At equilibrium, a 1.0 L reaction vessel contains 2.3 mol of Mg(OH)₂ and 0.170 mol of OH⁻. What is the equilibrium concentration of Mg²⁺ based on the reaction:
Mg(OH)₂ (s) ⇌ Mg²⁺ (aq) + 2 OH⁻ (aq) Kc = 1.80 x 10⁻¹¹

At equilibrium, a 1.0 L reaction vessel contains 2.3 mol of Mg(OH)₂ and 0.170 mol of OH⁻. The equilibrium concentration of Mg²⁺ based on the reaction is 0.622 × 10⁻⁹.

The equilibrium concentration relates to the summation of the initial concentration and the change gotten from its stoichiometric reaction.

From the reaction given, the equilibrium concentration can be represented as:

∴

Learn more about the equilibrium constant here:

link

The value of X for this chemical reaction is 1.65 x 10⁻⁴

Explanation:

Mg (OH)₂(s) <==> Mg²⁺+(aq) + 2OH₋₁(aq)  

Kc= 1.80x10⁻¹¹

KC = [Mg²⁺][OH-]²

Let us assume

Put X = [Mg²⁺] and  2x = [OH-]

1.80x10-11 = (X) (2X)²

1.80x10-11 = 4X³

X³ = 4.5x10⁻¹²

On simplifying the equation we get,

X = 1.65 x 10⁻⁴

The value of X for this chemical reaction is 1.65 x 10⁻⁴

c u c l 2 + 2 n a n o 3 → c u ( n o 3 ) 2 + 2 n a c l

  long weird answer but i hope this works

3 hours because it is a decent distance.