Answer
pH of buffer = 5.25
Hydronium ion concentration = 5.5 x 10^-6 M
Work Step by Step
1
(A) Given :
1. [conjugate base] = [CH3CH2COO-] = 0.35 M
2. [acid] = [CH3CH2COOH] = 0.15 M
3. Ka of propanoic acid = 1.3 X 10^-5
2
(B) Formulae :
1. Henderson - Hasselbalch equation of a buffer : pH = pKa + log [conjugate base] / [acid]
2. pKa = -log Ka
3. pH = -log [H+]
3
(C) Solution :
1. pKa = -log Ka
= - log (1.3 x 10^-5)
= 4.89 .... (1)
2. pH = pKa + log [conjugate base] / [acid]
= 4.89 + log (0.35 / 0.15)
= 5.25
Hence, pH of the buffer is 5.26.
3. pH of the solution = -log [H+]
So,
[H+] = 10^(-pH)
= 10^(-5.25)
= 5.6 X 10^-6 M
4
(D) Answer :
pH of buffer = 5.25
Hydronium ion concentration = 5.5 x 10^-6 M
