# ionic reactions single and double sphere models

There are many types of ionic reactions single and double sphere models chemical bonds and forces that bind molecules together. The two most basic types of bonds are characterized as either ionic or covalent. In ionic bonding, atoms transfer electrons to each other. Ionic bonds require at least one electron donor and one electron acceptor. In contrast, atoms with the same share electrons in covalent bonds, because neither atom preferentially attracts or repels the shared electrons.

### Introduction

Ionic bonding is the complete transfer of valence electron(s) between atoms. It is a type of chemical bond that generates two oppositely charged ions. In ionic bonds, the metal loses electrons to become a positively charged cation, whereas the nonmetal accepts those electrons to become a negatively charged anion. Ionic bonds require an electron donor, often a metal, and an electron acceptor, a nonmetal.

Ionic bonding is observed because metals have few electrons in their outer-most orbitals. By losing those electrons, these metals can achieve noble gas configuration and satisfy the octet rule. Similarly, nonmetals that have close to 8 electrons in their valence shells tend to readily accept electrons to achieve noble gas configuration. In ionic bonding, more than 1 electron can be donated or received to satisfy the octet rule. The charges on the anion and cation correspond to the number of electrons donated or received. In ionic bonds, the net charge of the compound must be zero.

This sodium molecule donates the lone electron in its valence orbital in order to achieve octet configuration. This creates a positively charged cation due to the loss of electron.

This chlorine atom receives one electron to achieve its octet configuration, which creates a negatively charged anion.

The predicted overall energy of the ionic bonding process, which includes the ionization energy of the metal and electron affinity of the nonmetal, is usually positive, indicating that the reaction is endothermic and unfavorable. However, this reaction is highly favorable because of the electrostatic attraction between the particles. At the ideal interatomic distance, attraction between these particles releases enough energy to facilitate the reaction. Most ionic compounds tend to dissociate in polar solvents because they are often polar. This phenomenon is due to the opposite charges on each ion.

Example $$\PageIndex{1}$$: Chloride Salts

In this example, the sodium atom is donating its 1 valence electron to the chlorine atom. This creates a sodium cation and a chlorine anion. Notice that the net charge of the resulting compound is 0.

In this example, the magnesium atom is donating both of its valence electrons to chlorine atoms. Each chlorine atom can only accept 1 electron before it can achieve its noble gas configuration; therefore, 2 atoms of chlorine are required to accept the 2 electrons donated by the magnesium. Notice that the net charge of the compound is 0.

### Covalent Bonding

Covalent bonding is the sharing of electrons between atoms. This type of bonding occurs between two atoms of the same element or of elements close to each other in the periodic table. This bonding occurs primarily between nonmetals; however, it can also be observed between nonmetals and metals.

If atoms have similar electronegativities (the same affinity for electrons), covalent bonds are most likely to occur. Because both atoms have the same affinity for electrons and neither has a tendency to donate them, they share electrons in order to achieve octet configuration and become more stable. In addition, the ionization energy of the atom is too large and the electron affinity of the atom is too small warum wird kennenlernen zusammen geschrieben for ionic bonding to occur. For example: carbon does not form ionic bonds because it has 4 valence electrons, half of an octet. To form ionic bonds, Carbon molecules must either gain or lose 4 electrons. This is highly unfavorable; therefore, carbon molecules share their 4 valence electrons through single, double, and triple bonds so that each atom can achieve noble gas configurations. Covalent bonds include interactions of the sigma and pi orbitals; therefore, covalent bonds lead to formation of single, double, triple, and quadruple bonds.

Example $$\PageIndex{2}$$: $$PCl_3$$

In this example, a phosphorous atom is sharing its three unpaired electrons with three chlorine atoms. In the end product, all four of these molecules have 8 valence electrons and satisfy the octet rule.

### Bonding in Organic Chemistry

Ionic and covalent bonds are the two extremes of bonding. Polar covalent is the intermediate type of bonding between the two extremes. Some ionic bonds contain covalent characteristics and some covalent bonds are partially ionic. For example, most carbon-based compounds are covalently bonded but can also be partially ionic. Polarity is a measure of the separation of charge in a compound. A compound's polarity is dependent on the symmetry of the compound and on differences in electronegativity between atoms. Polarity occurs when the electron pushing elements, found on the left side of the periodic table, exchanges electrons with the electron pulling elements, on the right side of the table. This creates a spectrum of polarity, with ionic (polar) at one extreme, covalent (nonpolar) at another, and polar covalent in the middle.

