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 0.315 M solution of pyridine? (Assume that the temperature is 25 ∘C.) Express the pH numerically to two decimal places. View Available Hint(s) pH = nothing Submit 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 0.63 M solution of benzoic acid, what percentage of the molecules are ionized?

Answer 1

A)

use:

pKb = -log Kb

8.75= -log Kb

Kb = 1.778*10^-9

C5H5N dissociates as:

C5H5N +H2O -----> C5H5NH+ + OH-

0.315 0 0

0.315-x x x

Kb = [C5H5NH+][OH-]/[C5H5N]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.778*10^-9)*0.315) = 2.367*10^-5

since c is much greater than x, our assumption is correct

so, x = 2.367*10^-5 M

So, [OH-] = x = 2.367*10^-5 M

use:

pOH = -log [OH-]

= -log (2.367*10^-5)

= 4.6258

use:

PH = 14 - pOH

= 14 - 4.6258

= 9.37

Answer: 9.37

B)

use:

pKa = -log Ka

4.2 = -log Ka

Ka = 6.31*10^-5

C6H5COOH dissociates as:

C6H5COOH -----> H+ + C6H5COO-

0.63 0 0

0.63-x x x

Ka = [H+][C6H5COO-]/[C6H5COOH]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((6.31*10^-5)*0.63) = 6.305*10^-3

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

6.31*10^-5 = x^2/(0.63-x)

3.975*10^-5 - 6.31*10^-5 *x = x^2

x^2 + 6.31*10^-5 *x-3.975*10^-5 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1

b = 6.31*10^-5

c = -3.975*10^-5

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 1.59*10^-4

roots are :

x = 6.273*10^-3 and x = -6.336*10^-3

since x can't be negative, the possible value of x is

x = 6.273*10^-3

So, [H+] = x = 6.273*10^-3 M

use:

pH = -log [H+]

= -log (6.273*10^-3)

= 2.2025

Answer: 2.20