For Mg(OH)₂,K_sp = 6 × 10−¹².What will occur if 1.0 L of 0.010 M Mg(NO₃)₂ is prepared at pH 10.00?

Question

Quizplus · Accepted Answer

First, let's write the balanced equation for the reaction:

Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH-(aq)

The expression for the solubility product constant is:

Ksp = [Mg2+][OH-]^2

Now, let's calculate the value of Q, the reaction quotient:

Q = [Mg2+][OH-]^2

We know that the initial concentration of Mg2+ is zero, because we are adding Mg(OH)2 to water. We also know that [OH-] = 10^-4 (since pH = 10.00, [H+] = 10^-10, and Kw = [H+][OH-] = 10^-14).

Therefore, Q = 0.

Since Q < Ksp, a precipitate will form. The best choice is A.

For Mg(OH)2,Ksp = 6 × 10−12

For Mg(OH)₂,K_sp = 6 × 10−¹²