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.)

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:

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:

The molar concentration of NH₃ required is:

[NH₃]⁶ = 0,0252

[NH₃] = ⁶√0,0252

[NH₃] = 0,54M

I hope it helps!

"hydrogen bonding is the attraction of h atoms in the water molecule (slight + charge) to o atoms in a neighboring water molecule (slight - charge); it is a weak interatomic force. but it is still strong enough to the water molecules "stick" together."