Answer:
The molar concentration of NH₃ required is 0,54M
Explanation:
In this problem you need to sum the two reactions, thus:
NiC₂O₄(s) ⇄ Ni²⁺ + C₂O₄²⁻ ksp: 4x10⁻¹⁰
Ni²⁺ + 6NH₃ ⇄ Ni(NH₃)₆²⁺ kf: 1,2x10⁹
The sum of the reactions is:
NiC₂O₄(s) + 6NH₃ ⇄ Ni(NH₃)₆²⁺ + C₂O₄²⁻ k' = kf×ksp = 0,48
The equation for this equilibrium is:
[tex]0,48 = \frac{[Ni(NH_{3})_{6}^{2+}][C_{2}O_{4}^{2-}]}{[NH_{3}]^6}[/tex]
If we are lloking for the complete dissolution of 0,11mol/L of NiC₂O₄(s), the molar concentrations of Ni(NH₃)₆²⁺ and C₂O₄²⁻ are 0,11M, replacing:
[tex]0,48 = \frac{[0,11]^2}{[NH_{3}]^6}[/tex]
The molar concentration of NH₃ required is:
[tex][NH_{3}]^6 = \frac{[0,11]^2}{0,48}[/tex]
[NH₃]⁶ = 0,0252
[NH₃] = ⁶√0,0252
[NH₃] = 0,54M
I hope it helps!
The concentration of NH₃ required to just dissolve 0.011 mol of NiC₂O₄ is; 0.25 M
We are given the reactions;
NiC₂O₄(s) ⇄ Ni²⁺ + C₂O₄²⁻
ksp = 4 * 10⁻¹⁰
Ni²⁺ + 6NH₃ ⇄ Ni(NH₃)₆²⁺
kf = 1.2 * 10⁹
The sum of the two given reactions is:
NiC₂O₄(s) + 6NH₃ ⇄ Ni(NH₃)₆²⁺ + C₂O₄²⁻
k' = kf × ksp
k' = 1.2 * 10⁹ * 4 * 10⁻¹⁰
k' = 0.48
The equation for this equilibrium reaction is:
k' = [Ni(NH₃)₆²⁺][C₂O₄²⁻ ]/[NH₃]⁶
The molar concentrations of Ni(NH₃)₆²⁺ and C₂O₄²⁻ are 0.011M. Thus;
0.48 = [0.011]²/[NH₃]⁶
NH₃ = 0.25 M
Read more about Molar concentration at; https://brainly.com/question/2201903
