In: Chemistry
If a buffer solution is 0.160 M in a weak acid (Ka = 1.0 × 10-5) and 0.470 M in its conjugate base, what is the pH?
There are two ways to solve this problem.
First method:
Firatwe should calculate pKa value
pKa = -log Ka = -log (1.0 × 10-5)
pKa = 5
We know pKa value, weak acid concentration (0.160 M) and conjugate base concentration (0.470 M)
According to Henderson-Haselbalch equation: pH = pKa + log [base]/[acid]
Substitute all value in top equation
pH = 5 + log [0.470]/[0.160] = 5 + log [2.9375] = 5+0.467
pH = 5.467
Second method:
pH = -log[H+]
we have to calculate H+ ion concentration
the equilibrium equation between the weak acid and its conjugate
acid is:
AH <--> H++ A-
we know Ka value (1.0 x 10-5)
Ka= [H+][A-]/[AH] = 1.0 X
10-5
we know weak acid concentration ([AH] = 0.160 M) and conjugate base
concentration ([A-]=[0.470 M) substitute this values in
top equation
Ka= [H+][0.47]/[0.16] = 1.0 X 10-5
1.0 X 10-5 = (0.470) [H+] / (0.160)
[H+] = 3.40 X 10-6
then pH = -log[H+]
pH = - log 3.40 X 10-6 = 5.46