Answer
pH of buffer = 3.27
Hydronium ion concentration = 5.4 x 10^-4 M
Work Step by Step
1
(A) Given :
1. [conjugate base] = [F-] = 0.25 M
2. [acid] = [HF] = 0.20 M
3. Ka of HF = 6.8 x 10^-4
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 (6.8 x 10^-4)
= 3.17 .... (1)
2. pH = pKa + log [conjugate base] / [acid]
= 3.17 + log (0.25 / 0.20)
= 3.17 + 0.10
= 3.27
Hence, pH of the buffer is 3.7
3. pH of the solution = -log [H+]
So,
[H+] = 10^(-pH)
= 10^(-3.27)
= 5.4 x 10^-4 M
4
(D) Answer :
pH of buffer = 3.27
Hydronium ion concentration = 5.4 x 10^-4 M
