Answer
See explanation below.
Work Step by Step
1
(A) Given :
1. [conjugate base] = [Y-] = 0.220 M
2. [acid] = [HY] = 0.110 M
3. pH of buffer = 8.77
2
(B) Formulae :
1. Henderson - Hasselbalch equation of a buffer :
pH = pKa + log [conjugate base] / [acid]
3
(C) Solution :
1. pH = pKa + log [conjugate base] / [acid]
So,
8.77 = pKa + log (0.220 / 0.110)
pKa = 8.77 - 0.30
pKa = 8.47 ...... (1)
2. [Ba(OH)2] added = Moles added / Volume
[Ba(OH)2] = 0.0015 moles / 0.350 L = 0.004286 M Ba(OH)2
[OH-] added = 2 x [Ba(OH)2] = 2 x 0.004286 M = 0.00857 M
3. Now, on addition of 0.00857 M OH-, the acid concentration will DECREASE by 0.00857 M and base concentration will increase by 0.00857 M.
So,
new [Y-] = 0.2200 + 0.00857 = 0.22857 M
new [HY] = 0.1100 - 0.00857 = 0.10143 M
new pH = pKa + log([Y-] / [HY])
= 8.47 + log(0.22857 / 0.10143)
= 8.47 + 0.35
= 8.82
Hence, the new pH of the buffer is 8.82