Question

A 52.0-ml volume of 0.35M CH3COOH (Ka= 1.8*10^-5) is titrated with 0.40M NaOH . Calculate the pH after the addition of 21.0ml of NaOH.

Answer 1

Ka= 1,8•10^-5 
pKa = - log(Ka) = 4,74 


Equivalence volume : 

Ca x Va = Cb x Vb 

Vb = Ca x Va / Cb 
Vb = 0,35 x 52,0 / 0,40 
Vb = 45,5 mL 


21,0 mL < 45,5 mL 

. . . then : 


pH = pKa / 2 - log( (Ca x Va - Cb x V) / (Va + V)) / 2 
pH = pKa / 2 - log( (0,35 x 52 - 0,40 x 21) / (52 + 21)) / 2 
pH = 3,24