Answer
pH of buffer = 4.13
Hydronium ion concentration = 7.413 x 10^-5 M
Work Step by Step
1
(A) Given :
1. [conjugate base] = [C6H5COO-] = 0.28 M
2. [acid] = [C6H5COOH] = 0.33 M
3. Ka of Benzoic acid = 6.3 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 (6.3 x 10^-5)
= 4.20 .... (1)
2. pH = pKa + log [conjugate base] / [acid]
= 4.20 + log (0.28 / 0.33)
= 4.13
Hence, pH of the buffer is 4.13.
3. pH of the solution = -log [H+]
So,
[H+] = 10^(-pH)
= 10^(-4.13)
= 7.413 x 10^-5 M
4
(D) Answer :
pH of buffer = 4.13
Hydronium ion concentration = 7.413 x 10^-5 M