Both of these bonds are important in organic chemistry. Ionic bonds are important because they allow the synthesis of specific organic compounds. Scientists can manipulate ionic properties and these interactions in order to form desired products. Covalent bonds are especially important since most carbon molecules interact primarily through covalent bonding. Covalent bonding allows molecules to share electrons with other molecules, creating long chains of compounds and allowing more complexity in life.

### References

1. Vollhardt, K. Peter C., and Neil E. Schore. Organic Chemistry Structure and Function. New York: W. H. Freeman, 2007.
2. Petrucci, Ralph H. General Chemistry: Principles and Modern Applications. Upper Saddle River, NJ: Pearson Education, 2007.
3. Brown, Theodore L., Eugene H. Lemay, and Bruce E. Bursten. Chemistry: The Central Science. 6th ed. Englewood Cliffs, NJ: Prentice Hall, 1994.

### Problems

1. Are these compounds ionic or covalent?

2. In the following reactions, indicate whether the reactants and products are ionic or covalently bonded.

a)

b) Clarification: What is the nature of the bond between sodium and amide? What kind of bond forms between the anion carbon chain and sodium?

c)

### Solutions

• 1) From left to right: Covalent, Ionic, Ionic, Covalent, Covalent, Covalent, Ionic.
• 2a) All products and reactants are ionic.
• 2b) From left to right: Covalent, Ionic, Ionic, Covalent, Ionic, Covalent, Covalent, Ionic.
• 2c) All products and reactants are covalent.

### Learning Objectives

1. ionic reactions single and double sphere models
2. Recognize chemical reactions as single-replacement reactions and double-replacement reactions.
3. Use the periodic table, an activity series, or solubility rules to predict whether single-replacement reactions or double-replacement reactions will occur.

single frauen aus saarbrücken Up to now, we have presented chemical reactions as a topic, but we have not discussed how the products of a chemical reaction can be predicted. Here we will begin our study of certain types of chemical reactions that allow us to predict what the products of the reaction will be.

A  is a chemical reaction in which one element is substituted for another element in a compound, generating a new element and a new compound as products. For example,

2 HCl(aq) + Zn(s) → ZnCl(aq) + H(g)

is an example of a single-replacement reaction. The hydrogen atoms in HCl are replaced by Zn atoms, and in the process a new element—hydrogen—is formed. Another example of a single-replacement reaction is

2 NaCl(aq) + F(g) → 2 NaF(s) + Cl(g)

Here the negatively charged ion changes from chloride to fluoride. A typical characteristic of a single-replacement reaction is that there is one element as a reactant and another element as a product.

Not all proposed single-replacement reactions will occur between two given reactants. This is most easily demonstrated with fluorine, chlorine, bromine, and iodine. Collectively, these elements are called the halogens and are in the next-to-last column on the periodic table (see ). The elements on top of the column will replace the elements below them on the periodic table but not the other way around. Thus, the reaction represented by

CaI(s) + Cl(g) → CaCl(s) + I(s)

will occur, but the reaction

CaF(s) + Br(ℓ) → CaBr(s) + F(g)

will not because bromine is below fluorine on the periodic table. This is just one of many ways the periodic table helps us understand chemistry.

Figure 4.1 Halogens on the Periodic Table

The halogens are the elements in the next-to-last column on the periodic table.

### Example 2

Will a single-replacement reaction occur? If so, identify the products.

1. MgCl + I → ?
2. CaBr + F → ?

Solution

1. Because iodine is below chlorine on the periodic table, a single-replacement reaction will not occur.
2. Because fluorine is above bromine on the periodic table, a single-replacement reaction will occur, and the products of the reaction will be CaF and Br.

Test Yourself

Will a single-replacement reaction occur? If so, identify the products.

FeI + Cl → ?

Yes; FeCl and I

Chemical reactivity trends are easy to predict when replacing anions in simple ionic compounds—simply use their relative positions on the periodic table. However, when replacing the cations, the trends are not as straightforward. This is partly because there are so many elements that can form cations; an element in one column on the periodic table may replace another element nearby, or it may not. A list called the  does the same thing the periodic table does for halogens: it lists the elements that will replace elements below them in single-replacement reactions. A simple activity series is shown below.

### Activity Series for Cation Replacement in Single-Replacement Reactions

• Li
• K
• Ba
• Sr
• Ca
• Na
• Mg
• Al
• Mn
• Zn
• Cr
• Fe
• Ni
• Sn
• Pb
• H
• Cu
• Hg
• Ag
• Pd
• Pt
• Au

Using the activity series is similar to using the positions of the halogens on the periodic table. An element on top will replace an element below it in compounds undergoing a single-replacement reaction. Elements will not replace elements above them in compounds.

