Question

Find the pH of 2.00 M HNO2 (aq) solution at 25 degrees C.

Given: HNO2 + H2O <--> NO2 + H30 Ka for HNO2 = 4.6 x 10^-4

The answer is pH=2.02. Please show me how to get this answer.

Answer 1

HNO2 dissociates as:

HNO2 -----> H+ + NO2-

2 0 0

2-x x x

Ka = [H+][NO2-]/[HNO2]

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 ((4.6*10^-4)*2) = 3.033*10^-2

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

4.6*10^-4 = x^2/(2-x)

9.2*10^-4 - 4.6*10^-4 *x = x^2

x^2 + 4.6*10^-4 *x-9.2*10^-4 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1

b = 4.6*10^-4

c = -9.2*10^-4

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 3.68*10^-3

roots are :

x = 3.01*10^-2 and x = -3.056*10^-2

since x can't be negative, the possible value of x is

x = 3.01*10^-2

so.[H+] = x = 3.01*10^-2 M

use:

pH = -log [H+]

= -log (3.01*10^-2)

= 1.52

I am sorry to say but answer ashould be 1.52