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Chapter 5: Problem 106

What is the net ionic equation for the reaction that occurs when an aqueous solution of \(\mathrm{KI}\) is added to an aqueous solution of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2} ?\)

Short Answer

Expert verified
The net ionic equation for the reaction that occurs when an aqueous solution of \(\mathrm{KI}\) is added to an aqueous solution of \(\mathrm{Pb(NO}_3)_2\) is \[2 \, \mathrm{I^-} + Pb^{2+} = \mathrm{PbI}_2(s)\]

Step by step solution

01

Write down the molecular equation

First, write down the molecular equation. Potassium iodide reacts with lead nitrate to produce lead iodide(which is a precipitate) and potassium nitrate. This translates to the following equation: \[2 \, \mathrm{KI} + \, \mathrm{Pb(NO}_3)_2 \, \rightarrow \, \mathrm{PbI}_2 + 2 \, \mathrm{KNO}_3\] This equation indicates that for each molecule of lead nitrate in solution, two molecules of potassium iodide are needed to produce one molecule of lead iodide and two molecules of potassium nitrate.
02

Write down the total ionic equation

The next step is to write down the ionic equation, which includes all of the ions present in the reaction: \[2 \, \mathrm{K^+} + 2 \, \mathrm{I^-} + Pb^{2+} + 2 \, \mathrm{NO}_3^- \, \rightarrow \, Pb \, \mathrm{I}_2(s) + 2 \, \mathrm{K^+} + 2 \, \mathrm{NO}_3^-\] The (s) next to \(PbI_2\) indicates that it is a solid (precipitate), while the other ions are all in aqueous solution.
03

Write down the net ionic equation

The net ionic equation is obtained by cancelling out the spectator ions that don鈥檛 participate in the reaction. In this case, the potassium ions \( \mathrm{K^+} \) and nitrate ions \( \mathrm{NO}_3^- \) are spectator ions. Removing these gives the net ionic equation: \[2 \, \mathrm{I^-} + Pb^{2+} \, \rightarrow \, \mathrm{PbI}_2(s)\] This equation indicates that in the reaction, lead ions combined with iodide ions to form the precipitate, lead iodide.

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Key Concepts

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

Molecular Equation
The molecular equation represents all reactants and products in their undissociated form as if they were compounds. In the reaction between potassium iodide (KI) and lead(II) nitrate \(\mathrm{Pb(NO}_3)_2\), we start by writing the balanced molecular equation:
  • \(2 \, \mathrm{KI} + \mathrm{Pb(NO}_3)_2 \rightarrow \mathrm{PbI}_2 + 2 \, \mathrm{KNO}_3\)
This equation shows that two molecules of potassium iodide react with one molecule of lead(II) nitrate. The result is one molecule of lead(II) iodide and two molecules of potassium nitrate. It is important to understand that while the molecular equation illustrates the compounds as whole entities, it doesn't show how these exist in solution.
To prepare for writing the total ionic equation, keep in mind the states of the compounds: \(\mathrm{PbI}_2\) is a solid precipitate formed in the solution, whereas the others remain dissolved and separate into ions.
Total Ionic Equation
The total ionic equation displays all the ions present in the aqueous solution as they actually exist in solution. In water, soluble ionic compounds dissociate into their constituent ions. Therefore, potassium iodide \((\mathrm{KI})\) and lead(II) nitrate \((\mathrm{Pb(NO}_3)_2)\) separate into ions, reflecting the true nature of the substances involved:
  • \(2 \, \mathrm{K^+} + 2 \, \mathrm{I^-} + \mathrm{Pb^{2+}} + 2 \, \mathrm{NO}_3^- \rightarrow \mathrm{PbI}_2(s) + 2 \, \mathrm{K^+} + 2 \, \mathrm{NO}_3^-\)
Here, you observe the distinction between solid and aqueous states. \(\mathrm{PbI}_2(s)\) remains as a solid due to its low solubility in water, forming a precipitate.
The total ionic equation lays the groundwork for identifying and canceling out ions on both sides of the equation that do not change and do not participate directly in the formation of the precipitate.
Spectator Ions
Spectator ions are ions that appear on both sides of the total ionic equation but do not participate in the actual chemical change. They essentially 'watch' as the other ions form a new product. In our equation, calcium ions \((\mathrm{K^+})\) and nitrate ions \((\mathrm{NO}_3^-)\) are present on both sides of the reaction:
  • They do not change and can be 'cancelled out' from the total ionic equation.
  • They remain in the solution and do not form part of the precipitate.
Once you remove these spectator ions, you arrive at the net ionic equation:
  • \(2 \, \mathrm{I^-} + \mathrm{Pb^{2+}} \rightarrow \mathrm{PbI}_2(s)\)
This net ionic equation highlights the actual chemical change taking place in the reaction, where only the reacting ions are shown forming the precipitate lead(II) iodide.

