Réponse :
pH= 11 : [H3O+]=10^-11 mol/L, [HO-] = 10^-14/10^-11 mol/L
pH= 7 : [H3O+] = 10^-7 mol/L, [HO-] = 10^-14/10^-7 mol/L
pH= 1 : [H3O+] = 10^-1 mol/L, [HO-] = 10^-14/10^-1 mol/L
Explications :
pH= -log[H3O+] <=> [H3O+]= 10^-pH
[HO-] = Ke/[H3O+] avec Ke= 10^-14