Question

Enter your answer in the provided box. What is the pH of a 0.157 M monoprotic acid whose K, is 9.598 x 10-3? pH =

Answer 1

HA dissociates as:

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

0.157 0 0

0.157-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 ((9.598*10^-3)*0.157) = 3.882*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)

9.598*10^-3 = x^2/(0.157-x)

1.507*10^-3 - 9.598*10^-3 *x = x^2

x^2 + 9.598*10^-3 *x-1.507*10^-3 = 0

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

a = 1

b = 9.598*10^-3

c = -1.507*10^-3

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 = 6.12*10^-3

roots are :

x = 3.432*10^-2 and x = -4.391*10^-2

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

x = 3.432*10^-2

So, [H+] = x = 3.432*10^-2 M

use:

pH = -log [H+]

= -log (3.432*10^-2)

= 1.4645

Answer: 1.4645