Problem 22 - Chemistry: The Central Science

Question

Identify the precipitate (if any) that forms when the following solutions are mixed, and write a balanced equation for each reaction. (a) \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{NaOH}\), (b) \(\mathrm{NaOH}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\), (c) \(\mathrm{Na}_{2} \mathrm{~S}\) and \(\mathrm{Cu}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}\).

Step-by-Step Solution

Answer

The precipitates formed when the given solutions are mixed are: (a) \(\mathrm{Ni}\left(\mathrm{OH}\right)_{2}(s)\), balanced net ionic equation: \(\mathrm{Ni}^{2+}(aq) + 2\mathrm{OH}^{-}(aq) \rightarrow \mathrm{Ni}\left(\mathrm{OH}\right)_{2}(s)\) (b) No precipitate formed, no reaction takes place. (c) \(\mathrm{Cu}_{\rm{S}}(s)\), balanced net ionic equation: \(\mathrm{Cu}^{2+}(aq) + \mathrm{S}^{2-}(aq) \rightarrow \mathrm{Cu}_{\rm{S}}(s)\)

Step 1: (a) Molecular Equation:

First, we need to write the balanced molecular equation for the reaction of \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{NaOH}\): \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(aq) + 2\mathrm{NaOH}(aq) \rightarrow \mathrm{Ni}\left(\mathrm{OH}\right)_{2}(s) + 2\mathrm{Na}\mathrm{NO}_{3}(aq)\)

Step 2: (a) Ionic Equation:

Now, write the complete ionic equation for the reaction: \(\mathrm{Ni}^{2+}(aq) + 2\mathrm{NO}_{3}^{-}(aq) + 2\mathrm{Na}^{+}(aq) + 2\mathrm{OH}^{-}(aq) \rightarrow \mathrm{Ni}\left(\mathrm{OH}\right)_{2}(s) + 2\mathrm{Na}^{+}(aq) + 2\mathrm{NO}_{3}^{-}(aq)\)

Step 3: (a) Precipitate and Net Ionic Equation:

We can identify that \(\mathrm{Ni}\left(\mathrm{OH}\right)_{2}(s)\) is formed as a precipitate. The balanced net ionic equation for the reaction is: \(\mathrm{Ni}^{2+}(aq) + 2\mathrm{OH}^{-}(aq) \rightarrow \mathrm{Ni}\left(\mathrm{OH}\right)_{2}(s)\)

Step 4: (b) Molecular Equation:

First, we need to write the balanced molecular equation for the reaction of \(\mathrm{NaOH}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\): \(2\mathrm{NaOH}(aq) + \mathrm{K}_{2} \mathrm{SO}_{4}(aq) \rightarrow 2\mathrm{KO}\mathrm{OH}(aq) + \mathrm{Na}_{2}\mathrm{SO}_{4}(aq)\)

Step 5: (b) Ionic Equation:

Now, write the complete ionic equation for the reaction: \(2\mathrm{Na}^{+}(aq) + 2\mathrm{OH}^{-}(aq) + 2\mathrm{K}^{+}(aq) + \mathrm{SO}_{4}^{2-}(aq) \rightarrow 2\mathrm{K}^{+}(aq) + 2\mathrm{OH}^{-}(aq) + 2\mathrm{Na}^{+}(aq) + \mathrm{SO}_{4}^{2-}(aq)\)

Step 6: (b) Precipitate and Net Ionic Equation:

In this case, there is no precipitate formed as all the species remain in solution. The net ionic equation will consist of all species remaining as ions in the solution, which indicates that there is no reaction taking place.

