Extra acid-base problems
1. The following is a list of weak acids and their Ka values:
HClO
hypochlorous acid
3.5×10−8
HCN
nitrous acid
4.5×10−4
H2S
HNO2
hydrocyanic acid 4.0×10−10
hydrogen sulfide 1.1×10−7
a. Which acid given above is the strongest? Explain your choice.
b. Write the Ka expression for the strongest acid.
c. Which acid has the strongest conjugate base? Explain, and write the Kb expression for that base.
d. Which side of the reaction, reactants or products, will be favored if: NaClO (aq) is combined with HCN (aq)
e. Which side of the reaction, reactants or products, will be favored if: NaClO (aq) is combined with H2S (aq)
2. A 0.150 M solution of butanoic acid, CH3CH2CH2COOH, was found to contain 1.51×10−3 M H3O+. What is the
Ka of butanoic acid?
3. Hypochlorous acid has a Ka of 3.5×10−8. Calculate the pH of the following 2 solutions:
a. 0.500 M HClO
b. 0.500 M NaClO
4. Calculate the pH of solutions of each of the following salts.
a. 0.150 M Ba(C2H3O2)2 (Acetic acid has Ka = 1.8×10–5)
b. 1.00 M (CH3)2NH2Cl (this is the chloride salt of dimethyl amine, Kb = 5.9×10–4)
5. Calculate the pH of a 1.20 M solution of the potassium salt of a weak acid in water at 40.0 °C. At this
temperature, the value of Ka for the weak acid is 4.5×10–5 and the pH of pure water is 6.77.
6. The ionization constant for butanoic acid (CH3CH2CH2COOH) has the value 1.5×10–5
(a) Write an equation to show the ionization of butanoic acid.
(b) What is the pH of a 0.0100 M solution of butanoic acid?
(c) How many milliliters of 0.0350 M NaOH would be required to completely react with 100.0 mL of
0.0100 M butanoic acid?
(d) Name and write the formula of the two products formed when butanoic acid reacts with the NaOH
solution.
7. Ethanol can be oxidized to ethanoic acid (with a 1:1 mole ratio). Calculate the mass of ethanol per liter that
has undergone oxidation in a wine a pH of 2.80. Assume the ethanoic acid is the only source of H+ and that
the Ka = 1.8×10–5 for the ethanoic acid.
8. What is the pH of 1.0×10–8 M HCl? Does your answer make sense?
9. What is the pH of 500.0 mL of a solution that is made to be 0.500 M acetic acid and 0.100 M sodium
acetate?
10. Calculate the pH of the above solution (in #8) after 10.0 mL of 2.00 M NaOH is added.
Key
1.
a.
b.
c.
d.
HNO2 is the strongest acid (highest Ka value)
Ka = [H3O+][NO2-]/[HNO2]
HCN has the strongest conjugate base Kb = [OH-][HCN]/[CN-]
The HClO and CN– products are the stronger acid and stronger base, so the reaction gets
shifted to the left (favors reactants).
e. The H2S and ClO– reactants are the stronger acid and stronger base, so the reaction is
shifted to the right (favors products
2. Ka = (0.00151)2/0.1485 = 1.54×10–5
3.
a. pH = 3.88
b. Kb = 2.86×10–7 , x = 3.780×10–4 pOH = 3.42, pH = 10.58
4.
a. Kb = 5.56×10–10 = x2/[0.30 – x] pOH = 4.89 pH = 9.11
b. Kb = 5.9×10–4 Ka = 1.69×10–11 = x2/1.00–x x = 4.1169×10–6 pH = 5.39
5. Kw at 40.0 C = 2.88×10–14 Kb = 6.41×10–10 [OH–] = 2.77×10–5 pOH = 4.56
find new Kw at 40.0 C = −log 2.88×10–14 = 13.54 pH = 13.54-4.56 = 8.98
6. a. CH3CH2CH2COOH + H2O ⇌ CH3CH2CH2COO− + H3O+
b. pH = 3.41
c. it would take 0.0286 L to neutralize
d. the products are water and sodium butanoate
7. calculate the initial ethanoic acid concentration using Ka and pH
1.8×10–5 = (0.00158)2/[HEth] [HEth] = 0.141 M
0.141 M times 46.08 g/mol ethanol = 6.50 g ethanoic acid/L
8. The total acid concentration is 1.1×10–7 M, when autoionization is included, pH = 6.96
9. 1.8×10–5 = 0.100 M · [H3O+]/0.500 M [H3O+] = pH 4.05
10. pH = 4.22 (there will be 0.023 moles of acetic acid and 0.070 moles of acetate).