Answer
See explanation below.
Work Step by Step
1
(A) Given :
1. [base] = [B] = 1.05 M
2. [conjugate acid] = [BH+] = 0.75 M
3. pH of buffer = 9.50
2
(B) Formulae :
1. Henderson - Hasselbalch equation of a buffer :
pH = pKa + log [base] / [acid]
3
(C) Solution :
1. pH = pKa + log [conjugate base] / [acid]
So,
9.50 = pKa + log (1.05 / 0.75)
pKa = 9.50 - 0.15
pKa = 9.35 ...... (1)
2. [HCl] added = Moles added / Volume
[HCl] = 0.0050 moles / 0.500 L = 0.010 M HCl
3. Now, on addition of 0.01 M HCl, the acid concentration will increase by 0.01 M and base concentration will decrease by 0.01 M.
So,
new [B] = 1.05 - 0.01 = 1.04 M
new [BH+] = 0.75 + 0.01 = 0.76 M
new pH = pKa + log([B] / [BH+])
= 9.35 + log(1.04 / 0.76)
= 9.35 + 0.14
= 9.49
Hence, the new pH of the buffer is 9.49