700mL * 1.1M = 1400mL * C.finale
da cui:
C.finale = 700mL * 1.1M / 1400mL = 0.55 M
1400mL di HCOOH 0.55M
+
500mL di NaOH 0.8M
TOTALE
1900mL di HCOOH = 0.55M * 1400ml/1900ml = 0.40 M
e
NaOH = 0.8M*500ml / 1900ml = 0.21 M

avviene:

HCOOH + NaOH --------> HCOONa + H2O
0.40_____0.21_________________
0.40-0.21__0___________0.21_____
___0.19_______________0.21

tampone:
[H3O+] = 2,1 x 10^-4*0.19/0.21 = 1.9*10^-4
pH = - log [H3O+]
pH= 3,7