Answer:
[tex]\large \boxed{\text{ 0.025 mol/L}}[/tex]
Explanation:
The solid Ca(OH)₂ has sunk to the bottom, so you have a saturated solution of calcium hydroxide.
The equation for the equilibrium is
Ca(OH)₂(s) ⇌ Ca²⁺(aq) + 2OH⁻(aq)
Let [Ca²⁺] = x
Then [OH⁻] = 2x
We can now use the solubility product constant expression.
It is difficult to read your value of Ksp. I shall use Ksp = 6.5 × 10⁻⁶. Then
[tex]K_{sp} =\text{[Ca$^{2+}$][OH$^{-}$]}^{2} = x\times (2x)^{2} = 4x^{3} = 6.5 \times 10^{-6}\\x^{3} = \dfrac{6.5\times 10^{-6}} {4} =1.62 \times 10^{-6}\\\\x = \sqrt[3]{1.62 \times 10^{-6}} = \textbf{0.025 mol/L}\\\text{The concentration of Ca$^{2+}$ ions is $\large \boxed{\textbf{ 0.025 mol/L}}$}[/tex]
