Question

If the concentration of Mg^2+ in the solution were 0.021 M, what minimum [OH^−] triggers precipitation of the Mg^2+ ion? (Ksp=2.06×10^−13.)

A) [OH^−] = 9.52 × 10^−13 M
B) [OH^−] = 4.76 × 10^−13 M
C) [OH^−] = 2.38 × 10^−13 M
D) [OH^−] = 1.19 × 10^−13 M

Answer 1

Using the Kₛₚ value for Mg(OH)₂ and the given Mg²⁺ concentration, the minimum [OH⁻] that causes precipitation is calculated to be approximately 4.94 × 10⁻· M, which is not among the provided options.

Step-by-step explanation:

To determine the minimum hydroxide ion concentration [OH−] that triggers the precipitation of Mg2+, we use the solubility product constant (Ksp) for Mg(OH)2, which is given as 2.06 × 10−13. The formula to calculate Ksp is [Mg2+][OH−]2. Given that the concentration of Mg2+ in the solution is 0.021 M, we can set up the equation 2.06 × 10−13 = (0.021 M)[OH−]2. Solving for [OH−], we get [OH−] = √(2.06 × 10−13 / 0.021 M), which gives us [OH−] = 9.88 × 10−7 M. To get the minimum concentration that causes precipitation, we divide this by 2 because the stoichiometry of OH− in the dissolution of Mg(OH)2 is 2:1. Thus, the minimum [OH−] is approximately 4.94 × 10−7 M. None of the given options are correct.