If 1.2 ·10^-14 moles of magnesium hydroxide, mg(oh)2, are dissolved in 1.0 l of aqueous solution, what are [oh-] and [h30+]?

1. [OH⁻] = 2.4 x 10⁻¹⁴ M.

2. [H₃O⁺] = 4.167 x 10⁻¹ M.

Explanation:

Firstly, we need to calculate the molarity of Mg(OH)₂:

M = (no. of moles of solute)/(V of the solution (L).

∴ M of MgCl₂ = (no. of moles of solute)/(V of the solution (L) = (1.2 x 10⁻¹⁴ mol)/(1.0 L) = 1.2 x 10⁻¹⁴ M.

1. [OH⁻]:

Mg(OH)₂ → Mg²⁺ + 2OH⁻.

∴ [OH⁻] = 2(1.2 x 10⁻¹⁴) = 2.4 x 10⁻¹⁴ M.

2. [H₃O⁺]:

∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.

∴ [H₃O⁺] = 10⁻¹⁴/[OH⁻] = 10⁻¹⁴/(2.4 x 10⁻¹⁴ M) = 4.167 x 10⁻¹ M.

answer: your mm

explanation:

where's the diagram?