Question

Consider the reaction: Mg(OH)₂ (s) ⇌ Mg²⁺ (aq) + 2 OH⁻ (aq)
At equilibrium, a 1.0 L reaction vessel contains 5.3 moles of Mg(OH)₂ and concentrations of 0.0080 M and 0.010 M of Mg²⁺ and OH⁻ respectively. What is Kc for this equilibrium?

Answer 1

Answer:

8.0 x 10⁻⁷

Explanation:

[Mg²⁺][OH⁻]² = [0.008] [0.010]² = 8.0 x 10⁻⁷

(Solids and Liquids do not appear in the expression)

Answer 2

The Kc of the system is  8 × 10^-7.

The reaction equation is written as follows; Mg(OH)₂ (s) ⇌ Mg²⁺ (aq) + 2 OH⁻ (aq)

We are told in the question that the Kc of the reaction is what we are to find.

Now;

Kc = [Mg²⁺] [ OH⁻ ]^2

Recall that Mg(OH)₂ does not come into the equilibrium expression because it is a pure solid.

Substituting values;

[Mg²⁺] = 0.0080 M

[ OH⁻ ] = 0.010 M

Kc = [0.0080 M] [0.010 M ]^2

Kc = 8 × 10^-7

Learn more: https://brainly.com/question/2510654