Question

1a. CH₃COOH (aq) + H₂O (l)  ⇌  CH₃COO^- (aq) + H₃O⁺ (aq)

If the K value for this equilibrium is 1.80 X 10^-5, what is the pH of a 1.45 M solution of CH₃COOH? (refer to above net equation)

1b. CH₃COO^- (aq) + H₂O (l)  ⇌  CH₃COOH (aq) + OH^- (aq)

If the K_eq for CH₃COOH is 1.80 X 10^-5, what is the pH of a 0.25 M solution of CH₃COONa? Hint: This K value is not the one you need.

Answer 1

a)
CH3COOH dissociates as:

CH3COOH          ----->     H+   + CH3COO-
1.45                 0         0
1.45-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)*1.45) = 5.109*10^-3

since c is much greater than x, our assumption is correct
so, x = 5.109*10^-3 M



so.[H+] = x = 5.109*10^-3 M


use:
pH = -log [H+]
= -log (5.109*10^-3)
= 2.29
Answer: 2.29

b)

use:
Kb = (1.0*10^-14)/Ka
Kb = (1.0*10^-14)/1.8*10^-5
Kb = 5.556*10^-10
CH3COO- dissociates as

CH3COO-        + H2O   ----->     CH3COOH +   OH-
0.25                        0         0
0.25-x                      x         x


Kb = [CH3COOH][OH-]/[CH3COO-]
Kb = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes

Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((5.556*10^-10)*0.25) = 1.179*10^-5

since c is much greater than x, our assumption is correct
so, x = 1.179*10^-5 M



use:
pOH = -log [OH-]
= -log (1.179*10^-5)
= 4.93


use:
PH = 14 - pOH
= 14 - 4.93
= 9.07

Answer: 9.07