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Chapter 4: Problem 42

Complete the following precipitation reactions with balanced molecular, total ionic, and net ionic equations: (a) \(\mathrm{CaCl}_{2}(a q)+\mathrm{Cs}_{3} \mathrm{PO}_{4}(a q) \longrightarrow\) (b) \(\mathrm{Na}_{2} \mathrm{~S}(a q)+\mathrm{ZnSO}_{4}(a q) \longrightarrow\)

Short Answer

Expert verified
(a) \[\text{Ca}_3(\text{PO}_4)_2(s) \rightarrow 3 \text{Ca}^{2+}(aq) + 2 (\text{PO}_4)^{3-}(aq) \], (b) \[(\text{S})^{2-}(aq) + \text{Zn}^{2+}(aq) \rightarrow \text{ZnS}(s) \]

Step by step solution

01

- Write the reactants

Identify and write down the reactants given in the problem:(a) \[\text{CaCl}_2(aq) + \text{Cs}_3\text{PO}_4(aq) \rightarrow\](b) \[\text{Na}_2\text{S}(aq) + \text{ZnSO}_4(aq) \rightarrow\]
02

- Determine the products

Use solubility rules to determine the products. Exchange the cations and anions to form new combinations:(a) \[\text{CaCl}_2(aq) + \text{Cs}_3\text{PO}_4(aq) \rightarrow \text{Ca}_3(\text{PO}_4)_2(s) + \text{CsCl}(aq) \](b) \[\text{Na}_2\text{S}(aq) + \text{ZnSO}_4(aq) \rightarrow \text{ZnS}(s) + \text{Na}_2\text{SO}_4(aq) \]
03

- Balance the molecular equations

Balance the equations by ensuring the same number of each type of atom on both sides:(a) \[\text{3 CaCl}_2(aq) + \text{2 Cs}_3\text{PO}_4(aq) \rightarrow \text{Ca}_3(\text{PO}_4)_2(s) + 6 \text{CsCl}(aq) \](b) \[\text{Na}_2\text{S}(aq) + \text{ZnSO}_4(aq) \rightarrow \text{ZnS}(s) + \text{Na}_2\text{SO}_4(aq) \]
04

- Write the total ionic equations

Disassociate all aqueous (soluble) compounds into their ions:(a) \[\text{3 Ca}^{2+}(aq) + 6 \text{Cl}^-(aq) + 6 \text{Cs}^+(aq) + 2 (\text{PO}_4)^{3-}(aq) \rightarrow \text{Ca}_3(\text{PO}_4)_2(s) + 6 \text{Cs}^+(aq) + 6 \text{Cl}^- (aq) \](b) \[\text{2 Na}^+(aq) + (\text{S})^{2-}(aq) + \text{Zn}^{2+}(aq) + (\text{SO}_4)^{2-}(aq) \rightarrow \text{ZnS}(s) + 2 \text{Na}^+(aq) + (\text{SO}_4)^{2-}(aq) \]
05

- Write the net ionic equations

Remove the spectator ions (ions that appear on both sides of the equation) to focus on the ions that form the precipitate:(a) \[\text{3 Ca}^{2+}(aq) + 2 (\text{PO}_4)^{3-}(aq) \rightarrow \text{Ca}_3(\text{PO}_4)_2(s) \](b) \[(\text{S})^{2-}(aq) + \text{Zn}^{2+}(aq) \rightarrow \text{ZnS}(s) \]

