Prima:
HBrO <-- --> H+ + BrO-
0.05__________0____0_all’inizio
0.05-X________X____X_all’equilibrio

Ka = [H+]*[BrO-] / [HBrO]
=
2*10^-9 = X^2 / (0.05-X)
…
X^2 + 2*10^*9*X – 10^-10 = 0
X = (-2*10^-9 + (4*10^-18 + 4*10^-10)^1/2)/2 = 10^-5
[H+] = X = 10^-5
=> pH = 5
Dopo:
[Ba(OH)2] = 0.25mol / 2L = 0.125M

Ba(OH)2 + 2HBrO --> Ba++ 2BrO- + 2H2O
0.125_____0.05_____________0________inizio
0.125-X___0.05-2X__________2X_______fiene

HBrO sembrerebbe in difetto => 0.05-2X = 0
=> X = 0.05/2 = 0.025
=> alla fine abbiamo:
Ba(OH)2 = 0.125-0.025 = 0.1M, base forte
e
BrO- = 2*0.025 = 0.05M, base debole
La base forte influisce negativamente sull’idrolisi basica dello ione ipobromito
Ba(OH)2 --> Ba++ + 2OH-
0.1_________________0_
0__________________0.2_

BrO- + H2O <- -> HBrO + OH-
0.05_____________0_____0.2_
0.05-X___________X____0.2+X_

Ki = 10^-14/Ka = X*(X+0.2) / (0.05-X)
=> X = 1.25*10^-6 trascurabile rispetto a 0.2
=> [OH-] = 0.2 => pOH = 0.7 => pH = 14-0.7 = 13.3