Question

The ?aKa of a weak monoprotic acid is 1.85×10−5.1.85×10−5. What is the pH of a 0.0841 M0.0841 M solution of this acid?

pH=

Answer 1

HA dissociates as:

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

8.41*10^-2 0 0

8.41*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 ((1.85*10^-5)*8.41*10^-2) = 1.247*10^-3

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

1.85*10^-5 = x^2/(8.41*10^-2-x)

1.556*10^-6 - 1.85*10^-5 *x = x^2

x^2 + 1.85*10^-5 *x-1.556*10^-6 = 0

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

a = 1

b = 1.85*10^-5

c = -1.556*10^-6

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 = 6.224*10^-6

roots are :

x = 1.238*10^-3 and x = -1.257*10^-3

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

x = 1.238*10^-3

So, [H+] = x = 1.238*10^-3 M

use:

pH = -log [H+]

= -log (1.238*10^-3)

= 2.9072

Answer: 2.91