### Example 3

Use the activity series to predict the products, if any, of each equation.

1. FeCl + Zn → ?
2. HNO + Au → ?

Solution

1. Because zinc is above iron in the activity series, it will replace iron in the compound. The products of this single-replacement reaction are ZnCl and Fe.
2. Gold is below hydrogen in the activity series. As such, it will not replace hydrogen in a compound with the nitrate ion. No reaction is predicted.

Test Yourself

Use the activity series to predict the products, if any, of this equation.

AlPO + Mg → ?

Mg(PO) and Al

A  occurs when parts of two ionic compounds are exchanged, making two new compounds. A characteristic of a double-replacement equation is that there are two compounds as reactants and two different compounds as products. An example is

CuCl(aq) + 2 AgNO(aq) → Cu(NO)(aq) + 2 AgCl(s)

There are two equivalent ways of considering a double-replacement equation: either the cations are swapped, or the anions are swapped. (You cannot swap both; you would end up with the same substances you started with.) Either perspective should allow you to predict the proper products, as long as you pair a cation with an anion and not a cation with a cation or an anion with an anion.

### Example 4

Predict the products of this double-replacement equation: BaCl + NaSO → ?

Solution

Thinking about the reaction as either switching the cations or switching the anions, we would expect the products to be BaSO and NaCl.

Test Yourself

Predict the products of this double-replacement equation: KBr + AgNO → ?

KNO and AgBr

Predicting whether a double-replacement reaction occurs is somewhat more difficult than predicting a single-replacement reaction. However, there is one type of double-replacement reaction that we can predict: the precipitation reaction. A  occurs when two ionic compounds are dissolved in water and form a new ionic compound that does not dissolve; this new compound falls out of solution as a solid. The formation of a solid precipitate is the driving force that makes the reaction proceed.

To judge whether double-replacement reactions will occur, we need to know what kinds of ionic compounds form precipitates. For this, we use, which are general statements that predict which ionic compounds dissolve (are soluble) and which do not (are not soluble or insoluble). lists some general solubility rules. We need to consider each ionic compound (both the reactants and the possible products) in light of the solubility rules in. If a compound is soluble, we use the (aq) label with it, indicating it dissolves. If a compound is not soluble, we use the (s) label with it and assume that it will precipitate out of solution. If everything is soluble, then no reaction will be expected.

Table 4.1 Some Useful Solubility Rules

 These compounds generally dissolve in water (are soluble): Exceptions: All compounds of Li+, Na+, K+, Rb+, Cs+, and NH+ None All compounds of NO− and CHO− None Compounds of Cl−, Br−, I− Ag+, Hg2+, Pb2+ Compounds of SO2 Hg2+, Pb2+, Sr2+, Ba2+ These compounds generally do not dissolve in water (are insoluble): Exceptions: Compounds of CO2− and PO3− Compounds of Li+, Na+, K+, Rb+, Cs+, and NH+ Compounds of OH− Compounds of Li+, Na+, K+, Rb+, Cs+, NH+, Sr2+, and Ba2+

For example, consider the possible double-replacement reaction between NaSO and SrCl. The solubility rules say that all ionic sodium compounds are soluble and all ionic chloride compounds are soluble except for Ag+, Hg2+, and Pb2+, which are not being considered here. Therefore, NaSO and SrCl are both soluble. The possible double-replacement reaction products are NaCl and SrSO. Are these soluble? NaCl is (by the same rule we just quoted), but what about SrSO? Compounds of the sulfate ion are generally soluble, but Sr2+ is an exception: we expect it to be insoluble—a precipitate. Therefore, we expect a reaction to occur, and the balanced hoe flirten via email chemical equation would be

NaSO(aq) + SrCl(aq) → 2 NaCl(aq) + SrSO(s)

You would expect to see a visual change corresponding to SrSO precipitating out of solution ().

Figure 4.2 Double-Replacement Reactions

Some double-replacement reactions are obvious because you can see a solid precipitate coming out of solution.

### Example 5

Will a double-replacement reaction occur? If so, identify the products.

