Calculate the pH of a 0.59 M CH3COONa solution.?

Question

need this for chemistry thanks

Answer 1

CH3COONa>>CH3COO- + Na+
When CH3COO- is put in water it picks up H+ to form CH3COOH. But as soon as some H+ is withdrawn from the solution this way, the water dissociates some more to try to replace that H+. The reaction would be
H2O>>H+ + OH-
If we assume that H2O thus furnishes all the H+ used up in converting CH3COO- in CH3COOh then we can add the reactions to get the net reaction
CH3COO- + H2O >> CH3COO- + OH-
The equilibrium constant callede hydrolysis constant can be written :
Kh= CH3COOH OH-/ CH3COO-= Kw/Kdiss
Let x=moles/l CH3COO- that hydrolyze by the net reaction.
This will give us at equilibrium x moles/l CH3COOH and x moles/l OH- while reducing the concentration of CH3COO- to 0.59-x
1.10^-14/1.8 10^-5=5.6 10^-10 =Kh= x^2/0.59-x
x=0.000018 = OH- So pOH=4.7 >>pH=9.25

Answer 2

From the Ka for ethanoic acid, work out the Kb for the ethanoate anion. Now use [OH-] = root (Kb x molarity). Then use Kw to calculate [H+], and hence the pH.

This type of calculation is now considered to be too difficult for British 18-year-olds, but used to be frequent in A-level examinations.