Question

4-Nitrobenzoic acid has a pKa of 3.442. What is the pH of a 0.25 M solution of 4-nitrobenzoic acid in pure water?

Answer 1

for simplicity lets write the acid as HA

we have below equation to be used:

pKa = -log Ka

3.442 = -log Ka

log Ka = -3.442

Ka = 10^(-3.442)

Ka = 3.614*10^-4

Lets write the dissociation equation of HA

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

0.25 0 0

0.25-x x x

Ka = [H+][A-]/[HA]

Ka = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((3.614*10^-4)*0.25) = 9.505*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)

3.614*10^-4 = x^2/(0.25-x)

9.035*10^-5 - 3.614*10^-4 *x = x^2

x^2 + 3.614*10^-4 *x-9.035*10^-5 = 0

Let's solve this quadratic equation

Comparing it with general form: (ax^2+bx+c=0)

a = 1

b = 3.614*10^-4

c = -9.035*10^-5

solution of quadratic equation is found by below formula

x = {-b + √(b^2-4*a*c)}/2a

x = {-b - √(b^2-4*a*c)}/2a

b^2-4*a*c = 3.615*10^-4

putting value of d, solution can be written as:

x = {-3.614*10^-4 + √(3.615*10^-4)}/2

x = {-3.614*10^-4 - √(3.615*10^-4)}/2

solutions are :

x = 9.326*10^-3 and x = -9.688*10^-3

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

x = 9.326*10^-3

so,

[H+] = x = 9.326*10^-3 M

we have below equation to be used:

pH = -log [H+]

= -log (9.326*10^-3)

= 2.03

Answer: 2.03