Question

A solution contains 0.25\(M \mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}\) and 0.25\(M \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) 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 \(\mathrm{Na}_{2} \mathrm{CO}_{3}.\)

Short answer

In summary, the metal ions cannot be separated by slowly adding sodium carbonate, as the requirement for successful separation i.e. 99% of nickel ions precipitation before copper ions start to precipitate is not met. This is because the carbonate concentration required for 99% precipitation of nickel (\( 5.2\times10^{-7} M \)) is higher than that required for copper precipitation to begin (\( 1.01\times10^{-9} M \)).

Step-by-step solution

Identify the precipitation reactions and write the solubility product expressions

When sodium carbonate is added, the precipitate of nickel carbonate and copper carbonate can form. The precipitation reactions are: 1) Ni²⁺(aq) + CO₃²⁻(aq) ⇌ NiCO₃(s) 2) Cu²⁺(aq) + CO₃²⁻(aq) ⇌ CuCO₃(s) Now, let's write the solubility product expressions for both reactions: For Nickel Carbonate: \(K_{sp1} = \mathrm{[Ni^{2+}][CO_3^{2-}]} \) For Copper Carbonate: \(K_{sp2} = \mathrm{[Cu^{2+}][CO_3^{2-}]} \)

Find the solubility product constants for the precipitates

From tabulated data, the solubility product constants (Ksp) for the nickel carbonate and copper carbonate are: \(K_{sp1} = \mathrm{1.3\times10^{-7}}\) (for Nickel Carbonate) \(K_{sp2} = \mathrm{2.5\times10^{-10}}\) (for Copper Carbonate)

Determine the concentration required to reach 99% precipitation

We need to find the concentration of carbonate ions required to precipitate 99% of each metal ion. Let's calculate the CO₃²⁻ concentration required for each metal ion to reach 99% precipitation: For Nickel Carbonate: \[[\mathrm{CO_3^{2-}}] = \frac{K_{sp1}}{[\mathrm{Ni^{2+}} \times 0.01]} = \frac{1.3\times10^{-7}}{\mathrm{0.25M \times 0.01}} = 5.2\times10^{-7} M\] For Copper Carbonate: \[ [\mathrm{CO_3^{2-}}] = \frac{K_{sp2}}{[\mathrm{Cu^{2+}} \times 0.99]} = \frac{2.5\times10^{-10}}{\mathrm{0.25M \times 0.99}} = 1.01\times10^{-9} M\]

Compare the carbonate concentrations for 99% precipitation

To achieve the successful separation of the metal ions, the carbonate concentration required for 99% precipitation of nickel should be lower than that for copper precipitation to begin. Comparing the calculated carbonate concentrations above: \[ 5.2\times10^{-7} M > 1.01\times10^{-9} M\] Since the carbonate concentration requirement for nickel precipitation (found in Step 3) is higher than that required for copper precipitation to begin, we can conclude that adding sodium carbonate slowly to the solution will not result in successful separation of 99% nickel ions before copper ions begin to precipitate. In summary, the metal ions cannot be separated by slowly adding sodium carbonate, as the requirement for successful separation i.e. 99% of nickel ions precipitation before copper ions start to precipitate is not met.

Key concepts

Solubility Product

Understanding the solubility product is essential when dealing with chemical separations, especially in solutions containing different metal ions. The solubility product, denoted as \( K_{sp} \), represents the equilibrium constant for a solid substance dissolving in an aqueous solution.
It quantifies the extent to which a compound can dissolve to form ions in a solution. In chemical terms, it is defined by the equation where the concentrations of the resulting ions are multiplied together, each raised to the power of their stoichiometric coefficients.
For example, when dealing with nickel carbonate and copper carbonate, we can write their solubility products as:

• Nickel Carbonate: \( K_{sp1} = \mathrm{[Ni^{2+}][CO_3^{2-}]} \)
• Copper Carbonate: \( K_{sp2} = \mathrm{[Cu^{2+}][CO_3^{2-}]} \)

Understanding the value of the solubility product helps us predict whether a compound will precipitate under certain concentrations. The lower the \( K_{sp} \) value, the less soluble the compound is in the solution.

Precipitation Reactions

Precipitation reactions play a crucial role in separating different metal ions from a solution. These reactions occur when two ions in an aqueous solution combine to form an insoluble solid, known as the precipitate.
During the addition of sodium carbonate \( \mathrm{Na_2CO_3} \), nickel ions \( \mathrm{Ni^{2+}} \) and copper ions \( \mathrm{Cu^{2+}} \) can react with carbonate ions \( \mathrm{CO_3^{2-}} \), leading to the formation of insoluble nickel carbonate \( \mathrm{NiCO_3} \) and copper carbonate \( \mathrm{CuCO_3} \).

• The reaction for nickel carbonate precipitation is: \( \mathrm{Ni^{2+}_{(aq)} + CO^{2-}_{3(aq)} \rightleftharpoons NiCO_{3(s)}} \)
• For copper carbonate, it is: \( \mathrm{Cu^{2+}_{(aq)} + CO^{2-}_{3(aq)} \rightleftharpoons CuCO_{3(s)}} \)

These reactions depend on the concentrations of the ions involved and their respective solubility products. In practice, to achieve effective separation, it's crucial to find a way to ensure one metal precipitates significantly before the other begins, which is a common challenge in such chemical processes.

Metal Ion Separation

Metal ion separation is a vital procedure in many chemical processes, including purification, recovery, and analysis of elements. The selective precipitation of metal ions is a common method to achieve this separation.
To separate ions like nickel \( \mathrm{Ni^{2+}} \) and copper \( \mathrm{Cu^{2+}} \), you must control the conditions so that one ion precipitates before the other. This involves adjusting the concentration of the precipitating agent (e.g., \( \mathrm{CO_3^{2-}} \) ions) so that nearly all of one metal ion precipitates out before the other's precipitation.
The effectiveness of this method relies on the significant difference between the metals' solubility products. When the conditions aren't met, meaning the less soluble compound precipitates first, separation becomes difficult. In our example, since \( K_{sp1} \) for nickel carbonate is much higher than \( K_{sp2} \) for copper carbonate, it indicates that copper will start to precipitate sooner, making separation ineffective in achieving the desired purity level.

Solubility Product Constant

The solubility product constant, \( K_{sp} \), is a fundamental concept in chemistry used to describe the solubility of ionic compounds. It combines the concentrations of the dissolved ions raised to their respective coefficients from the balanced chemical equation.
For example, using nickel carbonate \( \mathrm{NiCO_3} \) and copper carbonate \( \mathrm{CuCO_3} \):

• Nickel Carbonate: \( K_{sp1} = \mathrm{1.3 \times 10^{-7}} \)
• Copper Carbonate: \( K_{sp2} = \mathrm{2.5 \times 10^{-10}} \)

These values represent the maximum product of the concentrations of the dissociated ions that can exist in a saturated solution. A lower \( K_{sp} \) suggests that the compound is less soluble in water.
In practical applications, knowing the \( K_{sp} \) helps chemists predict and control the conditions required for precipitation reactions, and it can guide decisions about how to effectively separate different ions in a solution.