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 :
H2CO3(aq) + H2O(l) → HCO3-(aq) + H3O+(aq)
HCO3-(aq) + H2O(l) → (CO3)2-(aq) + H3O+(aq)
As the buffer contains KHCO3 and K2CO3, i.e. HCO3- and (CO3)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. [conjugate base] = [(CO3)2-] = 0.37 M
2. [acid] = [KHCO3] = 0.22 M
3. Ka2 = 4.7 X 10^-11
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 (4.7 x 10^-11)
= 10.33 .... (1)
2. pH = pKa + log [conjugate base] / [acid]
= 10.33 + log (0.37 / 0.22)
= 10.33 + 0.22
= 10.55
Hence, pH of the buffer is 10.55.
5
(D) Answer :
pH of buffer as calculated above = 10.55
