Question

find the pH a.) 0.1M propanoic acid (ka=1.3*10^-5) b.)0.1M sodium propanoate c.) 0.1M H20 d.) of a mixture with 0.1M Propionic acid and sodium propanoate

Answer 1

a)
HA is CH3CH2COOH
for simplicity lets write weak acid as HA

HA          ----->     H+   +   A-
0.1000                 0         0
0.1000-x               x         x

Ka = [H+][A-]/[HA]
Ka = x*x/(c-x)
since ka is small, x will be small and it can be ignored as compared to c
So, above expression becomes

Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((1.3E-5)*0.1000) = 1.14E-3

pH = -log [H+] = -log (1.14E-3) = 2.94

b)
A- is CH3CH2COO-

Kb = 10^-14/Ka
= 10^-14/(1.3*10^-5)
= 7.7*10^-10

for simplicity lets write weak base as A-

A-        + H2O   ----->     AH   +   OH-
0.1000                        0         0
0.1000-x                      x         x

Kb = [AH][OH-]/[B-]
Kb = x*x/(c-x)
since kb is small, x will be small and it can be ignored as compared to c
So, above expression becomes

Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((7.7E-10)*0.1000) = 8.77E-6

pOH = -log [OH-] = -log (8.77E-6) = 5.06

PH = 14 - pOH = 14 - 5.06 = 8.94

c)

since water is neutral, pH will be 7

d)
This is buffer.

Ka = 1.30E-5

pKa = - log (Ka)
= - log(1.30E-5)
=4.89

use:
pH = pKa + log {[conjugate base]/[acid]}
= 4.89+ log {0.10/0.10}
=4.89