UBC CHEM 123 PS01: Acid-Base Chemistry Page 1 of 15
For all questions, assume that the given values are good to 4 significant figures.
1. For each of the following reactions, identify the role each chemical is playing: Bronsted-Lowry Acid,
Bronsted-Lowry Base, conjugate acid, or conjugate base. Identify the conjugate acid-base pairs.
Base Acid conj acid conj base
Acid Base conj base conj acid
Acid Base conj base conj acid
, UBC CHEM 123 PS01: Acid-Base Chemistry Page 2 of 15
2. (a) Acetic acid, CH3COOH, is a weak acid with a Ka of 1.8 × 10–5 at T = 25°C. Write the products of the acid-
base reaction between acetic acid and water.
O
H
C
C
O
+ H3O+
H H
(b) All species in an equilibrium reaction are present (in varying amounts) in the same flask at equilibrium.
The equilibrium distribution of products and reactants is quantified by the equilibrium constant, K, which
depends on the reaction and reaction conditions. Write out the expression (in terms of concentrations) for
the equilibrium constant, K, of the reaction from part (a).
[CH3 OO− ][H3 O+ ]
K=
[CH3 OOH]
The Ka is an equilibrium constant for a particular type reaction. It applies to acid dissociation reactions
where the equilibrium reaction is written to produce H3O+ in solution.
(c) It is a bit more convenient to communicate (and compare) Ka values as pKa values. Calculate the pKa
value of acetic acid.
pKa = –log(Ka) = –log (1.8 × 10–5) = 4.74
(d) The pH of a solution depends on the concentration of the hydronium ion, [H3O+], and is calculated as
pH = –log[H3O+]. Aqueous solutions with pH < 7 are acidic and those with pH > 7 are basic. The useful
range of the pH scale for aqueous solutions is from approximately 0 to 14.
A 0.10 M aqueous solution of acetic acid has a pH of 2.87 at equilibrium. Calculate the equilibrium
concentrations of H3O+ and CH3COO– in this solution.
pH = –log[H3O+]
[H3O+] = 10–pH = 10–2.87 = 0.00135 M
In a solution of 0.10 M CH3COOH(aq),
[H3O+] = [CH3COO–] = 0.00135 M
For all questions, assume that the given values are good to 4 significant figures.
1. For each of the following reactions, identify the role each chemical is playing: Bronsted-Lowry Acid,
Bronsted-Lowry Base, conjugate acid, or conjugate base. Identify the conjugate acid-base pairs.
Base Acid conj acid conj base
Acid Base conj base conj acid
Acid Base conj base conj acid
, UBC CHEM 123 PS01: Acid-Base Chemistry Page 2 of 15
2. (a) Acetic acid, CH3COOH, is a weak acid with a Ka of 1.8 × 10–5 at T = 25°C. Write the products of the acid-
base reaction between acetic acid and water.
O
H
C
C
O
+ H3O+
H H
(b) All species in an equilibrium reaction are present (in varying amounts) in the same flask at equilibrium.
The equilibrium distribution of products and reactants is quantified by the equilibrium constant, K, which
depends on the reaction and reaction conditions. Write out the expression (in terms of concentrations) for
the equilibrium constant, K, of the reaction from part (a).
[CH3 OO− ][H3 O+ ]
K=
[CH3 OOH]
The Ka is an equilibrium constant for a particular type reaction. It applies to acid dissociation reactions
where the equilibrium reaction is written to produce H3O+ in solution.
(c) It is a bit more convenient to communicate (and compare) Ka values as pKa values. Calculate the pKa
value of acetic acid.
pKa = –log(Ka) = –log (1.8 × 10–5) = 4.74
(d) The pH of a solution depends on the concentration of the hydronium ion, [H3O+], and is calculated as
pH = –log[H3O+]. Aqueous solutions with pH < 7 are acidic and those with pH > 7 are basic. The useful
range of the pH scale for aqueous solutions is from approximately 0 to 14.
A 0.10 M aqueous solution of acetic acid has a pH of 2.87 at equilibrium. Calculate the equilibrium
concentrations of H3O+ and CH3COO– in this solution.
pH = –log[H3O+]
[H3O+] = 10–pH = 10–2.87 = 0.00135 M
In a solution of 0.10 M CH3COOH(aq),
[H3O+] = [CH3COO–] = 0.00135 M
