Question

Set up the equilibrium equation and determine the pH of the 0.15 M solution of potassium nitrate? (Ka value for nitrous acid= 4.0x 10^-4).

Answer 1

Potassium nitrate is salt of KOH and HNO3

KOH is strong base and HNO3 is strong acid.

So, Potassium nitrate is neutral and its pH = 7.00

But I think question has a typing error and the compound should be Potassium nitrite which is KNO2

use:

Kb = Kw/Ka

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

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

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

Kb = 2.5*10^-11

NO2- dissociates as

NO2- + H2O -----> HNO2 + OH-

0.15 0 0

0.15-x x x

Kb = [HNO2][OH-]/[NO2-]

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^-11)*0.15) = 1.936*10^-6

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

so, x = 1.936*10^-6 M

use:

pOH = -log [OH-]

= -log (1.936*10^-6)

= 5.713

use:

PH = 14 - pOH

= 14 - 5.713

= 8.287

Answer: 8.29