Question

A.HClO is a weak acid (Ka = 4.0 × 10–8) and so the salt NaClO acts as a weak base. What is the pH of a solution that is 0.079 M in NaClO at 25 °C?

ph=

B.NH3 is a weak base (Kb = 1.8 × 10–5) and so the salt NH4Cl acts as a weak acid. What is the pH of a solution that is 0.014 M in NH4Cl at 25 °C?

ph=

Answer 1

A)

use:

Kb = (1.0*10^-14)/Ka

Kb = (1.0*10^-14)/4*10^-8

Kb = 2.5*10^-7

ClO- dissociates as

ClO- + H2O -----> HClO + OH-

0.079 0 0

0.079-x x x

Kb = [HClO][OH-]/[ClO-]

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 ((2.5*10^-7)*7.9*10^-2) = 1.405*10^-4

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

so, x = 1.405*10^-4 M

use:

pOH = -log [OH-]

= -log (1.405*10^-4)

= 3.85

use:

PH = 14 - pOH

= 14 - 3.85

= 10.15

Answer: 10.15

B)

use:

Ka = (1.0*10^-14)/Kb

Ka = (1.0*10^-14)/1.8*10^-5

Ka = 5.556*10^-10

NH4+ + H2O -----> NH3 + H+

1.4*10^-2 0 0

1.4*10^-2-x x x

Ka = [H+][NH3]/[NH4+]

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 ((5.556*10^-10)*1.4*10^-2) = 2.789*10^-6

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

so, x = 2.789*10^-6 M

so.[H+] = x = 2.789*10^-6 M

use:

pH = -log [H+]

= -log (2.789*10^-6)

= 5.55

Answer: 5.55