Question

If the concentration of Mg2+ in the solution were 0.039 M, what minimum [OH−] triggers precipitation of the Mg2+ ion? (Ksp=2.06×10−13.) Express your answer to two significant figures and include the appropriate units. nothing nothing

Answer 1

The minimum concentration of OH- ions to trigger the precipitation of Mg2+ given a Ksp of 2.06 × 10−13 is 2.3 × 10−4 M.

Step-by-step explanation:

To determine the minimum concentration of OH- ions that will trigger the precipitation of Mg2+, we can use the solubility product constant (Ksp) for Mg(OH)2. The Ksp expression for Mg(OH)2 is:

Ksp = [Mg2+][OH-]2

Given that Ksp = 2.06 × 10−13 and [Mg2+] = 0.039 M, we can solve for [OH-]:

2.06 × 10−13 = (0.039)[OH-]2

[OH-] = sqrt(2.06 × 10−13 / 0.039)

After calculating, [OH-] = 2.3 × 10−4 M. Therefore, the minimum concentration of OH- ions to trigger precipitation is 2.3 × 10−4 M.

Answer 2

2.30 × 10⁻⁶ M

Step-by-step explanation:

Step 1: Given data

Concentration of Mg²⁺ ([Mg²⁺]): 0.039 M

Solubility product constant of Mg(OH)₂ (Ksp): 2.06 × 10⁻¹³

Step 2: Write the reaction for the solution of Mg(OH)₂

Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

Step 3: Calculate the minimum [OH⁻] required to trigger the precipitation of Mg²⁺ as Mg(OH)₂

We will use the following expression.

Ksp = 2.06 × 10⁻¹³ = [Mg²⁺] × [OH⁻]²

[OH⁻] = 2.30 × 10⁻⁶ M