1. Ca(NO) + KBr → ?
2. NaOH + FeCl → ?

Solution

1. According to the solubility rules, both Ca(NO) and KBr are soluble. Now we consider what the double-replacement products would be by switching the cations (or the anions)—namely, CaBr and KNO. However, the solubility rules predict that these two substances would also be soluble, so no precipitate would form. Thus, we predict no reaction in this case.
2. According to the solubility rules, both NaOH and FeCl are expected to be soluble. If we assume that a double-replacement reaction may occur, we need to consider the possible products, which would be NaCl and Fe(OH). NaCl is soluble, but, according to the solubility rules, Fe(OH) is not. Therefore, a reaction would occur, and Fe(OH)(s) would precipitate out of solution. The balanced chemical equation is

2NaOH(aq) + FeCl(aq) → 2NaCl(aq) + Fe(OH)(s)

Test Yourself

Will a double-replacement equation occur? If so, identify the products.

Sr(NO) + KCl → ?

No reaction; all possible products are soluble.

Precipitation reactions occur when cations and anions in aqueous solution combine to form an insoluble ionic solid called a precipitate. Whether or not such a reaction occurs can be determined by using the solubility rules for common ionic solids. Because not all aqueous reactions form precipitates, one must consult the solubility rules before determining the state of the products and writing a net ionic equation. The ability to predict these reactions allows scientists to determine which ions are present in a solution, and allows industries to form chemicals by extracting components from these reactions.

### Properties of Precipitates

Precipitates are insoluble ionic solid products of a reaction, formed when certain cations and anions combine in an aqueous solution. The determining factors of the formation of a precipitate can vary. Some reactions depend on temperature, such as solutions used for buffers, whereas others are dependent only on solution concentration. The solids produced in precipitate reactions are crystalline solids, and can be suspended throughout the liquid or fall to the bottom of the solution. The remaining fluid is called supernatant liquid. The two components of the mixture (precipitate and supernate) can be separated by various methods, such as filtration, centrifuging, or decanting.

Figure 1: Above is a diagram of the formation of a precipitate in solution.

### Precipitation and Double Replacement Reactions

The use of solubility rules require an understanding of the way that ions react. Most precipitation reactions are single replacement reactions or double replacement reactions. A double replacement reaction occurs when two ionic reactants dissociate and bond with the respective anion or cation from the other reactant. The ions replace each other based on their charges as either a cation or an anion. This can be thought of as "switching partners"; that is, the two reactants each "lose" their partner and form a bond with a different partner:

Figure 2: A double replacement reaction

A double replacement reaction is specifically classified as a precipitation reaction when the chemical equation in question occurs in aqueous solution and one of the of the products formed is insoluble. An example of a precipitation reaction is given below:

$CdSO_{4(aq)} + K_2S_{(aq)} \rightarrow CdS_{(s)} + K_2SO_{4(aq)}$

Both reactants are aqueous and one product is solid. Because the reactants are ionic and aqueous, they dissociate and are therefore soluble. However, there are six solubility guidelines used to predict which molecules are insoluble in water. These molecules form a solid precipitate in solution.

Solubility Rules

Whether or not a reaction forms a precipitate is dictated by the solubility rules. These rules provide guidelines that tell which ions form solids and which remain in their ionic form in aqueous solution. The rules are to be followed from the top down, meaning that if something is insoluble (or soluble) due to rule 1, it has precedence over a higher-numbered rule.

1. Salts formed with group 1 cations and $$NH_4^+$$ cations are soluble. There are some exceptions for certain $$Li^+$$ salts.
2. Acetates ($$C_2H_3O_2^-$$), nitrates ($$NO_3^-$$), and perchlorates ($$ClO_4^-$$) are soluble.
3. Bromides, chlorides, and iodides are soluble.
4. Sulfates ($$SO_4^{2-}$$) are soluble with the exception of sulfates formed with $$Ca^{2+}$$, $$Sr^{2+}$$, and $$Ba^{2+}$$.
5. Salts containing silver, lead, and mercury (I) are insoluble.
6. ($$CO_3^{2-}$$), phosphates ($$PO_4^{3-}$$), sulfides, oxides, and hydroxides ($$OH^-$$) are insoluble. Sulfides formed with cations and hydroxides formed with calcium, strontium, and barium are exceptions.

If the rules state that an ion is soluble, then it remains in its aqueous ion form. If an ion is insoluble based on the solubility rules, then it forms a solid with an ion from the other reactant. If all the ions in a reaction are shown to be soluble, then no precipitation reaction occurs.

