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
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Ответ:
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:
The concentration of the solid substance of [Mg(OH)₂] = 1.0 MConcentration of [OH]⁻ = no of moles/volume= 0.170 M/1.0 L = 0.170 M∴
Learn more about the equilibrium constant here:
link
Ответ:
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
Ответ:
Hope this would help~