Question

A 15.5 mL sample of a 0.373 M aqueous nitrous acid solution is titrated with a 0.437 M aqueous barium hydroxide solution. What is the pH at the start of the titration, before any barium hydroxide has been added?

Answer 1

ka of HNO2 = 4.5*10^-4

At start, there is only 0.373 M of HNO2

HNO2 dissociates as:

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

0.373 0 0

0.373-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 ((4.5*10^-4)*0.373) = 1.296*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)

4.5*10^-4 = x^2/(0.373-x)

1.678*10^-4 - 4.5*10^-4 *x = x^2

x^2 + 4.5*10^-4 *x-1.678*10^-4 = 0

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

a = 1

b = 4.5*10^-4

c = -1.678*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 = 6.716*10^-4

roots are :

x = 1.273*10^-2 and x = -1.318*10^-2

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

x = 1.273*10^-2

So, [H+] = x = 1.273*10^-2 M

use:

pH = -log [H+]

= -log (1.273*10^-2)

= 1.8951

Answer: 1.90