A solution contains \(0.25 \mathrm{M} \mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}\) and \(0.25 \mathrm{M} \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\). Can the metal ions be separated by slowly adding \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) ? Assume that for successful separation \(99 \%\) of the metal ion must be precipitated before the other metal ion begins to precipitate, and assume no volume change on addition of \(\mathrm{Na}_{2} \mathrm{CO}_{3} .\)

Precipitation reactions for Ni2+ and Cu2+ are given as follows, along with their solubility product constants (Ksp): \(NiCO_3 (s) \rightleftharpoons Ni^{2+} (aq) + CO_3^{2-} (aq)\) ; Ksp(NiCO3) = 1.3 × 10⁻¹⁶ \(CuCO_3 (s) \rightleftharpoons Cu^{2+} (aq) + CO_3^{2-} (aq)\) ; Ksp(CuCO3) = 2.0 × 10⁻¹₀

To calculate the CO3²⁻ concentration needed to initiate the precipitation of Ni²⁺ we need to use the solubility product constant (Ksp) for the NiCO3 reaction: \(Ksp = [Ni^{2+}][CO_3^{2-}]\) Given that the initial concentration of Ni2+ is 0.25 M: \([CO_3^{2-}] = \dfrac{Ksp(NiCO3)}{[Ni^{2+}]}\) Substitute values: \([CO_3^{2-}] = \dfrac{1.3 \times 10^{-16}}{0.25}\) \([CO_3^{2-}] = 5.2 \times 10^{-16}\)

To calculate the CO3²⁻ concentration needed to initiate the precipitation of Cu²⁺ we need to use the solubility product constant (Ksp) for the CuCO3 reaction: \(Ksp = [Cu^{2+}][CO_3^{2-}]\) Given that the initial concentration of Cu2+ is 0.25 M: \([CO_3^{2-}] = \dfrac{Ksp(CuCO3)}{[Cu^{2+}]}\) Substitute values: \([CO_3^{2-}] = \dfrac{2.0 \times 10^{-10}}{0.25}\) \([CO_3^{2-}] = 8.0 \times 10^{-10}\)

To check if separation is possible, we need to calculate the minimum concentration of CO32- ions required to precipitate 99% of Ni²⁺, before Cu²⁺ starts to precipitate. Initial [Ni²⁺] = 0.25 M [CO32-] required for 99% precipitation of Ni²⁺ = (0.25 x 0.99) / [CO32-] calculated in Step 2 Substitute values: [CO3²⁻] required for 99% precipitation of Ni²⁺ = (0.25 x 0.99) / (5.2 x 10⁻¹⁶) = 4.762 × 10⁻¹⁵ Comparing the required concentration of CO32- ions for 99% precipitation of Ni²⁺ to that required for Cu²⁺ precipitation: 4.762 × 10⁻¹⁵ < 8.0 × 10⁻¹⁰ Since the concentration of CO3²⁻ ions needed for 99% precipitation of Ni²⁺ is less than the concentration of CO3²⁻ ions needed for Cu²⁺ precipitation: Yes, the metal ions can be separated by slowly adding Na2CO3.