pH of (a)Weak Acid Solutions, (b) Bases, (c) Polyprotic

Acids, and (d) Salts

Chemistry 142 B

Autumn 2004

J. B. Callis, Instructor

Lecture 21

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:

solutions thehaswhich

form theof is polynomial The

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) + OH-(aq)

base acid conj. Acid conj. base

][

]][[

B

OHBHKb

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)

DMA = DMA+ OH-

Initial

Change

Equil.

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) + C2O4

2-(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 SaltsSalt – 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)

Ac- = HAc + OH-

Initial

Change

Equil.

Problem 21-7 (cont.)

Answers to Problems in Lecture 211. [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.

3.

4.

5. pH = 12.48

6. 5.0 x 10-12 M

7. pH = 9.07

5102.1 aK

78.11pH