Answer
See explanation below.
Work Step by Step
1
(A) Given :
1. [conjugate base] = [A-] = 0.1500 M
2. [acid] = [HA] = 0.2000 M
3. pH of buffer = 3.35
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,
3.35 = pKa + log (0.1500 / 0.2000)
pKa = 3.35 - (-.012)
pKa = 3.47 ...... (1)
2. [NaOH] added = Moles added / Volume
[NaOH] = 0.0015 moles / 0.500 L = 0.003 M NaOH
3. Now, on addition of 0.003 M NaOH, the acid concentration will increase by 0.003 M and base concentration will decrease by 0.003 M.
So,
new [A-] = 0.1500 + 0.003 = 0.1530 M
new [HA] = 0.2000 - 0.003 = 0.1970 M
new pH = pKa + log([A-] / [HA])
= 3.47 + log(0.1530 / 0.1970)
= 3.47 + (-0.06)
= 3.37
Hence, the new pH of the buffer is 3.37.