Answer
pH of buffer as calculated above = 10.55
Work Step by Step
1
The dissociation of Carbonic Acid (H2CO3) occurs in the following two steps :
H3PO4(aq) + H2O(l) → H2PO4-(aq) + H3O+(aq)
H2PO4-(aq) + H2O(l) → (CO3)2-(aq) + H3O+(aq)
(HPO4)2-(aq) + H2O(l) → (PO4)3-(aq) + H3O+(aq)
As the buffer contains NaH2PO4 and Na2HPO4, i.e. H2PO4- and (PO4)2-. And, because both of these are present in the second reaction, the Ka of 2nd step, that is Ka2 is more significant with respect to this reaction.
2
(A) Given :
1. [base] = [(PO4)2-] = 0.40 M
2. [acid] = [KHCO3] = 0.50 M
3. Ka2 = 6.3 x 10^-8
3
(B) Formulae :
1. Henderson - Hasselbalch equation of a buffer : pH = pKa + log [conjugate base] / [acid]
2. pKa = -log Ka
4
(C) Solution :
1. pKa = -log Ka
= - log (6.3 x 10^-8)
= 7.20 .... (1)
2. pH = pKa + log [conjugate base] / [acid]
= 7.20 + log (0.40 / 0.50)
= 7.20 + (-0.10)
= 7.10
Hence, pH of the buffer is 7.10
5
(D) Answer :
pH of buffer as calculated above = 10.55
