Answer :
pH = -log[H⁺]
Ca(OH)₂ <---> Ca²⁺ + 2OH⁻
Concentration = 10⁻² mol/l = 0,01 mol/l
pOH = -log[OH⁻]
pOH = -log(0,01) = 2
pH + pOH = 14
pH + 2 = 14
pH = 14 - 2
pH = 12
Ca(OH)₂ <---> Ca²⁺ + 2OH⁻
Concentration = 10⁻² mol/l = 0,01 mol/l
pOH = -log[OH⁻]
pOH = -log(0,01) = 2
pH + pOH = 14
pH + 2 = 14
pH = 14 - 2
pH = 12