Question

pH of a saturated solution of Ca(OH)₂ is 9. The solubility product (Ksp) of Ca(OH)₂ is .
(1) 0.5 × 10⁻¹⁵
(2) 0.25 x 10⁻¹⁰
(3) 0.125 × 10⁻¹⁵
(4) 0.5 × 10⁻¹⁰

Answer 1

The solubility product (Ksp) of Ca(OH)₂ is  0.5 × 10⁻¹⁵.

• The chemical equation is Ca(OH)2 ⇌ Ca2+ + 2OH¯
• The solubility product (Ksp) expression Ksp = [Ca2+] [OH¯]2
• Now using the pH we get the pOH = 14.00 - 9 = 5
• Using the pOH value to get the [OH¯]; we know  [OH¯] = $10^{-pOH}$ = $10^{-5}$  M
• From the chemical equation, we note that the [Ca2+] is half the value of the [OH¯], therefore: [Ca2+] = $10^{-5}$ M divided by 2 = 0.5 x 10^(-5) M
• We now have the necessary values to put into the Ksp expression.

• Ksp = $(10^{-5})^{2}*(0.5* 10^{-5})$
• Hence Ksp = 0.5 × 10⁻¹⁵