a). de acid tare 0.0001 moli/litru
Cm = 10^-4 mol/L
fiind acid tare [H3O+] = Cm
stim ca pH = -lg[H3O+] sau [H3O+] = 10^-pH
=> 10^-4 = 10^-pH => pH = 4
solutie cu caracter acid
b).
Cm = 10^-5 mol/L
fiind acid tare [HO-] = Cm
stim ca pH = -lg[H3O+] sau [HO-] = 10^-pOH
=> 10^-5 = 10^-pOH => pOH = 5
pH + pOH = 14
=> pH = 14 - 5 = 9
=> solutie bazica