Question

Net-Ionic Equation for Hydrolysis? Expression for equilibrium constant (Ka or Kb)? Value of Ka or Kb? for NaC2H3O3, Na2CO3, NH4Cl, ZnCl2, KAl(SO4)2

Answer 1

NaC₂H₃O₃ = NaC₂H₃O₂??? Is it sodium acetate?

First, when Sodium Acetate (CH₃COONa) is dissolved in water, you get the reaction

CH₃COONa     CH₃COO^- + Na⁺

Next, because Na+ is the conjugate acid of a strong base (NaOH being the base), Na+ is a negligible acid. Therefore, Na+ will only act as a spectator ion and can be removed from the net ionic equation.

Then, because we know that Na+ will not hydrolyze water because it's a neglible acid, and also because we know that CH₃COO^- will hydrolyze water (CH₃COO^- is the conjugate base of a weak acid and will thus hydrolyze water) we get the reaction

CH₃COO^- + H₂O CH₃COOH + OH^-

And this reaction is the hydrolysis reaction for Sodium Acetate in water.

Ka = ([CH₃COOH] [ OH^- ]) / [CH₃COO^- ]

Na₂CO₃

Carbonate ion is derived from carbonic acid, the acid obtained when carbon dioxide gas is dissolved in water. This acid has two acidic protons, so that means there are two hydrolysis reactions: The equilibrium constants are Kb, which are the reciprocal of the corresponding Ka:

First hydrolysis:
CO₃^2- + H₂O HCO₃^- + OH^-
Kb = [HCO₃^-] [OH^-] / [CO₃^2-]
Kb = 1/Ka₂ = 1 / (5.6x10^-11) = 1.8 x 10¹⁰

Second Hydrolysis:
HCO₃^- + H₂O H₂CO₃ + OH^-

Kb = [H₂CO₃] [OH^-] / [HCO₃^-]
Kb = 1/Ka₁ = 1 / (2.55 x 10^-4) = 4 x 10³

Overall:
CO₃^2- + 2H₂O → H₂CO₃ + 2OH^-
Kb = [H2CO3] [OH^-]² / [CO₃^2-]

Kb = 1/Ka₁Ka₂

NH₄Cl

NH₄Cl + H₂O NH₄OH + HCl

NH₄Cl dissociates. All you are interested in is the ammonium ion
NH₄⁺ + H₂O NH₃ + H₃O⁺
Ka = [NH₃] [H₃O⁺ ] / [NH₄⁺]
Ka = 1 x 10^-14 / Kb = 1 x 10^-14 / 1.8 x 10^-5 = 5.5 x 10^-10

ZnCl₂
Zn²⁺ + H₂O ZnO + 2 H⁺

Only the Zn²⁺ hydrolyzes
Zn²⁺ + H₂O ZnOH⁺ + H⁺
Kb = [ZnOH⁺] [H⁺] / [Zn²⁺]

Zinc ions don't just precipitate as zinc hydroxide in water.
Also, ZnO won't form in aqueous solution, it would be Zn(OH)2, but only if there was an excess of OH-, not for aqueous zinc ions.

KAl(SO₄)₂
Al³⁺ + 3 OH^- Al(OH)₃

It's the aluminum ion that hydrolyzes
Al³⁺ + H₂O AlOH²⁺ + H⁺
Kb = [AlOH²⁺] [H⁺] / [Al³⁺]
There is no excess of OH- to make Al(OH)3. Dissolving aluminum chloride in water won't produced aluminum hydroxide precipitate. What you get are hydrated aluminum ions: [Al(H2O)₆]³⁺.

These [Al(H2O)₆]³⁺ react with water to make [Al(H₂O)5OH]²⁺ that's why I said [AlOH]²⁺. It's simply a "stripped down", simplified version.

Also:
SO₄^2- + H₂O   HSO₄^- + OH^-
Ka = [HSO4^-] [OH^-] / [SO₄^2-] = 1.2 x 10^-2