Question

15. Calculate the pH of a solution containing 2.5 x 10–​ 2 ​mol of nicotinic acid (a monoprotic acid) dissolved in 350 mL of water. ( K​a​ = 1.1 x 10-​ 5​). The answer is 3.05. Please Help

Answer 1

Let write the acid as HA

[HA] = mol / volume in L

= 2.5*10^-2 mol / 0.350 L

= 0.0714 M

HA dissociates as:

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

7.14*10^-2 0 0

7.14*10^-2-x x x

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

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 ((1.1*10^-5)*7.14*10^-2) = 8.862*10^-4

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

1.1*10^-5 = x^2/(7.14*10^-2-x)

7.854*10^-7 - 1.1*10^-5 *x = x^2

x^2 + 1.1*10^-5 *x-7.854*10^-7 = 0

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

a = 1

b = 1.1*10^-5

c = -7.854*10^-7

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 = 3.142*10^-6

roots are :

x = 8.807*10^-4 and x = -8.917*10^-4

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

x = 8.807*10^-4

So, [H+] = x = 8.807*10^-4 M

use:

pH = -log [H+]

= -log (8.807*10^-4)

= 3.05

Answer: 3.05