Answer
pH of buffer = 3.28
Hydronium ion concentration = 5.2 x 10^-4 M
Work Step by Step
1
(A) Given :
1. [conjugate base] = [NO2-] = 0.75 M
2. [acid] = [HNO2] = 0.55 M
3. Ka of HNO2 = 7.1 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 (7.1 X 10^-4)
= 3.15 .... (1)
2. pH = pKa + log [conjugate base] / [acid]
= 3.15 + log (0.75 / 0.55)
= 3.15 + 0.13
= 3.28
Hence, pH of the buffer is 3.28.
3. pH of the solution = -log [H+]
So,
[H+] = 10^(-pH)
= 10^(-3.28)
= 5.2 x 10^-4 M
4
(D) Answer :
pH of buffer = 3.28
Hydronium ion concentration = 5.2 x 10^-4 M
