Question

Determine the pH of each of the following solutions.

(a) 0.492 M carbonic acid (weak acid with Ka = 4.3e-07).

(b) 0.770 M benzoic acid (weak acid with Ka = 6.3e-05).

(c) 0.697 M pyridine (weak base with Kb = 1.7e-09).

Answer 1

a)

H2CO3 dissociates as:

H2CO3 -----> H+ + HCO3-

0.492 0 0

0.492-x x x

Ka = [H+][HCO3-]/[H2CO3]

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 ((4.3*10^-7)*0.492) = 4.6*10^-4

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

so, x = 4.6*10^-4 M

So, [H+] = x = 4.6*10^-4 M

use:

pH = -log [H+]

= -log (4.6*10^-4)

= 3.3373

Answer: 3.34

b)

C6H5COOH dissociates as:

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

0.77 0 0

0.77-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 ((6.3*10^-5)*0.77) = 6.965*10^-3

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

so, x = 6.965*10^-3 M

So, [H+] = x = 6.965*10^-3 M

use:

pH = -log [H+]

= -log (6.965*10^-3)

= 2.1571

Answer: 2.16

c)

C5H5N dissociates as:

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

0.697 0 0

0.697-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 ((1.7*10^-9)*0.697) = 3.442*10^-5

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

so, x = 3.442*10^-5 M

So, [OH-] = x = 3.442*10^-5 M

use:

pOH = -log [OH-]

= -log (3.442*10^-5)

= 4.4632

use:

PH = 14 - pOH

= 14 - 4.4632

= 9.5368

Answer: 9.54