Question

Calculate the pH at 25c of a 0.035M solution of a weak acid that has a Ka=1.3x10⁵

Answer 1

Lets write the acid as HA

HA dissociates as:

HA -----> H+ + A-

3.5*10^-2 0 0

3.5*10^-2-x x x

Ka = [H+][A-]/[HA]

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 ((1.3*10^-5)*3.5*10^-2) = 6.745*10^-4

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

1.3*10^-5 = x^2/(3.5*10^-2-x)

4.55*10^-7 - 1.3*10^-5 *x = x^2

x^2 + 1.3*10^-5 *x-4.55*10^-7 = 0

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

a = 1

b = 1.3*10^-5

c = -4.55*10^-7

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 = 1.82*10^-6

roots are :

x = 6.681*10^-4 and x = -6.811*10^-4

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

x = 6.681*10^-4

so.[H+] = x = 6.681*10^-4 M

use:

pH = -log [H+]

= -log (6.681*10^-4)

= 3.2

Answer: 3.2