Question

Unlike strong acids and bases that ionize completely in solution, weak acids or bases partially ionize. The tendency of a weak acid or base to ionize can be quantified in several ways including

Ka or Kb,

pKa or pKb,

and percent ionization*.

*This assumes that species of equal concentrations are being compared as percent ionization is affected by concentration.

Part A

Pyridine is a weak base that is used in the manufacture of pesticides and plastic resins. It is also a component of cigarette smoke. Pyridine ionizes in water as follows:

C5H5N+H2O⇌C5H5NH++OH−

The pKb of pyridine is 8.75. What is the pH of a 0.400 M solution of pyridine? (Assume that the temperature is 25 ∘C.)

Express the pH numerically to two decimal places.

Part B

Benzoic acid is a weak acid that has antimicrobial properties. Its sodium salt, sodium benzoate, is a preservative found in foods, medications, and personal hygiene products. Benzoic acid ionizes in water:

C6H5COOH⇌C6H5COO−+H+

The pKa of this reaction is 4.2. In a 0.80 M solution of benzoic acid, what percentage of the molecules are ionized?

Express your answer to two significant figures and include the appropriate units.

Answer 1

A)

The pKb of pyridine is 8.75. What is the pH of a 0.400 M solution of pyridine? (Assume that the temperature is 25 ∘C.)

This is a base in water so, let the base be CH3NH2 = "B" and CH3NH3+ = HB+ the protonated base "HB+"

there are free OH- ions so, expect a basic pH

B + H2O <-> HB+ + OH-

The equilibrium Kb:

Kb = [HB+][OH-]/[B]

initially:

[HB+] = 0

[OH-] = 0

[B] = M

the change

[HB+] = x

[OH-] = x

[B] = - x

in equilibrium

[HB+] = 0 + x

[OH-] = 0 + x

[B] = M - x

Now substitute in Kb

Kb = [HB+][OH-]/[B]

Kb = x*x/(M-x)

x^2 + Kbx - M*Kb = 0

x^2 + (10^-8.75)x - (0.4)(10^-8.75) = 0

solve for x

x = 2.66*10^-5

substitute:

[HB+] = 0 + x = 2.66*10^-5M

[OH-] = 0 + x = 2.66*10^-5M

[B] = M - x = 0.40-2.66*10^-5= 0.3999734 M

%ionization of B = BH+ / B * 100% =  2.66*10^-5)/(0.4)*100 = 0.00665 %

B)

for benzoic acid:

First, assume the acid:

HF

to be HA, for simplicity, so it will ionize as follows:

HA <-> H+ + A-

where, H+ is the proton and A- the conjugate base, HA is molecular acid

Ka = [H+][A-]/[HA]; by definition

initially

[H+] = 0

[A-] = 0

[HA] = M;

the change

initially

[H+] = + x

[A-] = + x

[HA] = - x

in equilbrirum

[H+] = 0 + x

[A-] = 0 + x

[HA] = M - x

substitute in Ka

Ka = [H+][A-]/[HA]

Ka = x*x/(M-x)

x^2 + Kax - M*Ka = 0

if M = 0.3 M; then

x^2 + (10^-4.2)x - 0.8*(10^-4.2) = 0

solve for x

x =0.0071

substitute

[H+] = 0 + 0.0071= 0.0071 M

[A-] = 0 + 0.0071= 0.0071 M

[HA] = M - x = 0.8-0.0071= 0.7929 M

% ionization = A- / HA * 100% = 0.0071 / 0.8 *100 = 0.89%