The dissociation of Ca(OH)₂ in aqueous medium is given below: Ca(OH)₂(aq) ⇄ Ca⁺²(aq) + 2 OH⁻ The molarity of Ca(OH)₂ = [tex] \frac{6.5 x 10^{-9} mole }{10.0 L} [/tex] = 6.5 x 10⁻¹⁰ M Ca(OH)₂ The solubility product (Ksp) of Ca(OH)₂ = 6.5 x 10⁻⁶ The solubility (S) of Ca(OH)₂ is : Ksp = [Ca⁺²][OH⁻]² = S (2S)² = 4 S³ 6.5 x 10⁻⁶ = 4 S³ S = 0.0117 M [OH⁻] = 2 S = 2 * (0.0117 M) = 0.0234 M Each mole of Ca(OH)₂ gives two moles of OH⁻ ions therefore; [OH⁻] = 2 * (6.5 x 10⁻¹⁰ M) = 1.3 x 10⁻⁹ M So the concentration of [OH⁻] ions less then the saturated [OH⁻] conc. ⇒ The solution is not precipitated Since [OH⁻] < 10⁻⁷ Total conc. [OH⁻] = (10⁻⁷ due to ionization of water + 1.3 x 10⁻⁹ M) = 1.013 x 10⁻⁷ M pOH = - log (1.013 x 10⁻⁷ M) = 6.957 pH = 14 - pOH = 7.043
