Sol.
As Conc. of HA = [HA] = 1.50 M
Conc. of NaA = [NaA] = 2.25 M
Ka = 8.90 × 10-6
pKa = - log(Ka) = - log( 8.90 × 10-6 ) = 5.05
So , Using Henderson - Hasselbalch equation ,
pH = pKa + log ( [NaA] / [HA] )
= 5.05 + log ( 2.25 / 1.50 )
= 5.23
Also , Hydronium Concentration at equilibrium
= [H3O+] = 10-pH
= 10-5.23
= 5.89 × 10-6 M
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