Question

Calculate the pH of a 2.41 M solution of NaNO2

Answer 1

ka of HNO2 = 4.5*10^-4

NO2- is base

Find Kb of NO2-

use:

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

Kb = (1.0*10^-14)/4.5*10^-4

Kb = 2.222*10^-11

NO2- dissociates as

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

2.41 0 0

2.41-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 ((2.222*10^-11)*2.41) = 7.318*10^-6

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

so, x = 7.318*10^-6 M

use:

pOH = -log [OH-]

= -log (7.318*10^-6)

= 5.14

use:

PH = 14 - pOH

= 14 - 5.14

= 8.86

Answer: 8.86