Using Solubility Rules to Predict Precipitate Formation

Using Solubility Rules to Predict Precipitate Formation 1.15 Double displacement reactions occur in aqueous solutions. Every substance dissolves in ...
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Using Solubility Rules to Predict Precipitate Formation

1.15

Double displacement reactions occur in aqueous solutions. Every substance dissolves in water to a certain degree. Solubility is a measure of the amount of a substance that dissolves in water at a given temperature and pressure. A substance that remains solid in water is said to have low solubility. An example of a substance with low solubility is chalk (calcium carbonate, CaCO3(s)). An example of a soluble substance is table salt (sodium chloride, NaCl(s)). In this section, you will learn how to predict the products of double displacement reactions using solubility rules.

solubility a measure of the extent to which a solute dissolves in a solvent at a given temperature and pressure

The Solubility Rules Being able to predict the products of a chemical reaction is very useful to chemists and other professionals. For example, water quality engineers can test water samples for the presence of different ions. If they know the product of a chemical reaction, then they can determine which reactants they need to add to the water in order to determine the presence of certain ions. A vast amount of information has been collected on the solubility of various compounds in water. This information helps chemists and chemical technicians identify the contents of solutions. Table 1

Solubility Rules for Ionic Compounds in Water 

Anions



Cations (Li,

Na,

K,

Rb,

Cs,

Fr)

Solubility of compounds

most

alkali ions

most

hydrogen ion, H (aq)

soluble

ammonium ion, NH4

soluble

most

soluble

most 

nitrate, NO3



acetate, C2H3O2

soluble

Ag

low solubility

most others

soluble

Ag, Pb2, Hg22, Cu, TI

low solubility

all others

soluble

Ca2, Sr2, Ba2, Pb2, Ra2

low solubility

chloride, Cl bromide, Br iodide,

I

sulfate, SO42

all others sulfide,

S2

alkali ions,

hydroxide, OH

soluble H(aq),

,

NH4

Be2,

Mg2,

Ca2,

Sr2,

Ba2,

Ra2

soluble

all others

low solubility

alkali ions, H(aq), NH4, Sr2, Ba2, Ra2, TI

soluble

all others

low solubility

alkali ions, H(aq), NH4

soluble

all others

low solubility

3

phosphate, PO4

carbonate, CO32 sulfite, SO32

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The solubility of different ionic compounds in water is summarized in Table 1. You can use this table to predict whether or not an ionic compound formed during a double displacement reaction is soluble in water (no precipitate forms) or has low solubility in water (a precipitate forms).

Predicting the Formation of a Precipitate In the previous section, you learned that some double displacement reactions in aqueous solution result in the formation of a precipitate and an ionic compound in aqueous state (Figure 1). You looked at the following two examples: aqueous silver nitrate  aqueous calcium chloride → solid silver chloride  2 AgNO3(aq)



CaCl2(aq)



2 AgCl(s)



aqueous calcium nitrate Ca(NO3)2(aq)

aqueous sodium hydroxide  aqueous calcium nitrate → solid calcium hydroxide  aqueous sodium nitrate 2 NaOH(aq)



Ca(NO3)2(aq)



Ca(OH)2(s)



2 NaNO3(aq)

When two ionic compounds in aqueous state are mixed together, you can use the solubility rules to predict whether or not a chemical reaction will result in the formation of a precipitate.

SAMPLE problem 1 Predicting Precipitate Formation (a) Determine the products (if any) when a solution of sodium chloride is mixed with a solution of silver nitrate. If a reaction occurs, summarize the reaction as a balanced chemical equation. Step 1: Identify Type of Reaction and Possible Products Sodium chloride and silver nitrate are both ionic compounds. Therefore, if a reaction occurs, it will be a double displacement reaction. Write the possible double displacement reaction as a word equation: sodium chloride  silver nitrate → silver chloride  sodium nitrate Step 2: Look Up Solubility of Both Products silver chloride: low solubility sodium nitrate: soluble Step 3: Indicate States of Reactants and Products aqueous sodium chloride  aqueous silver nitrate → solid silver chloride  aqueous sodium nitrate

Figure 1 In this double displacement reaction, two aqueous ionic compounds react to form a new solid compound and a new aqueous ionic compound. A solution of sodium iodide, NaI(aq), reacts immediately when mixed with a solution of lead(II) nitrate, Pb(NO3)2(aq), to produce a bright yellow precipitate.

