The solubility product constant of calcium hydroxide is 6.5x10-6. if 0.30 mol of sodium hydroxide is added to 2 l of 0.0010m ca(oh)2, what is the final concentration of the calcium ion? show your work.

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

A) the enthralling of formation for pure element in its standard state is always positivehope this : )