Question

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

Answer 1

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

Answer 2

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 $\mathbf{K_c}$ 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:

$\mathbf{K_c = \dfrac{[Mg^{2+}_{(aq)}] [OH^-]^2}{[Mg(OH)_2_{(s)}]} }$

• The concentration of the solid substance of  [Mg(OH)₂] =  1.0 M
• Concentration of [OH]⁻ = no of moles/volume
• = 0.170 M/1.0 L
• = 0.170 M

∴

$\mathbf{1.8 \times 10^{-11} = \dfrac{[Mg^{2+}_{(aq)}] (0.170)^2}{1} }$

$\mathbf{ [Mg^{2+}_{(aq)}] = \dfrac{ 1 .8 \times 10^{-11} }{(0.170)^2} }$

$\mathbf{ [Mg^{2+}_{(aq)}] =6.22 \times 10^{-10}}$

$\mathbf{ [Mg^{2+}_{(aq)}] =0.622 \times 10^{-9}}$

Learn more about the equilibrium constant here:

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