Since silver chloride has low solubility, it will precipitate out of solution. Sodium chloride, silver nitrate, and sodium nitrate are all soluble. Step 4: Write Chemical Equation for Reaction NaCl(aq)  AgNO3(aq) → AgCl(s)  NaNO3(aq)

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DID YOU

KNOW

?

Detecting Fingerprints Silver nitrate, AgNO3(s), is used by forensic investigators to reveal fingerprints. After the moisture from perspiration has evaporated, fingerprint residue consists of a large portion of sodium chloride, NaCl(s). When sodium chloride comes in contact with silver nitrate, one of the products is solid silver chloride, AgCl(s). The amount of silver chloride is so small that it is hard to see. When placed in UV light, however, it appears black or reddish-brown.

LEARNING

TIP

Precipitate Means Reaction For a reaction to occur, two of the ions must precipitate out of solution. Otherwise, there is no reaction—only four types of ions floating in solution.

Step 5: Balance Equation The equation is already balanced. NaCl(aq)  AgNO3(aq) → AgCl(s)  NaNO3(aq) (b) Determine the products (if any) when a solution of sodium chloride is mixed with a solution of potassium bromide. If a reaction occurs, summarize the reaction as a balanced chemical equation. Step 1: Identify Type of Reaction and Possible Products Sodium chloride and potassium bromide are both ionic compounds. If a reaction occurs, it will be a double displacement reaction. Write the possible double displacement reaction as a word equation: sodium chloride  potassium bromide → sodium bromide  potassium chloride Step 2: Look Up Solubility of Both Products sodium bromide: soluble potassium chloride: soluble Step 3: Indicate States of Reactants and Products Since both sodium bromide and potassium chloride are soluble in water, all ions stay in solution and no reaction takes place. Step 4: Write Chemical Equation for Reaction sodium chloride  potassium bromide → no reaction Example 1 Determine the products (if any) when a solution of sodium sulfate is mixed with a solution of lead(II) nitrate. If a reaction occurs, summarize the reaction as a balanced chemical equation. Solution Double displacement reaction: sodium sulfate  lead(II) nitrate → lead(II) sulfate  sodium nitrate lead(II) sulfate: low solubility sodium nitrate: soluble aqueous sodium sulfate  aqueous lead(II) nitrate → solid lead(II) sulfate  aqueous sodium nitrate Na2SO4(aq)  Pb(NO3)2(aq) → PbSO4(s)  NaNO3(aq) (unbalanced) (balanced) Na2SO4(aq)  Pb(NO3)2(aq) → PbSO4(s)  2 NaNO3(aq)

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Example 2 Determine the products (if any) when a solution of sodium nitrate is mixed with a solution of potassium chloride. If a reaction occurs, summarize the reaction as a balanced chemical equation. Solution Double displacement reaction: sodium nitrate  potassium chloride → sodium chloride  potassium nitrate sodium chloride: soluble potassium nitrate: soluble sodium nitrate  potassium chloride → no reaction

Practice Understanding Concepts 1. Determine the products (if any). If a reaction occurs, summarize the reaction as a balanced chemical equation. (a) a solution of lead(II) nitrate mixed with a solution of sodium chloride (b) a solution of sodium sulfate mixed with a solution of calcium chloride (c) a solution of magnesium acetate mixed with a solution of silver nitrate (d) a solution of sodium acetate mixed with a solution of potassium chloride

Total Ionic and Net Ionic Equations When a solution of sodium sulfate is combined with a solution of calcium chloride, calcium sulfate precipitate forms and sodium chloride remains in solution. The chemical reaction can be represented by the following equation: Na2SO4(aq)  CaCl2(aq) → CaSO4(s)  2 NaCl(aq)

