Answer
pH of buffer = 10.57
Work Step by Step
1
(A) Given :
1. [conjugate acid] = [CH3NH3+] = 0.60 M
2. [base] = [CH3NH2] = 0.50 M
3. pKb of Base CH3NH2 = 3.35
2
(B) Formulae :
1. Henderson - Hasselbalch equation of a buffer : pH = pKa + log [base] / [acid]
2. pKa + pKb = pKw = 14
3
(C) Solution :
1. pKa = pKw - pKb
= 14 - 3.35
= 10.65 .... (1)
2. pH = pKa + log [base] / [acid]
= 10.65 + log (0.50 / 0.60)
= 10.65 + (-0.08)
= 10.57
Hence, pH of the buffer is 10.57
4
(D) Answer :
pH of buffer = 10.57
