NaHC2O4 = 1,80g : 112g/mol = 1,6*10^-2 mol

HCl = 0,025M * 0,400L = 10^-2 mol

NaHC2O4 + HCl -----> H2C2O4 + NaCl
1,6*10^-2__10^-2____________________all'inizio
1,6*10^-2-10^-2______10^-2____/____alla fine
=
6*10^-3________________10^-2
=
tampone H2C2O4 / HC2O4-

[H+] = Ka1 * H2C2O4 / HC2O4-

[H+] = 5,6*10^-2 * 10^-2mol / 6*10^-3mol = 9,2*10^-2

pH = 1,04