Mg(OH)2 is a sparingly soluble compound, in this case a base, with a solubility product, Ksp, of 5.61×10−11. It is used to control the pH and provide nutrients in the biological (microbial) treatment of municipal wastewater streams.

Based on the given value of the Ksp, what is the molar solubility of Mg(OH)2 in pure H2O?

1.27 x 10⁻⁴M

Explanation:

For a 2 to 1 ionization ratio, solubility in pure water can be calculated using the formula S = ∛(Ksp/27) = ∛(5.61 x 10⁻¹¹/27 = 1.27 x 10⁻⁴M.

Mg(OH)₂ ⇄ Mg⁺² + 2OH⁻ => Ksp = [Mg⁺²][OH⁻]² = (x)(2x)² = 4x³

Solve for 'x' => x = Solubility = ∛Ksp/4

c. 75.7

i think that is the right

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error