Question

Write a molecular equation for the precipitation reaction that occurs (if any) when each pair of aqueous solutions is mixed. If no reaction occurs, write "NO REACTION." a. potassium carbonate and lead(II) nitrate b. lithium sulfate and lead(II) acetate c. copper(II) nitrate and magnesium sulfide d. strontium nitrate and potassium iodide

Short answer

\( \text{K}_2\text{CO}_3(\text{aq}) + \text{Pb(NO}_3)_2(\text{aq}) \rightarrow 2 \text{KNO}_3(\text{aq}) + \text{PbCO}_3(s)\), \( \text{Li}_2\text{SO}_4(\text{aq}) + \text{Pb(C}_2\text{H}_3\text{O}_2)_2(\text{aq}) \rightarrow 2 \text{LiC}_2\text{H}_3\text{O}_2(\text{aq}) + \text{PbSO}_4(s)\), \( \text{Cu(NO}_3)_2(\text{aq}) + \text{MgS}(\text{aq}) \rightarrow \text{CuS}(s) + \text{Mg(NO}_3)_2(\text{aq})\), NO REACTION for \( \text{Sr(NO}_3)_2 \text{and} \text{KI} \) reaction.

Step-by-step solution

Predicting the Reaction for Potassium Carbonate and Lead(II) Nitrate

Write down the formulas for potassium carbonate (\( \text{K}_2\text{CO}_3 \text{(aq)} \rightarrow 2 \text{K}^+ \text{(aq)} + \text{CO}_3^{2-} \text{(aq)} \) and lead(II) nitrate (\( \text{Pb(NO}_3)_2 \text{(aq)} \rightarrow \text{Pb}^{2+} \text{(aq)} + 2 \text{NO}_3^- \text{(aq)} \) and check the solubility rules. Potassium ions (\( \text{K}^+ \) and nitrate ions (\( \text{NO}_3^- \) are always soluble, but lead(II) carbonate (\( \text{PbCO}_3 \) is insoluble. Therefore, a precipitation reaction occurs.

Molecular Equation for Potassium Carbonate and Lead(II) Nitrate

Using the fact that lead(II) carbonate is insoluble, write a balanced molecular equation: \(\text{K}_2\text{CO}_3(\text{aq}) + \text{Pb(NO}_3)_2(\text{aq}) \rightarrow 2 \text{KNO}_3(\text{aq}) + \text{PbCO}_3(s)\)

Predicting the Reaction for Lithium Sulfate and Lead(II) Acetate

Write down the formulas for lithium sulfate (\( \text{Li}_2\text{SO}_4 \text{(aq)} \rightarrow 2 \text{Li}^+ \text{(aq)} + \text{SO}_4^{2-} \text{(aq)} \) and lead(II) acetate (\( \text{Pb(C}_2\text{H}_3\text{O}_2)_2 \text{(aq)} \rightarrow \text{Pb}^{2+} \text{(aq)} + 2 \text{C}_2\text{H}_3\text{O}_2^- \text{(aq)} \) and check solubility rules. As lead(II) sulfate (\( \text{PbSO}_4 \) is insoluble, a precipitation reaction will occur.

Molecular Equation for Lithium Sulfate and Lead(II) Acetate

Since lead(II) sulfate is insoluble, we can write the balanced molecular equation: \(\text{Li}_2\text{SO}_4(\text{aq}) + \text{Pb(C}_2\text{H}_3\text{O}_2)_2(\text{aq}) \rightarrow 2 \text{LiC}_2\text{H}_3\text{O}_2(\text{aq}) + \text{PbSO}_4(s)\)

Predicting the Reaction for Copper(II) Nitrate and Magnesium Sulfide

Considering the formulas for copper(II) nitrate (\( \text{Cu(NO}_3)_2 \text{(aq)} \rightarrow \text{Cu}^{2+} \text{(aq)} + 2 \text{NO}_3^- \text{(aq)} \) and magnesium sulfide (\( \text{MgS} \text{(aq)} \rightarrow \text{Mg}^{2+} \text{(aq)} + \text{S}^{2-} \text{(aq)} \) and solubility rules, copper(II) sulfide (\( \text{CuS} \) is insoluble. This indicates a precipitation reaction will happen.

