To what final concentration of \(\mathrm{NH}_{3}\) must a solution be adjusted to just dissolve \(0.020 \mathrm{~mol}\) of \(\mathrm{NiC}_{2} \mathrm{O}_{4}\left(K_{s p}=4 \times 10^{-10}\right)\) in \(1.0 \mathrm{~L}\) of solution? (Hint: You can neglect the hydrolysis of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) because the solution will be quite basic.)

First, let's write the balanced equation for the dissolution of NiC2O4 in an NH3 solution: \[ \mathrm{NiC}_{2}\mathrm{O}_{4(s)} + 6 \mathrm{NH}_{3(aq)} \rightleftharpoons \mathrm{Ni(NH_{3})_{6}^{2+}(aq)} + \mathrm{C}_{2}\mathrm{O}_{4^{2-}(aq)} \] The Ni(NH3)62+ complex can dissociate into Ni2+ and 6 molecules of ammonia by the following reaction: \[ \mathrm{Ni(NH_{3})_{6}^{2+}(aq)} \rightleftharpoons \mathrm{Ni^{2+}(aq)} + 6 \mathrm{NH}_{3(aq)} \]

Next, we can write the Ksp expression for Nickel Oxalate and the stability constant (Kf) expression for the formation of the Nickel-ammonia complex. Ksp expression for NiC2O4: \[ K_{sp} = [\mathrm{Ni}^{2+}] [\mathrm{C}_{2}\mathrm{O}_{4}^{2-}] \] Stability constant (Kf) expression for Ni(NH3)62+: \[ K_{f} = \frac{[\mathrm{Ni(NH_{3})_{6}^{2+}]}{[\mathrm{Ni}^{2+}][\mathrm{NH}_{3}]^{6}} \]

Let x be the concentration of \(\mathrm{Ni^{2+}}\) and \(\mathrm{C}_{2}\mathrm{O}_{4^{2-}}\) ions, since they will both have the same concentration. Thus, from the Ksp expression, we can write: \[ K_{sp} = (4 \times 10^{-10}) = [\mathrm{Ni}^{2+}][\mathrm{C}_{2}\mathrm{O}_{4^{2-}}] = x^2 \] Now, we have 0.020 mol of \(\mathrm{NiC}_{2}\mathrm{O}_{4}\), so the concentration of \(\mathrm{Ni(NH_{3})_{6}^{2+}}\) will be 0.020 mol/L. Let y be the concentration of \(\mathrm{NH}_{3}\) required to dissolve the given amount of \(\mathrm{NiC}_{2}\mathrm{O}_{4}\). From the stability constant expression, we can write: \[ K_{f} = \frac{[\mathrm{Ni(NH_{3})_{6}^{2+}]}{[\mathrm{Ni}^{2+}][\mathrm{NH}_{3}]^{6}} = \frac{0.020}{x (y - x)^6} \] Now, we need to eliminate x from this equation. From Ksp, we know that \(x^2 = 4 \times 10^{-10}\). Therefore, \(x = \sqrt{4 \times 10^{-10}}\) and we can substitute x into the stability constant expression. \[ K_{f} = \frac{0.020}{\sqrt{4 \times 10^{-10}} (y - \sqrt{4 \times 10^{-10}})^6} \] Finally, we can solve for the concentration of NH3 needed (y) by substituting the Kf value for Nickel-ammonia complex. This value can be found in literature or tables, for example: Kf = 5.0 × 10^8. \[ 5.0 \times 10^8 = \frac{0.020}{\sqrt{4 \times 10^{-10}} (y - \sqrt{4 \times 10^{-10}})^6} \] Solving for y, we obtain: \[ y \approx 0.475 \, M \] Thus, the concentration of \(\mathrm{NH}_{3}\) required to dissolve \(0.020 \mathrm{~mol}\) of \(\mathrm{NiC}_{2}\mathrm{O}_{4}\) in \(1.0 \mathrm{~L}\) of solution is approximately 0.475 M.