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Most popular questions from this chapter

The electrolyte in a lead storage battery must have a concentration between 4.8 and \(5.3 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) if the battery is to be most effective. A \(5.00 \mathrm{mL}\) sample of a battery acid requires \(49.74 \mathrm{mL}\) of \(0.935 \mathrm{M} \mathrm{NaOH}\) for its complete reaction (neutralization). Does the concentration of the battery acid fall within the desired range? [Hint: Keep in mind that the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) produces two \(\mathrm{H}^{+}\) ions per formula unit.]

For use in titrations, we want to prepare \(20 \mathrm{L}\) of \(\mathrm{HCl}(\mathrm{aq})\) with a concentration known to four significant figures. This is a two-step procedure beginning with the preparation of a solution of about \(0.10 \mathrm{M}\) HCl. A sample of this dilute HCl(aq) is titrated with a NaOH(aq) solution of known concentration. (a) How many milliliters of concentrated \(\mathrm{HCl}(\mathrm{aq})\) \((d=1.19 \mathrm{g} / \mathrm{mL} ; 38 \% \mathrm{HCl}, \text { by mass })\) must be diluted with water to 20.0 L to prepare \(0.10 \mathrm{M} \mathrm{HCl}\) ? (b) \(\mathrm{A} 25.00\) \(\mathrm{mL}\) sample of the approximately \(0.10\) \(\mathrm{M}\) HCl prepared in part (a) requires \(20.93\) \(\mathrm{mL}\) of \(0.1186\) \(\mathrm{M}\) NaOH for its titration. What is the molarity of the \(\mathrm{HCl}(\mathrm{aq}) ?\) (c) Why is a titration necessary? That is, why not prepare a standard solution of \(0.1000\) \(\mathrm{M} \mathrm{HCl}\) simply by an appropriate dilution of the concentrated HCl(aq)?

The highest \(\left[\mathrm{H}^{+}\right]\) will be found in an aqueous solution that is (a) \(0.10 \mathrm{M} \mathrm{HCl} ;\) (b) \(0.10 \mathrm{M} \mathrm{NH}_{3} ;\) (c) \(0.15 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{COOH} ;(\mathrm{d}) 0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\).

A neutralization reaction between an acid and a base is a common method of preparing useful salts. Give net ionic equations showing how the following salts could be prepared in this way: (a) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4};\) (b) \(\mathrm{NH}_{4} \mathrm{NO}_{3} ;\) and \((\mathrm{c})\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\).

Chile saltpeter is a natural source of \(\mathrm{NaNO}_{3}\); it also contains \(\mathrm{NaIO}_{3} .\) The \(\mathrm{NaIO}_{3}\) can be used as a source of iodine. Iodine is produced from sodium iodate in a two-step process occurring under acidic conditions: \(\begin{aligned} \mathrm{IO}_{3}^{-}(\mathrm{aq})+\mathrm{HSO}_{3}^{-}(\mathrm{aq}) & \longrightarrow \mathrm{I}^{-}(\mathrm{aq}) +\mathrm{SO}_{4}^{2-}(\mathrm{aq}) \end{aligned} \quad\) ( not balanced) \(\mathrm{I}^{-}(\mathrm{aq})+\mathrm{IO}_{3}^{-}(\mathrm{aq}) \longrightarrow\) \(\mathrm{I}_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad(\text { not balanced })\) In the illustration, a 5.00 L sample of a \(\mathrm{NaIO}_{3}(\mathrm{aq})\) solution containing \(5.80 \mathrm{g} \mathrm{NaIO}_{3} / \mathrm{L}\) is treated with the stoichiometric quantity of \(\mathrm{NaHSO}_{3}\) (no excess of either reactant). Then, a further quantity of the initial \(\mathrm{NaIO}_{3}(\mathrm{aq})\) is added to the reaction mixture to bring about the second reaction. (a) How many grams of NaHSO \(_{3}\) are required in the first step? (b) What additional volume of the starting solution must be added in the second step?
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