Question

Find the pH for the following solutions:

0.01M CH3COOH (acetic acid) in a 50 mL Falcon tube

0.01M HCl in a 50 mL Falcon tube

0.01M NaOH in a 50 mL Falcon tube

Answer 1

1)

Ka for acetic acid is 1.8*10^-5

CH3COOH dissociates as:

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

1*10^-2 0 0

1*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)*1*10^-2) = 4.243*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/(1*10^-2-x)

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

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

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

a = 1

b = 1.8*10^-5

c = -1.8*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 = 7.203*10^-7

roots are :

x = 4.154*10^-4 and x = -4.334*10^-4

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

x = 4.154*10^-4

so.[H+] = x = 4.154*10^-4 M

use:

pH = -log [H+]

= -log (4.154*10^-4)

= 3.38

Answer: 3.38

2)

HCl is strong acid

It will dissociate completely to give, [H+] = 0.01 M

use:

pH = -log [H+]

= -log (1*10^-2)

= 2

Answer: 2.0

3)

NaOH is strong base

It will dissociate completely to give, [OH-] = 0.01 M

use:

pOH = -log [OH-]

= -log (1*10^-2)

= 2

use:

PH = 14 - pOH

= 14 - 2

= 12

Answer: 12.0