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Chapter 8: Problem 50

Lead ions can be removed from solution by precipitation with sulfate ions. Suppose that a solution contains lead(II) nitrate. Write complete ionic and net ionic equations to show the reaction of aqueous lead(II) nitrate with aqueous potassium sulfate to form solid lead(II) sulfate and aqueous potassium nitrate.

Short Answer

Expert verified
Complete ionic equation: Pb^{2+}(aq) + 2NO3^{-}(aq) + 2K^{+}(aq) + SO4^{2-}(aq) -> PbSO4 (s) + 2K^{+}(aq) + 2NO3^{-}(aq). Net ionic equation: Pb^{2+}(aq) + SO4^{2-}(aq) -> PbSO4 (s).

Step by step solution

01

Write the balanced molecular equation

First, write the balanced molecular equation for the reaction between lead(II) nitrate and potassium sulfate. The chemical formula for lead(II) nitrate is Pb(NO3)2, for potassium sulfate is K2SO4, for lead(II) sulfate is PbSO4, and for potassium nitrate is KNO3. The balanced molecular equation is: Pb(NO3)2 (aq) + K2SO4 (aq) -> PbSO4 (s) + 2KNO3 (aq).
02

Write the complete ionic equation

Next, write the complete ionic equation by breaking all soluble compounds into their constituent ions. Lead(II) nitrate and potassium nitrate are soluble in water and dissociate into ions, while lead(II) sulfate is insoluble and remains as a solid:Pb^{2+}(aq) + 2NO3^{-}(aq) + 2K^{+}(aq) + SO4^{2-}(aq) -> PbSO4 (s) + 2K^{+}(aq) + 2NO3^{-}(aq).
03

Write the net ionic equation

To write the net ionic equation, cancel out the spectator ions that appear on both sides of the complete ionic equation. The spectator ions in this reaction are the potassium ions (K^{+}) and the nitrate ions (NO3^{-}). This leaves the equation with the ions that participate in the reaction:Pb^{2+}(aq) + SO4^{2-}(aq) -> PbSO4 (s).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Precipitation Reactions
Precipitation reactions occur when two soluble salts react in solution to form one or more insoluble products, known as precipitates. These reactions are essential in various applications like water treatment, where harmful ions are removed by precipitating them out of the water.

For instance, when aqueous lead(II) nitrate reacts with aqueous potassium sulfate, a precipitation reaction occurs. Here, lead(II) sulfate, an insoluble salt, forms and precipitates out of the solution. The solid precipitate can be observed as a cloudiness or a solid deposit at the bottom of the mixing container.
Solubility Rules
Understanding solubility rules is crucial for predicting whether a substance will dissolve in water and is key to writing correct ionic equations. These rules are based on empirical observations and aid in identifying which compounds form precipitates.

  • Most nitrate (NO3−) salts are soluble.
  • Most sulfate (SO42−) salts are soluble, with exceptions including lead(II) sulfate (PbSO4), which is insoluble.
  • Most salts of alkali metals (e.g., sodium, potassium) are soluble.

By applying these rules, one can predict that lead(II) nitrate and potassium sulfate are soluble, while lead(II) sulfate will precipitate out of the solution.
Net Ionic Equation
A net ionic equation shows the chemical species that are directly involved in the chemical reaction, eliminating the spectator ions. To write a net ionic equation, one must first write the complete ionic equation and then cancel out the ions that do not participate in the formation of the precipitate.

After canceling the spectator ions, the net ionic equation for the reaction between lead(II) nitrate and potassium sulfate is:
\[Pb^{2+}(aq) + SO4^{2-}(aq) \rightarrow PbSO4 (s)\].
This equation highlights the actual chemistry that is occurring in the solution - the combination of lead(II) ions with sulfate ions to form the precipitate lead(II) sulfate.
Spectator Ions
Spectator ions are ions that are present in the solution but do not directly participate in the precipitation reaction. They remain in the aqueous phase both before and after the reaction. Identifying these ions is a significant step in simplifying the complete ionic equation to the net ionic equation.

In the given reaction, potassium (K+) and nitrate (NO3−) ions are spectator ions. They are unchanged and present on both sides of the complete ionic equation, which allows them to be omitted in the net ionic equation. This focus on the reacting ions simplifies the understanding of the chemical process taking place.

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Most popular questions from this chapter

Complete and balance each equation. If no reaction occurs, write NO REACTION. a. LiI(aq) + BaS(aq) b. KCl(aq) + CaS(aq) c. CrBr2(aq) + Na2CO3(aq) d. NaOH(aq) + FeCl3(aq)

Write balanced complete ionic and net ionic equations for each acid-base reaction. a. HI(aq) + RbOH(aq) b. HCHO2(aq) + NaOH(aq) c. HC2H3O2(aq) + LiOH(aq)

A solution contains one or more of the following ions: Ag+, Ca2+, and Cu2+. When you add sodium chloride to the solution, no precipitate forms. When you add sodium sulfate to the solution, a white precipitate forms. You filter off the precipitate and add sodium carbonate to the remaining solution, producing another precipitate. Which ions were present in the original solution? Write net ionic equations for the formation of each of the precipitates observed.

Consider the generic ionic compounds with the formulas A2X and BY2 and the following solubility rules: A2X soluble; BY2 soluble; AY insoluble; BX soluble. Let circles represent A+ ions, squares represent B2+ ions, triangles represent X2- ions, and diamonds represent Y- ions. Solutions of the two compounds (A2X and BY2) can be represented as follows: Draw a molecular-level representation showing the result of mixing the two solutions (A2X and BY2) and write an equation to represent the reaction.

A laboratory procedure calls for making 400.0 mL of a 1.1 M NaNO3 solution. What mass of NaNO3 (in g) do you need?
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