Question

Suppose a 0.050 M benzoic acid, HC7H5O2 solution is prepared. The Ka for benzoic acid is 6.3×10 −5. Calculate the pH of the benzoic acid solution.

pH = ?

Answer 1

HC7H5O2 dissociates as:

HC7H5O2 -----> H+ + C7H5O2-

5*10^-2 0 0

5*10^-2-x x x

Ka = [H+][C7H5O2-]/[HC7H5O2]

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 ((6.3*10^-5)*5*10^-2) = 1.775*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)

6.3*10^-5 = x^2/(5*10^-2-x)

3.15*10^-6 - 6.3*10^-5 *x = x^2

x^2 + 6.3*10^-5 *x-3.15*10^-6 = 0

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

a = 1

b = 6.3*10^-5

c = -3.15*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 = 1.26*10^-5

roots are :

x = 1.744*10^-3 and x = -1.807*10^-3

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

x = 1.744*10^-3

So, [H+] = x = 1.744*10^-3 M

use:

pH = -log [H+]

= -log (1.744*10^-3)

= 2.7586

Answer: 2.76