NaOH fully dissociates into its ions as Na⁺ and OH⁻.
NaOH(s) → Na⁺(aq) + OH⁻(aq)
Molarity of NaOH = moles / volume
= 0.30 / 2 L
= 0.15 mol/L
Ca(OH)2 partially dissociates into its ions as Ca²⁺ and OH⁻.
Ca(OH)₂(S) ⇄ Ca²⁺(aq) + 2OH⁻(aq)
When adding OH⁻ moles into the system, the backward reaction is promoted.
The total [OH⁻(aq)] = [OH⁻(aq)] from Ca(OH)₂ + [OH⁻(aq)] from NaOH
By applying ICE table to the equilibrium,
Ca(OH)₂(s) ⇄ Ca²⁺(aq) + 2OH⁻(aq)
Initial 0.0010 - 0.15
Change -X +X +2X
Equilibrium 0.0010 - X X 0.15 + 2X
Ksp = [Ca²⁺(aq)] [OH⁻.(aq)]²
6.5 x 10⁻⁶ = X * (0.15 + 2X)²
since Ksp is too small, 0.15 + 2X = 0.15
6.5 x 10⁻⁶ = X * (0.15)²
X = 6.5 x 10⁻⁶ / (0.15)²
X = 2.8 x 10⁻⁴ M
Hence,final [Ca²⁺(aq)] is 2.8 x 10⁻⁴ M