2.51 x 10-6 = 0
Use quadratic formula to obtain x = [H3O+] = 1.55 x 10-3 M
pH = -log(1.55 x 10-3) = 2.81
Or assume x << F to obtain x = 1.58 x 10-3 M and pH = 2.80. Good assumption since 1.58 x 10-3 <<
0.040.
(B) OH- + C6H5CO2H  C6H5CO2- + H2O
Strong base reacts with weak acid completely.
Initial mol C6H5CO2H = (0.250 L)(0.040 mol/L) = 1.0 x 10-2 mol C6H5CO2H (neglecting dissociation)
Initial mol OH- = (2.50 x 10-3 L)(2.0 mol/L) = 5.0 x 10-3 mol OHInitial mol C6H5CO2- = 0 (neglecting dissociation)
Final mol C6H5CO2H = 1.0 x 10-2 – 5.0 x 10-3 = 5.0 x 10-3 mol
Final mol C6H5CO2- = 5.0 x 10-3 mol
Buffer: pH = 4.20 + log(5.0 x 10-3/5.0 x 10-3) = 4.20
3. (A) What is the pH of 200 mL of a 0.0250 M sodium hypochlorite (NaOCl) solution? (This is the
active ingredient in most bleaches.) (B) Suppose 10.0 mL of 0.14 M HNO3 is added to this solution.
What would be the pH? (C) The solution in B is then diluted to a total volume of 500 mL. What would
be the pH?
(A) Weak base: Kb = Kw/Ka = 1.00 x 10-14/3.0 x 10-8 = 3.3 x 10-7 = x2/(F – x) = x2/(0.0250 – x)
Multiply out to get: x2 + 3.3 x 10-7x – 8.25 x 10-9 = 0
x = [OH-] = 9.07 x 10-5 M; pOH = 4.04; pH = 9.96. Or assume Kb = x2 / F to get same pH.
(B) Add (10.0 x 10-3 L)(0.14 mol H3O+/L) = 1.4 x 10-3 mol H3O+ (HNO3 is a strong acid.)
Strong acid reacts with weak base completely. pKa of HOCl = 7.53.
Initial mol OCl- = (200 x 10-3 L)(0.0250 mol/L) = 5.00 x 10-3 mol (neglecting association)
Initial mol HOCl = 0 (neglecting association)
Initial mol H3O+ = 1.4 x 10-3 mol
Final mol OCl- = 5.00 x 10-3 –1.4 x 10-3 = 3.6 x 10-3 mol
Final mol HOCl = 1.4 x 10-3 mol
Buffer: pH = 7.53 + log(3.6 x 10-3/1.4 x 10-3) = 7.94
(C) Moderate dilution will have no effect on pH because volume cancels out in numerator and
denominator of term after the log in H-H equation.
4. (A) What is the pH of 250.0 mL of 1.00 M hydroxyacetic acid (also known as glycolic acid)? (B)
What volume of 2.00 M NaOH must be added to 250.0 mL of 1.00 M hydroxyacetic acid to produce a
buffer solution of pH 4.00?
(A) Weak acid: Ka = 1.48 x 10-4 = x2 / (F – x) = x2 / (1.00 – x).
Multiply out to obtain x2 + 1.48 x 10-4x – 1.48 x 10-4 = 0.
Use quadratic formula to obtain x = [H3O+] = 1.21 x 10-2 M
pH = -log(1.21 x 10-2) = 1.92
Or assume x << F to obtain x = 1.22 x 10-2 M; pH = 1.91. Good assumption since x << F.
(B) Strong base reacts with weak acid completely.
Initial mol HA = (250.0 x 10-3 L)(1.00 mol/L) = 0.250 mol HA (neglecting dissociation)
Initial mol OH- (from NaOH) = x
Final mols of A- after reaction = x (neglecting dissociation)
Final mols of HA after reaction = 0.250 – x (neglecting dissociation)
A buffer has been formed, so use H-H equation:
pH = pKa + log(mol A-/mol HA); pKa of hydroxyacetic acid = 3.83.
4.00 = 3.83 + log (mol A-/mol HA); 4.00 – 3.83 = log (mol A-/mol HA)
0.17 = log(mol A-/mol HA); 100.17 = mol A-/mol HA = x/(0.250 – x) = 1.48.
Multiply out to get: 0.370 – 1.48x = x; 2.48x – 0.370 = 0; x = 0.149 mol of OH- needed.
(0.149 mol OH-)(L/2.00 mol OH-) = 7.45 x 10-2 L or 74.5 mL of NaOH needed.
Check work with H-H eqn.: pH = 3.83 + log(0.149/(0.250 – 0.149)) = 4.00
5. What mass of sodium formate (HCO2Na; FW = 68.01) must be added to 500.0 mL of 1.00 M formic
acid to produce a buffer solution that has a pH of 3.65?
