Question

Which of the following processes will likely result in a precipitation reaction? (a) Mixing a \(\mathrm{NaNO}_{3}\) solution with a \(\mathrm{CuSO}_{4}\) solution. (b) Mixing a \(\mathrm{BaCl}_{2}\) solution with a \(\mathrm{K}_{2} \mathrm{SO}_{4}\) solution. Write a net ionic equation for the precipitation reaction.

Short answer

Process (b) will result in a precipitation reaction. The net ionic equation is \( \mathrm{Ba^{2+}} + \mathrm{SO_{4}^{2-}} \rightarrow \mathrm{BaSO}_{4(s)} \).

Step-by-step solution

Identify the Potential Precipitation Reactions

A precipitation reaction occurs when two solutions are mixed to form an insoluble product or a precipitate. Using the solubility rules, it can be determined that: \n (a) \(\mathrm{NaNO}_{3}\) and \(\mathrm{CuSO}_{4}\) are both soluble compounds. There won't be any precipitation reaction.\n (b) \(\mathrm{BaCl}_{2}\) and \(\mathrm{K}_{2}\mathrm{SO}_{4}\) react to form \(\mathrm{BaSO}_{4}\) and \(\mathrm{2KCl}\). According to solubility rules, \(\mathrm{BaSO}_{4}\) is insoluble and so a precipitation reaction will occur.

Write the total ionic equation

The total ionic equation shows all the particles in the solution as they exist. \n \( \mathrm{Ba^{2+}} +2\mathrm{Cl^{-}} + 2\mathrm{K^{+}} + \mathrm{SO_{4}^{2-}} \rightarrow \mathrm{BaSO}_{4(s)} + 2\mathrm{K^{+}} + 2\mathrm{Cl^{-}} \)

Write the net ionic equation

The net ionic equation can be obtained by eliminating the spectator ions, ions which appear on both reactant and product side, from the total ionic equation.\n The net ionic equation will be: \n \( \mathrm{Ba^{2+}} + \mathrm{SO_{4}^{2-}} \rightarrow \mathrm{BaSO}_{4(s)} \)

Key concepts

Net Ionic Equation

When dealing with chemical reactions, specifically in solutions, the net ionic equation provides a clear view of the actual chemical change happening. It excludes ions that don't participate in the formation of the insoluble product, known as spectator ions. In a precipitation reaction, the net ionic equation represents the formation of the insoluble compound, which is the solid that precipitates from the solution.
To create a net ionic equation from a full chemical equation, follow these simple steps:

• Write down the balanced chemical equation and the total ionic equation, which shows all soluble ionic species dissociated into ions.
• Identify the spectator ions in the equation; these are ions that appear unchanged on both the reactant and the product side.
• Subtract these spectator ions from the total ionic equation to write the net ionic equation.

For example, when the solution of \( \mathrm{BaCl}_2 \) is mixed with \( \mathrm{K_2SO_4} \), the resulting net ionic equation is \( \mathrm{Ba^{2+}} + \mathrm{SO_{4}^{2-}} \rightarrow \mathrm{BaSO_{4(s)}} \). In this reaction, \( \mathrm{BaSO_4} \) forms as a precipitate, while \( \mathrm{K^+} \) and \( \mathrm{Cl^-} \) act as spectator ions.

Solubility Rules

Understanding solubility rules is crucial for predicting whether a precipitate will form in a reaction. These rules offer a simplified guide to determine if compounds are soluble (able to dissolve) or insoluble (form a solid) in water.
Here are some basic rules that might help:

• Most nitrate (\(\mathrm{NO_3^-}\)) salts are soluble.
• Salts of alkali metals (like \(\mathrm{Na^+}\), \(\mathrm{K^+}\)) and ammonium (\(\mathrm{NH_4^+}\)) are soluble.
• Chloride (\(\mathrm{Cl^-}\)), bromide (\(\mathrm{Br^-}\)), and iodide (\(\mathrm{I^-}\)) salts are soluble, except those of silver, lead, and mercury.
• Most sulfate (\(\mathrm{SO_4^{2-}}\)) salts are soluble, except for barium sulfate (\(\mathrm{BaSO_4}\)), calcium sulfate, and lead sulfate.
• Carbonates (\(\mathrm{CO_3^{2-}}\)), phosphates (\(\mathrm{PO_4^{3-}}\)), and sulfides are generally insoluble, with exceptions like those of sodium, potassium, and ammonium.

Using these rules simplifies predictions of precipitation and helps in writing net ionic equations, like when \(\mathrm{BaCl_{2}}\) and \(\mathrm{K_{2}SO_{4}}\) are mixed, leading to the formation of \(\mathrm{BaSO_{4}}\), an insoluble compound.

Insoluble Compounds

In chemical reactions involving solutions, some compounds form solids instead of dissolving. These solids are commonly known as insoluble compounds or precipitates. Understanding how and why they form is key to mastering precipitation reactions.
An insoluble compound is formed when the ionic product of the constituent ions (from two reacting solutions) exceeds the solubility product. This is the point at which the solution can no longer hold all of the ions in the dissolved state, resulting in some ions being forced together to form a solid.
Some examples of common insoluble compounds include:

• Barium sulfate \((\mathrm{BaSO_{4}})\), often used in medical imaging.
• Lead(II) chloride \((\mathrm{PbCl_2})\), which precipitates from solutions containing lead and chloride ions.
• Calcium carbonate \((\mathrm{CaCO_{3}})\), found in limestone and marble.

In equations, writing an insoluble compound as a solid indicates its formation. For instance, in a reaction where \( \mathrm{BaSO_{4}} \) appears, it cues that a precipitation occurs, leading to a solid precipitate.