Question

27.

Determine the pH of each of the following solutions.

(a) 0.354 M hypoiodous acid (weak acid with K_a = 2.3e-11).

= _________



(b) 0.156 M carbonic acid (weak acid with K_a = 4.3e-07).

= ________



(c) 0.476 M pyridine (weak base with K_b = 1.7e-09).

= ________

Answer 1

a)

HOI dissociates as:

HOI -----> H+ + IO-

0.354 0 0

0.354-x x x

Ka = [H+][IO-]/[HOI]

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 ((2.3*10^-11)*0.354) = 2.853*10^-6

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

so, x = 2.853*10^-6 M

So, [H+] = x = 2.853*10^-6 M

use:

pH = -log [H+]

= -log (2.853*10^-6)

= 5.5446

Answer: 5.54

b)

H2CO3 dissociates as:

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

0.145 0 0

0.145-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)*0.145) = 2.497*10^-4

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

so, x = 2.497*10^-4 M

So, [H+] = x = 2.497*10^-4 M

use:

pH = -log [H+]

= -log (2.497*10^-4)

= 3.6026

Answer: 3.60

c)

C5H5N dissociates as:

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

0.476 0 0

0.476-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)*0.476) = 2.845*10^-5

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

so, x = 2.845*10^-5 M

So, [OH-] = x = 2.845*10^-5 M

use:

pOH = -log [OH-]

= -log (2.845*10^-5)

= 4.546

use:

PH = 14 - pOH

= 14 - 4.546

= 9.454

Answer: 9.45