Remember that in a buffer, initial concentrations of A- and HA do not change significantly from what you
start with. pKa of formic acid = 3.74.
Initial mols HA = (500.0 x 10-3 L)(1.00 mol/L) = 0.500 mol HA
Use H-H eqn.: 3.65 = 3.74 + log(mol A-/0.500 mol HA); -0.090 = log(mol A-/0.500 mol HA);
10-0.090 = mol A-/0.500 mol HA = 0.813 mol A- = (0.813)(0.500 mol) = 0.406mol A- needed.
(0.406mol HCO2Na)(68.01 g HCO2Na/mol HCO2Na) = 27.6 g HCO2Na needed.
Refer to Table 9-2 in your textbook (Harris, 2007) to answer the following two questions.
6. A mass of 100.0 g of ACES is dissolved in H2O and brought to a volume of 500 mL. (A) What is the
pH of this solution? (B) Suppose 12.0 mL of 12.0 M HCl is added to this solution. What would be the
pH?
(A) ACES is a weak acid; Ka = 10-6.85 = 1.41 x 10-7.
F = (100.0 g ACES)(1 mol ACES/182.200 g ACES)(1/0.500 L) = 1.01 mol/L
Ka = x2/(F – x) = 1.41 x 10-7 = x2/(1.01 – x). Assume x << 1.01, so Ka = 1.41 x 10-7 = x2/1.01.
x = [H3O+] = 3.77 x 10-4 M (assumption was good since 3.77 x 10-4 << 1.01); pH = 3.42
(B) Adding strong acid to weak acid. SA dominates H3O+ concentration.
(12.0 x 10-3 L)(12.0 mol H3O+/L) = 0.144 mol H3O+ (this dwarfs the H3O+ contributed by the WA)
(0.144 mol H3O+/512 x 10-3 L) = 2.81 x 10-1 mol H3O+/L; pH = 0.55
7. Which of the substances in Table 9-2 would be suitable for preparing buffers with the following pH
ranges: (A) 6.7  0.2; (B) 7.0  0.1; (C) 8.0  0.2?
Looking for pH  pKa. (A) ADA, BIS-TRIS propane, PIPES, ACES (B) MOPSO, imidazole HCl
(C) HEPPS, TRICINE, glycine amide hydrochloride, TRIS HCl
8. (A) What is the pH of 250 mL of a solution consisting of 0.15 M acetic acid and 0.26 M sodium
acetate (CH3CO2Na)? (B) Suppose 3.75 mL of 5.0 M NaOH was added to the solution in A. What
would be the pH? (C) Suppose another 3.75 mL of 5.0 M NaOH was added to the solution in B. What
would be the pH?
(A) Buffer:
mols CH3CO2H = (0.250 L)(0.15 mol/L) = 3.8 x 10-2 mol HA
mols CH3CO2- = (0.250 L)(0.26 mol/L) = 6.5 x 10-2 mol ApH = 4.76 + log(6.5 x 10-2/3.8 x 10-2) = 4.99
(B) Add strong base (OH-):
Initial mol HA = 3.8 x 10-2 mol HA (neglecting dissociation)
Initial mol A- = 6.5 x 10-2 mol A- (neglecting dissociation)
Initial mol OH- = (3.75 x 10-3 L)(5.0 mol/L) = 1.9 x 10-2 mol OHFinal mol HA = 3.8 x 10-2 – 1.9 x 10-2 = 1.9 x 10-2 mol HA
Final mol A- = 6.5 x 10-2 + 1.9 x 10-2 mol = 8.4 x 10-2 mol
Still a buffer: pH = 4.76 + log(8.4 x 10-2/1.9 x 10-2) = 5.41
(C) Add more strong base:
Initial mol HA = 1.9 x 10-2 mol HA
Initial mol A- = 8.4 x 10-2 mol AInitial mol OH- = (3.75 x 10-3 L)(5.0 mol/L) = 1.9 x 10-2 mol OHFinal mol HA = 1.9 x 10-2 – 1.9 x 10-2 = 0 mol HA (neglecting association)
Final mol A- = 8.4 x 10-2 + 1.9 x 10-2 = 1.0 x 10-1 mol A- (neglecting association)
This is no longer a buffer; it is a solution of A- (weak base, CH3CO2-).
Kb = Kw/Ka = 1.00 x 10-14/1.75 x 10-5 = 5.71 x 10-10 = x2/(F – x) = x2/(1.0 x 10-1 – x).
Multiply out to get x2 + 5.71 x 10-10x – 5.71 x 10-11 = 0; x = [OH-] = 7.56 x 10-6 M; pOH = 5.12;
pH = 8.88. Or assume Kb = x2/F to get same answer.