### Net Ionic Equations

To understand the definition of a net ionic equation, recall the equation for the double replacement reaction. Because this particular reaction is a precipitation reaction, states of matter can be assigned to each variable pair:

AB(aq) + CD(aq)AD(aq) + CB(s)

The first step to writing a net ionic equation is to separate the soluble (aqueous) reactants and products into their respective cations and anions. Precipitates do not dissociate in water, so the solid should not be separated. The resulting equation looks like that below:

A+(aq) + B-(aq) + C+(aq) + D-(aq)A+(aq) + D-(aq) + CB(s)

In the equation above, A+and D- ions are present on both sides of the equation. These are called spectator ions because they remain unchanged throughout the reactionSince they go through the equation unchanged, they can be eliminated to show the net ionic equation:

C+ (aq)+ B- (aq) → CB (s)

The net ionic equation only shows the precipitation reaction. A net ionic equation must be balanced on both sides not only in terms of atoms of elements, but also in terms of electric charge. Precipitation reactions are usually represented solely by net ionic equations. If all products are aqueous, a net ionic equation cannot be written because all ions are canceled out as spectator ions. Therefore, no precipitation reaction occurs.

### Applications and Examples

Precipitation reactions are useful in determining whether a certain element is present in a solution. If a precipitate is formed when a chemical reacts with lead, for example, the presence of lead in water sources could be tested by adding the chemical and monitoring for precipitate formation. In addition, precipitation reactions can be used to extract elements, such as magnesium from seawater. Precipitation reactions even occur in the human body between antibodies and antigens; however, the environment in which this occurs is still being studied.

Example 1

Complete the double replacement reaction and then reduce it to the net ionic equation.

$NaOH_{(aq)} + MgCl_{2 \;(aq)} \rightarrow$

First, predict the products of this reaction using knowledge of double replacement reactions (remember the cations and anions “switch partners”).

$2NaOH_{(aq)} + MgCl_{2\;(aq)} \rightarrow 2NaCl + Mg(OH)_2$

Second, consult the solubility rules to determine if the products are soluble. Group 1 cations ($$Na^+$$) and chlorides are soluble from rules 1 and 3 respectively, so $$NaCl$$ will be soluble in water. However, rule 6 states that hydroxides are insoluble, and thus $$Mg(OH)_2$$ will form a precipitate. The resulting equation is the following:

$2NaOH(aq) + MgCl_{2\;(aq)} \rightarrow 2NaCl_{(aq)} + Mg(OH)_{2\;(s)}$

Third, separate the reactants into their ionic forms, as they would exist in an aqueous solution. Be sure to balance both the electrical charge and the number of atoms:

$2Na^+_{(aq)} + 2OH^-_{(aq)} + Mg^{2+}_{(aq)} + 2Cl^-_{(aq)} \rightarrow Mg(OH)_{2\;(s)} + 2Na^+_{(aq)} + 2Cl^-_{(aq)}$

Lastly, eliminate the spectator ions (the ions that occur on both sides of the equation unchanged). In this case, they are the sodium and chlorine ions. The final net ionic equation is:

$Mg^{2+}_{(aq)} + 2OH^-_{(aq)} \rightarrow Mg(OH)_{2(s)}$

Example 2

Complete the double replacement reaction and then reduce it to the net ionic equation.

$CoCl_{2\;(aq)} + Na_2SO_{4\;(aq)} \rightarrow$

SOLUTION

The predicted products of this reaction are $$CoSO_4$$ and $$NaCl$$. From the solubility rules, $$CoSO_4$$ is soluble because rule 4 states that sulfates ($$SO_4^{2-}$$) are soluble. Similarly, we find that $$NaCl$$ is soluble based on rules 1 and 3. After balancing, the resulting equation is as follows:

$CoCl_{2\;(aq)} + Na_2SO_{4\;(aq)} \rightarrow CoSO_{4\;(aq)} + 2 NaCl_{(aq)}$

Separate the species into their ionic forms, as they would exist in an aqueous solution. Balance the charge and the atoms. Cancel out all spectator ions (those that appear as ions on both sides of the equation.):

Co2- (aq) + 2Cl-(aq) + 2Na+ (aq) + SO42-(aq)Co2- (aq) + SO42-(aq) + 2Na+ (aq) + 2Cl-(aq)

No precipitation reaction

This particular example is important because all of the reactants and the products are aqueous, meaning they cancel out of the net ionic equation. There is no solid precipitate formed; therefore, no precipitation reaction occurs.

### Practice Problems

Write the net ionic equation for the potentially double displacement reactions. Make sure to include the states of matter and balance the equations.

