Question

Calculate the pH of a 0.19 M CH₃COOLi solution. (K_a for acetic acid = 1.8 × 10⁻⁵.)

Answer 1

use:
Kb = Kw/Ka
Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC
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-
0.19                        0         0
0.19-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)*0.19) = 1.027*10^-5

since c is much greater than x, our assumption is correct
so, x = 1.027*10^-5 M



use:
pOH = -log [OH-]
= -log (1.027*10^-5)
= 4.9883


use:
PH = 14 - pOH
= 14 - 4.9883
= 9.0117
Answer: 9.01