Titration curve:

Let us calculate the next point of the titration curve for the moment at which 12.5 cm³ of NaOH solution will be poured in:

c_acid = (0.05 × (25 - 12.5))/ 25 + 12.5= 0.017 M

Csalt = (0.05 × 12.5)/ 37.5 =0.017 M;  [H⁺] = (1.75 × 10^-5  × 0.017/0.017)M

pH = -log [H⁺]  =  - log 1.75 ×  10^-5

=  4.76

That is pH = pK_a

We still calculate the pH of the third intermediate point for V = 15 cm³

c_acid =(0.05 × (25 - 15))/(25 + 15)= 0.014 M

c_salt =  (0.05 ×15/35) 0.021 M [H⁺] = (1.75 × 10^-5  × 0.014)0.021

 pH = - log [H⁺] = - log 11.67 × 10^-6   = 5.9

After equivalence point (on addition of 25.01 NaOH): In this situation both the excess NaOH and acetate are source of hydroxide ion. We may consider that the contribution from the acetate ion will be small, because the excess of strong base repress the reaction of acetate with water. We have after that

[OH^- ] ≈ c_NaOH = 0.05 (25.01 - 25 )/25.01+ 25 = 9.99 ×  10 6

pH = 14.00  - [ - log (  9.99 ×  10 ^-6 )   =  8.99

On additional titration, the pH of the solution will be determined just through excess 0.05 M NaOH solution.