Answer
See explanation below.
Work Step by Step
1
(A) Given :
1. [base] = [B] = 0.40 M
2. [conjugate acid] = [BH+] = 0.25 M
3. pH of buffer = 8.88
2
(B) Formulae :
1. Henderson - Hasselbalch equation of a buffer :
pOH = pKb + log [conjugate acid] / [base]
2. pH = 14 - pOH
3
(C) Solution :
1. pH = 14 - pOH
So, pOH = 14 - 8.88 = 5.12
2. pOH = pKb + log [conjugate acid] / [base]
So, 5.12 = pKb + log (0.25 / 0.40)
pKb = 5.12 - (-0.20) = 5.32
3. [HCl] added = Moles added / Volume
[HCl] = 0.0020 moles / 0.25 L = 0.008 M HCl
4. Now, on addition of 0.008 M HCl, the acid concentration will increase by 0.008 M and base concentration will decrease by 0.008 M.
So,
new [B] = 0.40 - 0.008 = 0.392 M
new [HB+] = 0.25 + 0.003 = 0.258 M
new pOH = pKb + log [conjugate acid] / [base]
= 5.32 + log(0.258 / 0.392)
= 5.32 + (-0.182)
= 5.14
pH = 14 - pOH = 14 - 5.14 = 8.86
Hence, the new pH of the buffer is 8.86.
