Answer
pH of buffer = 9.47
Work Step by Step
1
(A) Given :
1. [conjugate acid] = [NH4+] = 0.15 M
2. [base] = [NH3] = 0.25 M
3. pKb of Base NH3 = 4.75
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 - 4.75
= 9.25 .... (1)
2. pH = pKa + log [base] / [acid]
= 9.25 + log (0.25 / 0.15)
= 9.25 + 0.22
= 9.47
Hence, pH of the buffer is 9.47.
4
(D) Answer :
pH of buffer = 9.47