Lecture 21
pH of (a)Weak Acid Solutions,
(b) Bases, (c) Polyprotic
Acids, and (d) Salts
Chemistry 142 B
Autumn 2004
J. B. Callis, Instructor
Calculation of the pH of Weak Acid
Solutions - A Systematic Approach (1)
• Problem 21-1: What is the pH of a solution of
1.00 M nitrous acid, Ka = 4.0 x 10-4
• Major species in solution: HNO2 and H2O
• Which species can generate H+ ions?
– HNO2(aq) = H+(aq) + NO2-(aq) Ka = 4.0 x 10-4
– H2O(aq) = H+(aq) + OH-(aq)
Kw = 1.0 x 10-14
– Ignore contribution from water, Ka >> Kw
Calculation of the pH of Weak Acid
Solutions - A Systematic Approach (2)
• The equilibrium expression is
The initial concentration are
Calculation of the pH of Weak Acid
Solutions - A Systematic Approach (3)
Let x be the change in concentration of HNO2
that is required to achieve equilibrium. Then the
equilibrium concentrations are:
Calculation of the pH of Weak Acid
Solutions - A Systematic Approach (4)
We rearrange this equation to yield a second
order polynomial:
The polynomial is of the form
which has the solutions
Calculation of the pH of Weak Acid
Solutions - A Systematic Approach (5)
For this example, the solutions are:
x=
Only the first solution is valid because it leads to
all positive concentrations.
[H+] = [NO2-] =
[HNO2] =
Calculation of the pH of Weak Acid
Solutions - A Systematic Approach (6)
Question 1(a): What is the pH of this solution?
ans:
Question 1(b): Were we correct to neglect H+ from
the water
ans:
Question 1(c): What % of the acid is ionized?
ans:
Problem 21-2: Calculate of the pH of a
Mixture of Weak Acids
Calculate the pH of a mixture of 1.00 M of phenol
(Ka = 1.6 x 10-10 and 5.00 M acetic acid (Ka = 1.8
x 10-5).
•Major Species in Solution: phenol (HPhe), acetic
acid (HAc) and H2O
•Which Species Can Generate H+ ions?
–HAc(aq) = H+(aq) + Ac-(aq) KHAc = 1.8 x 10-5
–HPhe(aq) = H+(aq) + Phe-(aq) KHPhc = 1.8 x 10-10
–H2O(aq) = H+(aq) + OH-(aq)
Kw = 1.0 x 10-14
–Ignore contribution from water and phenol,
– KAc >> KHPhe >> Kw
Problem 21-2: Calculate of the pH of a
Mixture of Weak Acids (2)
Focusing on the Acetic Acid equilibrium:
K
x=
pH =
Problem 21-2: Calculate of the pH of a
Mixture of Weak Acids (3)
How much Phe- is generated?
Problem 21-3: Find the Ka of a
weak acid from % dissociation.
If 0.10 M propanoic acid dissociates 1.1%, what is Ka?
Bases
Definition (Bronsted-Lowry) – a proton
acceptor. Strong bases dissociate completely.
(e.g. metal hydroxides from Groups 1A and
1B.
NaOH(s) -> Na+(aq) + OH-(aq)
Problem 21-4: Calculate the
pH of a solution of 3.0 x 10-3
M Ca(OH)2(aq)
Note: A base doesn’t have to contain OH-, it just
needs to be able to accept H+, e.g. aqueous
ammonium.
NH3(aq) + H2O(l) = NH4+(aq) + OH-(aq)
Many nitrogen containing compounds are bases.
General reaction:
B(aq) + H2O(l) = BH+(aq) +
base
acid
OH-(aq)
conj. Acid conj. base


[ BH ][OH ]
Kb 
[ B]
Problem 21-5: What is the pH of 1.5 M
Dimethylamine (CH3)2NH (Kb = 5.9 x 10-4).
Let x be the amount of dimethylamine that has dissociated.
Conc.
(M)
Initial
Change
Equil.
DMA =
DMA+
OH-
Problem 21-5 (cont.)
Polyprotic Acids
Can furnish more than one proton per molecule of acid.
They do this in a step-wise manner.
Example: oxalic acid:
H2C2O4(aq) = H+(aq) + HC2O4-(aq)
Ka1 = 5.6 x 10-2
HC2O4-(aq) = H+(aq) + C2O42-(aq)
Ka2 = 5.4 x 10-5
Problem 21-6 – Calculate the pH of 0.050 M
Ascorbic Acid (Ka1 = 1.0 x 10-5; Ka2 = 5.0 x 10-12).
Let x be the amount of [H+] that is produced. Assume that it all
comes from the first ionization. Then let the [H+] determine the
amount of the doubly ionized base, Asc2-.
Acid-Base Properties of Salts
Salt – an ionic compound that dissolves in H2O to give
ions. Sometimes the ions can behave as acids or
bases.
(a) Anions that correspond to strong acids, e.g. Cl- and
NO3- are weak, weak bases. Also, cations from
strong bases, e.g. Na+, K+ are weak, weak acids.
(b) Salts that consist of cations from strong bases and
anions from strong acids produce neutral solutions
(pH= 7).
(c) Salts of weak acids produce basic solutions.
(d) Salts of weak bases produce acid solutions.
Problem 21-7: What is the pH of 0.25 M
Sodium Acetate (NaAc)?
Let x be the amount of acetic acid that is formed by the
following reaction: Ac-(aq) + H2O(l) = HAc(aq) + OH-(aq).
Conc.
(M)
Initial
Change
Equil.
Ac- =
HAc +
OH-
Problem 21-7 (cont.)
Answers to Problems in Lecture 21
1.
[H+] = [NO2-] = 0.020 M; [HNO2] = 0.98 M (a) pH = 1.70
(b) Yes (c) 2.0 %
2. pH = 2.023; Very little dissociation of HPhe ;Very
little H+ from HPhe.
5
K

1
.
2

10
3. a
4. pH  11.78
5. pH = 12.48
6. 5.0 x 10-12 M
7. pH = 9.07