To what final concentration of nh3 must a solution be adjusted to just dissolve 0.060 mol of nic2o4 (ksp = 4×10−10) in 1.0 l of solution? the kf for ni(nh3)62+ = 1.2 x 109

Given:
0.060 mol of NiC2O4
Ksp = 4 x 10⁻¹⁰
1.0 L of solution
Kf of Ni(NH3)6 2⁺ = 1.2 x 10⁹
NiC2O4 + 6NH3 ⇋ Ni(NH3)6 2+ + 2O4 2- 
NiC2O4 ⇋ Ni 2+ + C2O4 2- ...Ksp 
Ni2+ + 6NH3 ⇋ Ni(NH3)6 2+...Kf 

Ksp * Kf = (4 x 10⁻¹⁰) * (1.2 x 10⁹) = 0.48

K = 0.48 = [Ni(NH3)6 2+][C2O4 2-] / [NH3]⁶ 
0.48 = (0.060)² / [NH3]⁶ ... (dissolved C2O4 2- = 0.060M) 
[NH3]⁶ = (0.060)² / 0.48 = 0.0036 / 0.48 = 0.0075
NH3 = ⁶√0.0075 

NH3 = 0.44 M

Hno3(aq) → h+(aq) + no3−(aq). picture ii best represents the strong acid hno3. strong acids are strong electrolytes