Question

What is the pH, [H+], [OH-], and [C6H5CO2-] of a 0.05 M C6H5CO2H solution? pKa=4.202

Answer 1

use:

pKa = -log Ka

4.202 = -log Ka

Ka = 6.281*10^-5

C6H5CO2H dissociates as:

C6H5CO2H -----> H+ + C6H5CO2-

5*10^-2 0 0

5*10^-2-x x x

Ka = [H+][C6H5CO2-]/[C6H5CO2H]

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 ((6.281*10^-5)*5*10^-2) = 1.772*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)

6.281*10^-5 = x^2/(5*10^-2-x)

3.14*10^-6 - 6.281*10^-5 *x = x^2

x^2 + 6.281*10^-5 *x-3.14*10^-6 = 0

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

a = 1

b = 6.281*10^-5

c = -3.14*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 = 1.257*10^-5

roots are :

x = 1.741*10^-3 and x = -1.804*10^-3

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

x = 1.741*10^-3

So, [H+] = x = 1.741*10^-3 M

use:

pH = -log [H+]

= -log (1.741*10^-3)

= 2.76

[C6H5CO2-] = x = 1.741*10^-3 M

use:

[OH-] = Kw/[H+]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[OH-] = (1.0*10^-14)/[H+]

[OH-] = (1.0*10^-14)/(1.741*10^-3)

[OH-] = 5.744*10^-12 M

pH = 2.76

[H+] = 1.74*10^-3

[OH-] = 5.74*10^-12 M

[C6H5CO2-] = 1.74*10^-3 M