Question

Question 5 A solution of 0.05M hexanoic acid is prepared. Ka = 1.41x10-5 a. What is the pH of the solution? (write answer to the hundredths place) b. What is the % ionization of the solution? ( write answer to the hundredths place) A Click Submit to complete this assessment.

Answer 1

a)

HA dissociates as:

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

5*10^-2 0 0

5*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 ((1.41*10^-5)*5*10^-2) = 8.396*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)

1.41*10^-5 = x^2/(5*10^-2-x)

7.05*10^-7 - 1.41*10^-5 *x = x^2

x^2 + 1.41*10^-5 *x-7.05*10^-7 = 0

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

a = 1

b = 1.41*10^-5

c = -7.05*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 = 2.82*10^-6

roots are :

x = 8.326*10^-4 and x = -8.467*10^-4

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

x = 8.326*10^-4

So, [H+] = x = 8.326*10^-4 M

use:

pH = -log [H+]

= -log (8.326*10^-4)

= 3.0796

Answer: 3.08

b)

% ionisation = x * 100 / initial concentration

= (8.326*10^-4)*100 / 0.05

= 1.67 %

Answer: 1.67 %