Question

A. You have 100.0 mL of 0.050 M acetic acid in water at 25 C. What are thre pH and pOH of this solution.

B.What is the pKa for acetic acid?

C.The pKa of trifluoroacetic acid (CF3CO2H) is 0.23. Which is a stronger acid? that or acetic acid?

Answer 1

A)

Ka of CH3COOH = 1.8*10^-5

CH3COOH dissociates as:

CH3COOH -----> H+ + CH3COO-

5*10^-2 0 0

5*10^-2-x x x

Ka = [H+][CH3COO-]/[CH3COOH]

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 ((1.8*10^-5)*5*10^-2) = 9.487*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)

1.8*10^-5 = x^2/(5*10^-2-x)

9*10^-7 - 1.8*10^-5 *x = x^2

x^2 + 1.8*10^-5 *x-9*10^-7 = 0

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

a = 1

b = 1.8*10^-5

c = -9*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 = 3.6*10^-6

roots are :

x = 9.397*10^-4 and x = -9.577*10^-4

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

x = 9.397*10^-4

So, [H+] = x = 9.397*10^-4 M

use:

pH = -log [H+]

= -log (9.397*10^-4)

= 3.027

pOH = 14 - pH

= 14 - 3.027

= 10.97

pH = 3.03

pOH = 10.97

B)

use:

pKa = -log Ka

= -log (1.8*10^-5)

= 4.7447

Answer: 4.74

C)

The acid with smaller pKa is stronger

pKa is smaller for CF3CO2H

So, CF3CO2H is stronger acid

Answer: CF3CO2H