Question

From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.011 mol of NiC2O4 (Ksp = 4×10−10) in 1.00 L of solution. (Hint: You can neglect the hydrolysis of C2O42− because the solution will be quite basic.)

Answer 1

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:

$0,48 = \frac{[Ni(NH_{3})_{6}^{2+}][C_{2}O_{4}^{2-}]}{[NH_{3}]^6}$

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:

$0,48 = \frac{[0,11]^2}{[NH_{3}]^6}$

The molar concentration of NH₃ required is:

$[NH_{3}]^6 = \frac{[0,11]^2}{0,48}$

[NH₃]⁶ = 0,0252

[NH₃] = ⁶√0,0252

[NH₃] = 0,54M

I hope it helps!