Question

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 0.500 M solution of pyridine? 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 dissociates in water: C6H5COOH⇌C6H5COO−+H+ The pKa of this reaction is 4.2. In a 0.67 M solution of benzoic acid, what percentage of the molecules are ionized? Express the percentage numerically using two significant figures.

Answer 1

This is a base in water so, let the base be Pyridine = "B" and Pyridine H+ = 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 - (0.5)(10^-8.75) = 0

solve for x

x = 2.98*10^-5

substitute:

[HB+] = 0 + x =  2.98*10^-5M

[OH-] = 0 + x =  2.98*10^-5M

pH = 14 + pOH = 14 + log( 2.98*10^-5) = 9.47421

B)

First, assume the acid:

Hbenzoic

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 = 0.67M; then

x^2 + (10^-4.2)x - 0.67*(10^-4.2) = 0

solve for x

x =0.00647

substitute

[H+] = 0 + 0.00647= 0.00647M

[A-] = 0 + 0.00647= 0.00647 M

[HA] = M - x = 0.67-0.00647= 0.66353 M

pH = -log(H+) = -log(0.00647 ) = 2.18909