Unlike strong acids and bases that ionize completely in solution, weak acids or

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+H2OC5H5NH++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:

C6H5COOHC6H5COO+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.

Explanation / Answer

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%

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