Question

8. What is the pH of 0.040 M HC3H5O2, propanoic acid? a. 8.68 b. 6.89 c. 5.27 d. 3.14

Answer 1

Ka of propanoic acid = 1.34*10^-5

HC3H5O2 dissociates as:

HC3H5O2 -----> H+ + C3H5O2-

4*10^-2 0 0

4*10^-2-x x x

Ka = [H+][C3H5O2-]/[HC3H5O2]

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 ((1.34*10^-5)*4*10^-2) = 7.321*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)

1.34*10^-5 = x^2/(4*10^-2-x)

5.36*10^-7 - 1.34*10^-5 *x = x^2

x^2 + 1.34*10^-5 *x-5.36*10^-7 = 0

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

a = 1

b = 1.34*10^-5

c = -5.36*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 = 2.144*10^-6

roots are :

x = 7.255*10^-4 and x = -7.389*10^-4

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

x = 7.255*10^-4

So, [H+] = x = 7.255*10^-4 M

use:

pH = -log [H+]

= -log (7.255*10^-4)

= 3.1394

Answer: 3.14