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>> Hydrolysis of Amphiprotic Anion-Part1
Hydrolysis of Amphiprotic Anion:
Let us consider hydrolysis of amphiprotic anion only, i.e., when counter cation is not hydrolysed, example of some salts of this category are NaHCO3, NaHS, Na2HPO4, NaH2PO4.
Here, H2PO-4 and HPO2-4 are amphiprotic anions. pH after their hydrolysis can be calculated as,
pH of H2PO-4 in aqueous medium = (pka1 + pka2)/2
pH of H2PO2-4 in aqueous medium = (pka2 + pka3)/2
Here, H2PO2-4 is conjugate base of H2PO-4 and H3PO4 is conjugate acid of H2PO-4.
Similarly, PO3-4 is conjugate base of HPO-24 and HPO-4 is conjugate acid of PO3-4 .
(iv) Let us consider amphiprotic bicarbonate anion.
pH HCO-3of ion after hydrolysis in aqueous medium
= (pka1 + pka2)/2
(v) Let us consider the hydrolysis of amphiprotic anion along with cation, e.g., NH4HCO3, NH4HS.
In above examples both cations and anions are derived from weak base and weak acids respectively hence, both will undergo hydrolysis in aqueous medium.
When these salts are dissolved in water, [H3O+] concentration can be determined as,
[H3O+] = √ka1[kw/kb + ka2]
pH = -log = √ka1[kw/kb + ka2]
Hydrolysis at a Glance
(Strong acid + Strong Base)
2. Ch3COONa (Weak acid + Strong base)
(Strong acid + Weak base)
4. CH3COONH4 (Weak acid + Weak base)
h = √kw/Cka
h = √kw/Ckb
h = √kw/(ka + kb)
Kh = kw/ka
Kh = kw/Ckb
Kh = kw/(ka + kb)
pH=1/2[pkw + pka + logC]
pH=1/2[pkw- pkb - logC]
pH=1/2[pkw + pka - pkb]
In the case of salt of weak acid and weak base, nature of medium after hydrolysisis decided in the following manner:
(i) If Ka = Kb, the medium will be neutral.
(ii) If Ka > Kb, the medium will be acidic.
(iii) If Ka < Kb, the medium will be basic.
The degree of hydrolysis of salts of weak acids and weak bases is unaffected by dilution because there is no concentration term in the expression of degree ofhydrolysis.
Note : Degree of hydrolysis always increases with increase in temperature because at elevated temperature increase in Kw is greater as compared to Ka and Kb.