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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 the reactants and products in their undissociated form. It shows all the species in the reaction as whole formulas, not as ions.
For example, in the precipitation reaction between calcium chloride and cesium phosphate:
\[\begin{equation}\text{3 CaCl}_2(aq) + \text{2 Cs}_3\text{PO}_4(aq) \rightarrow \text{Ca}_3(\text{PO}_4)_2(s) + 6 \text{CsCl}(aq)\end{equation}\]
This equation shows the compounds reacting and the precipitate forming.
Total ionic equation
The total ionic equation disassociates all the soluble compounds into their respective ions, showing every ion participating in the reaction.
For calcium chloride and cesium phosphate, the total ionic equation is:
\[\begin{equation}\text{3 Ca}^{2+}(aq) + 6 \text{Cl}^-(aq) + 6 \text{Cs}^+(aq) + 2 (\text{PO}_4)^{3-}(aq) \rightarrow \text{Ca}_3(\text{PO}_4)_2(s) + 6 \text{Cs}^+(aq) + 6 \text{Cl}^- (aq)\end{equation}\]
This format helps us see which ions are unchanged during the reaction, known as spectator ions.
Net ionic equation
The net ionic equation removes the spectator ions to focus only on the ions that participate in the formation of the precipitate. This provides a clearer picture of the actual chemical change taking place.
For the example above, removing the spectator ions, we get:
\[\begin{equation}\text{3 Ca}^{2+}(aq) + 2 (\text{PO}_4)^{3-}(aq) \rightarrow \text{Ca}_3(\text{PO}_4)_2(s)\end{equation}\]
Here, only the ions that form the solid product are shown.
Chemical balancing
Balancing chemical equations is crucial since it ensures that the law of conservation of mass is maintained.
All atoms on both sides of the equation should be equal.
Using the grand example, let's balance it:
\[\begin{equation}\text{CaCl}_2(aq) + \text{Cs}_3\text{PO}_4(aq) \rightarrow \text{Ca}_3(\text{PO}_4)_2(s) + \text{CsCl}(aq)\end{equation}\]
Balance the calcium and phosphate ions:
\[\begin{equation}\text{3 CaCl}_2(aq) + \text{2 Cs}_3\text{PO}_4(aq) \rightarrow \text{Ca}_3(\text{PO}_4)_2(s) + 6 \text{CsCl}(aq)\end{equation}\]
This maintains the number of atoms for each element on both sides.
Solubility rules
Solubility rules guide us on whether a compound dissolves in water or forms a precipitate. These rules help predict the outcome of reactions.
Generally:
  • Nitrates (\text{NO}_3^-), acetates (\text{CH}_3\text{COO}^-), and alkali metal (group 1) salts are soluble.
  • Most sulfates (\text{SO}_4^{2-}) are soluble, except those of barium, calcium, and lead.
  • Chlorides (\text{Cl}^-), bromides (\text{Br}^-), and iodides (\text{I}^-) are soluble, except those of silver, lead, and mercury.
Using these rules, we identify the insoluble products leading to precipitation, such as:\[\begin{equation}\text{Ca}_3(\text{PO}_4)_2(s)\end{equation}\] and \[\begin{equation}\text{ZnS}(s)\end{equation}\]

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Most popular questions from this chapter

Calculate each of the following quantities: (a) Mass (g) of solute needed to make \(475 \mathrm{~mL}\) of \(5.62 \times 10^{-2} \mathrm{M}\) potassium sulfate (b) Molarity of a solution that contains \(7.25 \mathrm{mg}\) of calcium chloride in each milliliter (c) Number of \(\mathrm{Mg}^{2+}\) ions in each milliliter of \(0.184 \mathrm{M}\) magnesium bromide

Why do decomposition redox reactions typically have compounds as reactants, whereas combination redox and displacement redox reactions have one or more elements as reactants?

Balance each of the following redox reactions and classify it as a combination, decomposition, or displacement reaction: (a) \(\mathrm{Mg}(s)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{Mg}(\mathrm{OH})_{2}(s)+\mathrm{H}_{2}(g)\) (b) \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{Al}(s) \longrightarrow \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{Cr}(s)\) (c) \(\mathrm{PF}_{3}(g)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{PF}_{5}(g)\)

The mass percent of \(\mathrm{Cl}^{-}\) in a seawater sample is determined by titrating \(25.00 \mathrm{~mL}\) of seawater with \(\mathrm{AgNO}_{3}\) solution, causing a precipitation reaction. An indicator is used to detect the end point, which occurs when free \(\mathrm{Ag}^{+}\) ion is present in solution after all the \(\mathrm{Cl}^{-}\) has reacted. If \(53.63 \mathrm{~mL}\) of \(0.2970 \mathrm{M} \mathrm{AgNO}_{3}\) is required to reach the end point, what is the mass percent of \(\mathrm{Cl}^{-}\) in the seawater \((d\) of seawater \(=1.024 \mathrm{~g} / \mathrm{mL}) ?\)

Predict the product(s) and write a balanced equation for each of the following redox reactions: (a) \(\operatorname{Sr}(s)+\operatorname{Br}_{2}(l) \longrightarrow\) (b) \(\mathrm{Ag}_{2} \mathrm{O}(s) \stackrel{\Delta}{\longrightarrow}\) (c) \(\operatorname{Mn}(s)+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow\)
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