Question

Calculate the pH of a 0.150 M solution of benzoic acid (C₆H₅CO₂H) if  

K_a= 6.3 x 10^-5for the acid.    C₆H₅CO₂H(aq)     ⇌    H⁺(aq) + C₆C₅CO₂^-(aq)

Answer 1

C6H5COOH dissociates as:

C6H5COOH -----> H+ + C6H5CO2-

0.15 0 0

0.15-x x x

Ka = [H+][C6H5CO2-]/[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 ((6.3*10^-5)*0.15) = 3.074*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/(0.15-x)

9.45*10^-6 - 6.3*10^-5 *x = x^2

x^2 + 6.3*10^-5 *x-9.45*10^-6 = 0

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

a = 1

b = 6.3*10^-5

c = -9.45*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 = 3.78*10^-5

roots are :

x = 3.043*10^-3 and x = -3.106*10^-3

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

x = 3.043*10^-3

So, [H+] = x = 3.043*10^-3 M

use:

pH = -log [H+]

= -log (3.043*10^-3)

= 2.5167

Answer: 2.52