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