Question

Enter your answer in the provided box. The K, for benzoic acid is 6.5 x 10-5. Calculate the pH of a 0.11 M benzoic acid solution.

Answer 1

C6H5COOH dissociates as:

C6H5COOH -----> H+ + C6H5COO-

0.11 0 0

0.11-x x x

Ka = [H+][C6H5COO-]/[C6H5COOH]

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 ((6.5*10^-5)*0.11) = 2.674*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)

6.5*10^-5 = x^2/(0.11-x)

7.15*10^-6 - 6.5*10^-5 *x = x^2

x^2 + 6.5*10^-5 *x-7.15*10^-6 = 0

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

a = 1

b = 6.5*10^-5

c = -7.15*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 = 2.86*10^-5

roots are :

x = 2.642*10^-3 and x = -2.707*10^-3

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

x = 2.642*10^-3

So, [H+] = x = 2.642*10^-3 M

use:

pH = -log [H+]

= -log (2.642*10^-3)

= 2.5781

Answer: 2.58