Respuesta :

Louli
The given acid is HCl of 0.28 M which means that:
[H+] = 0.28
pH = -log[H+] = -log[0.28] = 0.5528
pOH = 14 - pH
pOH = 14 - 0.5528 = 13.447
pOH = -log[OH-]
log[OH-] = -13.447
[OH-] = 10^-13.447 = 3.5727*10^-14