Molecular Equation for Copper(II) Nitrate and Magnesium Sulfide

Write the balanced molecular equation knowing copper(II) sulfide is insoluble: \(\text{Cu(NO}_3)_2(\text{aq}) + \text{MgS}(\text{aq}) \rightarrow \text{CuS}(s) + \text{Mg(NO}_3)_2(\text{aq})\)

Predicting the Reaction for Strontium Nitrate and Potassium Iodide

Analyze the solubility of compounds formed from strontium nitrate (\( \text{Sr(NO}_3)_2 \text{(aq)} \rightarrow \text{Sr}^{2+} \text{(aq)} + 2 \text{NO}_3^- \text{(aq)} \) and potassium iodide (\( \text{KI} \text{(aq)} \rightarrow \text{K}^+ \text{(aq)} + \text{I}^- \text{(aq)} \) reactions. Since all resulting compounds, including strontium iodide (\( \text{SrI}_2 \)) and potassium nitrate (\( \text{KNO}_3 \) are soluble, there is no precipitation reaction.

Molecular Equation for Strontium Nitrate and Potassium Iodide

No precipitation reaction occurs, therefore the response is: NO REACTION.

Key concepts

Chemical Solubility

Chemical solubility is a fundamental concept in chemistry that describes the ability of a substance to dissolve in a solvent. It's pivotal to understand that solubility is not just a 'yes' or 'no' situation; rather, it's a scale of how well substances mix at a given temperature and pressure. Soluble compounds will form homogeneous mixtures, while insoluble ones tend to form precipitates.

In the exercise provided, for instance, potassium carbonate (\( \text{K}_2\text{CO}_3 \text{(aq)} \rightleftharpoons 2 \text{K}^+ \text{(aq)} + \text{CO}_3^{2-} \text{(aq)} \) is soluble in water, as its ions dissociate completely in solution. The scenario with lead(II) nitrate is similar up to the point of it coming together with carbonate ions. The formation of lead(II) carbonate (\( \text{PbCO}_3 \) as a solid suggests its poor solubility in water—this low solubility leads to the solid precipitating out of the solution. Understanding these nuances is crucial in predicting whether a precipitation reaction will occur when two aqueous solutions are mixed.

Molecular Equations

Molecular equations provide a concise way to represent chemical reactions, especially precipitation reactions. These equations show the complete chemical formulas of the reactants and products without indicating their ionic nature. This depiction is particularly helpful when you want to emphasize the actual substances involved in a reaction.

In the case of our exercises, a molecular equation was used to describe the reaction between potassium carbonate and lead(II) nitrate. The equation \( \text{K}_2\text{CO}_3(\text{aq}) + \text{Pb(NO}_3)_2(\text{aq}) \rightarrow 2 \text{KNO}_3(\text{aq}) + \text{PbCO}_3(s) \) provides insight into the products of our reaction, which includes a precipitate, \( \text{PbCO}_3(s) \) (the ‘s’ stands for solid). Molecular equations are valuable when looking at the balance of matter and can be converted into complete ionic and net ionic equations for a deeper analysis of the reaction.

Solubility Rules

Solubility rules are a set of guidelines that help us predict the solubility of different ionic compounds in water. These rules are based on empirical observations and trends within groups of compounds. Generally, nitrates, acetates, and almost all alkali metal (Group 1) compounds are soluble, which helps us understand why potassium nitrate is soluble in our exercise.

However, there are exceptions, such as silver and lead compounds, which are often insoluble unless paired with certain ions. For example, lead(II) sulfate and lead(II) carbonate are insoluble, leading to precipitation when they form in a solution. Applying solubility rules is a skill that requires practice, but they are a powerful tool for predicting and explaining the outcomes of mixing different solutions as we saw with our lead(II) nitrate reactions. A balanced understanding of these rules is fundamental in the study of inorganic chemistry and is instrumental in tasks such as the one provided in the exercise.