1. $$Fe(NO_3)_{3\;(aq)} + NaOH_{(aq)} \rightarrow$$
2. $$Al_2(SO_4)_{3\;(aq)} + BaCl_{2\;(aq)} \rightarrow$$
3. $$HI_{(aq)} + Zn(NO_3)_{2\;(aq)} \rightarrow$$
4. $$CaCl_{2\;(aq)} + Na_3PO_{4\;(aq)} \rightarrow$$
5. $$Pb(NO_3)_{2\;(aq)} + K_2SO_{4 \;(aq)} \rightarrow$$

#### Solutions

1. Regardless of physical state, the products of this reaction are $$Fe(OH)_3$$ and $$NaNO_3$$. The solubility rules predict that $$NaNO_3$$ is soluble because all nitrates are soluble (rule 2). However, $$Fe(OH)_3$$ is insoluble, because hydroxides are insoluble (rule 6) and $$Fe$$ is not one of the cations which results in an exception. After dissociation, the ionic equation is as follows:

$Fe^{3+}_{(aq)} + NO^-_{3\;(aq)} + Na^+_{(aq)} + 3OH^-_{(aq)} \rightarrow Fe(OH)_{3\;(s)} + Na^+_{(aq)} + NO^-_{3\;(aq)}$

Canceling out spectator ions leaves the net ionic equation:

$Fe^{3+}_{(aq)} + OH^-_{(aq)} \rightarrow Fe(OH)_{\;3(s)}$

2. From the double replacement reaction, the products are $$AlCl_3$$ and $$BaSO_4$$. $$AlCl_3$$ is soluble because it contains a chloride (rule 3); however, $$BaSO_4$$ is insoluble: it contains a sulfate, but the $$Ba^{2+}$$ ion causes it to be insoluble because it is one of the cations that causes an exception to rule 4. The ionic equation is (after balancing):

$2Al^{3+}_{(aq)} + 6Cl^-_{(aq)} + 3Ba^{2+}_{(aq)} + 3SO^{2-}_{4\;(aq)} \rightarrow 2 Al^{3+}_{(aq)} +6Cl^-_{(aq)} + 3BaSO_{4\;(s)}$

Canceling out spectator ions leaves the following net ionic equation:

$Ba^{2+}_{(aq)} + SO^{2-}_{4\;(aq)} \rightarrow BaSO_{4\;(s)}$

3. From the double replacement reaction, the products $$HNO_3$$ and $$ZnI_2$$ are formed. Looking at the solubility rules, $$HNO_3$$ is soluble because it contains nitrate (rule 2), and $$ZnI_2$$ is soluble because iodides are soluble (rule 3). This means that both the products are aqueous (i.e. dissociate in water), and thus no precipitation reaction occurs.

4. The products of this double replacement reaction are $$Ca_3(PO_4)_2$$ and $$NaCl$$. Rule 1 states that $$NaCl$$ is soluble, and according to solubility rule 6, $$Ca_3(PO_4)_2$$ is insoluble. The ionic equation is:

$Ca^{2+}_{(aq)}+ Cl^-_{(aq)} + Na^+_{(aq)} + PO^{3-}_{4\;(aq)} \rightarrow Ca_3(PO_4)_{2\;(s)} + Na^+_{(aq)} + Cl^-_{(aq)}$

After canceling out spectator ions, the net ionic equation is given below:

$Ca^{2+}_{(aq)} + PO^{3-}_{4\;(aq)} \rightarrow Ca_3(PO_4)_{2\;(s)}$

5. The first product of this reaction, $$PbSO_4$$, is soluble according to rule 4 because it is a sulfate. The second product, $$KNO_3$$, is also soluble because it contains nitrate (rule 2). Therefore, no precipitation reaction occurs.

### References

1. Campbell, Dan, Linus Pauling, and Davis Pressman. "The Nature of the Forces Between Antigen and Antibody and of the Precipitation Reaction." Physiological Reviews 23.3 (1943): 203-219..

2. Harwood, William, F Herring, Jeffry Madura, and Ralph Petrucci. General Chemistry. 9th ed. Upper Saddle River: Pearson Pretence Hall, 2007. Print.
3. Freeouf, J.L, Grischkowsky, D., McInturff, D.T., Warren, A.C., & Woodall, J.M. (1990). Arsenic precipitates and the semi-insulating properties of gaas buffer layers grown by low-temperature molecular beam epitaxy. Applied Physics Letters, 57(13)
4. Petrucci, et al. General Chemistry: Principles & Modern Applications. 9th ed. Upper Saddle River, New Jersey 2007.

### Contributors

• Julie Schaffer (UCD), Corinne Herman (UCD)

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