Step 7: (c) Molecular Equation:

First, we need to write the balanced molecular equation for the reaction of \(\mathrm{Na}_{2}\mathrm{S}\) and \(\mathrm{Cu}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}\): \(\mathrm{Na}_{2}\mathrm{S}(aq) + \mathrm{Cu}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}(aq) \rightarrow \mathrm{Cu}_{\rm{S}}(s) + 2 \mathrm{Na}\mathrm{CH_3COO}(aq)\)

Step 8: (c) Ionic Equation:

Now, write the complete ionic equation for the reaction: \(2\mathrm{Na}^{+}(aq) + \mathrm{S}^{2-}(aq) + \mathrm{Cu}^{2+}(aq) + 2\mathrm{CH}_{3} \mathrm{COO}^{-}(aq) \rightarrow \mathrm{Cu}_{\rm{S}}(s) + 2 \mathrm{Na}^{+}(aq) + 2\mathrm{CH}_{3} \mathrm{COO}^{-}(aq)\)

Step 9: (c) Precipitate and Net Ionic Equation:

We can identify that \(\mathrm{Cu}_{\rm{S}}(s)\) is formed as a precipitate. The balanced net ionic equation for the reaction is: \(\mathrm{Cu}^{2+}(aq) + \mathrm{S}^{2-}(aq) \rightarrow \mathrm{Cu}_{\rm{S}}(s)\)

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is essential to accurately represent what occurs in a chemical reaction. The law of conservation of mass dictates that the number of atoms of each element must be the same on both sides of the equation. To balance a chemical equation, like the ones provided in the exercise, follow these steps:

• Write down the unbalanced equation.
• List the number of atoms of elements present in the reactants and products.
• Adjust the coefficients, which are the numbers in front of the compounds, to balance the atoms for each element on both sides.
• Ensure that all coefficients are in the lowest possible ratio.

For example, in the reaction between \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{NaOH}\), we balanced the equation by adding a coefficient of 2 in front of \(\mathrm{NaOH}\) and \(\mathrm{NaNO}_{3}\) to equate the number of \(\mathrm{Na}\) and \(\mathrm{OH}\) ions on both sides.

Solubility Rules

Solubility rules help predict the outcome of precipitation reactions. Knowing which compounds are soluble and which are not allows us to predict whether a precipitate will form when two solutions are mixed. A precipitate is an insoluble solid that emerges from a liquid solution. The general guidelines for solubility are:

• Salts containing alkali metal ions and the ammonium ion are typically soluble.
• Nitrates, acetates, and most perchlorates are soluble.
• Chlorides, bromides, and iodides are soluble, except when paired with silver, lead(II), or mercury(I).
• Most sulfates are soluble, with exceptions such as calcium sulfate, lead(II) sulfate, and barium sulfate.
• Hydroxides are generally insoluble, with the notable exceptions of sodium, potassium, and ammonium hydroxides.
• Carbonates, phosphates, sulfides, and oxides are mostly insoluble.

For instance, \(\mathrm{Ni}\left(\mathrm{OH}\right)_{2}\) and \(\mathrm{Cu}_{\rm{S}}\) are insoluble and thus form precipitates, as indicated by the solubility rules when \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{NaOH}\), \(\mathrm{Na}_{2}\mathrm{S}\) and \(\mathrm{Cu}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}\) are mixed.

Ionic and Molecular Equations

Chemical reactions in aqueous solutions can be represented by ionic and molecular equations. Molecular equations show the complete compounds as they appear in the solution. On the other hand, ionic equations break down soluble ionic compounds into their individual ions. This depiction is more precise because it demonstrates the actual forms that the reactants and products take in solution.

For instance, in step 1 and step 4 of the exercise, the reactions were initially written as molecular equations. However, to more accurately reflect the behavior in solution, in step 2 and step 5, the equations are rewritten to show the ions that are involved in the reaction. It becomes clear which ions come together to form the precipitate and which stay in solution as spectator ions.

Net Ionic Equations

Net ionic equations represent the essence of a chemical reaction by showing only the chemical species that are directly involved in the change. Spectator ions, which do not participate in the reaction, are omitted from the net ionic equation. The net ionic equation provides clarity on the actual chemistry taking place in solution.

In the provided steps 3, 6, and 9, the net ionic equations were determined by removing the spectator ions from the complete ionic equations. This isolates the formation of the insoluble precipitates, such as \(\mathrm{Ni}\left(\mathrm{OH}\right)_{2}\) and \(\mathrm{Cu}_{\rm{S}}\), which are the critical part of the chemical process taking place. Particularly in steps where no reaction occurs, like in step 6, the net ionic equation reflects no change since no precipitate is formed.