Question

For
the nitrous acid, HNO2, Ka= 4.0 x 10^-4. Calculate the ph of 0.27 M
HNO2

Answer #1

HNO2 dissociates as:

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

0.27 0 0

0.27-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*10^-4)*0.27) = 1.039*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*10^-4 = x^2/(0.27-x)

1.08*10^-4 - 4*10^-4 *x = x^2

x^2 + 4*10^-4 *x-1.08*10^-4 = 0

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

a = 1

b = 4*10^-4

c = -1.08*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 = 4.322*10^-4

roots are :

x = 1.019*10^-2 and x = -1.059*10^-2

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

x = 1.019*10^-2

So, [H+] = x = 1.019*10^-2 M

use:

pH = -log [H+]

= -log (1.019*10^-2)

= 1.9916

Answer: 1.99

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