Question

A solution contains 0.25 M Ni(NO₃)₂ and 0.25 M Cu(NO₃)₂. Can the metal ions be separated by slowly adding Na₂CO₃? Assume that for successful separation, 99% of the metal ion must be precipitated before the other metal ion begins to precipitate, and assume that no volume change occurs on addition of Na₂CO₃.

Short answer

Theseparation of metal through Na₂CO₃ slow addition is possible.

Step-by-step solution

State given information

The solution contains $0.25 M Ni(N{O}_3{)}_2\&0.25 M Cu(N{O}_3{)}_2.$

$$\begin{gathered} K_{sp }of NiC{O}_3=1.4\times {10}^{-7} \\ {K}_{sp }of CuC{O}_3=2.5\times {10}^{-10} \end{gathered}$$

CuCO₃ has smaller K_sp compared to NiCO₃. The first precipitation occurs in CuCO₃;then Na₂CO₃is added.

Calculate precipitation

The precipitationbegins when

$$\begin{gathered} K_{sp},CuC{O}_3=\left[C{u}^{2+}\right]\left[C{O_3}^{2-}\right] \\ \left[C{O_3}^{2-}\right]=\frac{K_{sp},CuC{O}_3}{\left[C{u}^{2+}\right]} \\ \left[C{O_3}^{2-}\right]=1\times {10}^{-9}M \\ {K}_{sp},NiC{O}_3=\left[N{i}^{2+}\right]\left[C{O_3}^{2-}\right] \\ \left[C{O_3}^{2-}\right]=\frac{K_{sp},NiC{O}_3}{\left[N{i}^{2+}\right]} \\ \left[C{O_3}^{2-}\right]=5.6\times {10}^{-7}M \\ {K}_{sp},CuC{O}_3=\left[C{u}^{2+}\right]\left[C{O_3}^{2-}\right] \\ \left[C{u}^{2+}\right]=\frac{K_{sp},CuC{O}_3}{\left[N{i}^{2+}\right]} \\ \left[C{u}^{2+}\right]=4.5\times {10}^{-4}M \end{gathered}$$

The percentage of presence of Cu²⁺ in the solution is$\frac{4.5\times {10}^{-4}M}{0.25 M}\times 100=0.18\%.$

The separation of metal through Na₂CO₃ slow addition is possible.