What this equation does not tell you is which ions are actually involved in the chemical reaction. A net ionic equation depicts only the ions that are involved in the chemical reaction. To write the net ionic equation for a given reaction, you must first determine the total ionic equation. A total ionic equation illustrates the separation of soluble ionic compounds into their respective ions. In the sodium sulfate and calcium chloride reaction, all the compounds dissociated into anions and cations. The total ionic equation is 2    2 2 Na (aq)  SO4(aq)  Ca(aq)  2 Cl (aq) → CaSO4(s)  2 Na (aq)  2 Cl (aq)

This equation indicates which ions are in solution and which ions are found in precipitate form. Notice that the sodium and chloride ions are found as both reactants and products. They have not changed their state during the reaction and appear exactly the same on both sides of the equation arrow. They are called spectator ions and can be crossed out: 2    2 2 Na (aq)  SO4(aq)  Ca(aq)  2 Cl (aq) → CaSO4(s)  2 Na (aq)  2 Cl (aq)

NEL

net ionic equation an equation that depicts only the ions that are involved in a chemical reaction total ionic equation a chemical equation that illustrates all soluble ionic compounds in their ionic form spectator ion an ion that is present during a chemical reaction but does not participate in the reaction

LEARNING

TIP

Subscripts and Coefficients When a compound dissociates into ions, the number of ions is denoted by the subscript. When writing the total ionic equation, the subscript becomes the coefficient. For example, in Na2SO4(aq), there are two sodium ions. Therefore, when this compound dissociates, it is represented as 2 2 Na (aq)  SO4(aq)

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The remaining ions are the ions that participate in the chemical reaction and are directly responsible for the formation of the precipitate. For the reaction of sodium sulfate with calcium chloride, the net ionic equation is 2  SO 2 → CaSO Ca(aq) 4(aq) 4(s)

SAMPLE problem 2 Writing Total and Net Ionic Equations (a) Write the total ionic equation and net ionic equation for the reaction between barium sulfide and sodium sulfate. Step 1: Identify Type of Reaction and Possible Products Barium sulfide and sodium sulfate are both ionic compounds. Therefore, if a reaction takes place, it will be a double displacement reaction. Write the possible double displacement reaction as a word equation: barium sulfide  sodium sulfate → barium sulfate  sodium sulfide Step 2: Look Up Solubility of Both Products barium sulfate: low solubility sodium sulfide: soluble Step 3: Indicate States of Reactants and Products aqueous barium sulfide  aqueous sodium sulfate → solid barium sulfate  aqueous sodium sulfide Step 4: Write Chemical Equation for Reaction BaS(aq)  Na2SO4(aq) → BaSO4(s)  Na2S(aq) Step 5: Balance Equation The equation is already balanced. BaS(aq)  Na2SO4(aq) → BaSO4(s)  Na2S(aq) Step 6: Write Total Ionic Equation Using your knowledge of nomenclature and solubility, rewrite the equation with all the ionic compounds that are soluble in water in ionic form. The total ionic equation is 2  S2  2 Na  SO 2 → BaSO  2 Ba(aq) 4(aq) 4(s)  2 Na (aq)  S(aq) (aq) (aq)

Step 7: Write Net Ionic Equation First cancel identical amounts of identical ions that appear on both sides of the equation: 2  S2  2 Na  SO 2 → BaSO  2 Ba(aq) 4(aq) 4(s)  2 Na (aq)  S(aq) (aq) (aq)

Then write the net ionic equation with the remaining ions and precipitate. Ensure that any coefficients are reduced, if applicable. 2  SO 2 → BaSO Ba(aq) 4(aq) 4(s)

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(b) Write the total ionic equation and net ionic equation for the reaction between ammonium hydroxide and sodium nitrate. Step 1: Identify Type of Reaction and Possible Products Ammonium hydroxide and sodium nitrate are both ionic compounds. If a reaction takes place, it will be a double displacement reaction. Write the possible double displacement reaction as a word equation: ammonium hydroxide  sodium nitrate → ammonium nitrate  sodium hydroxide Step 2: Look Up Solubility of Both Products ammonium nitrate: soluble sodium hydroxide: soluble Step 3: Indicate States of Reactants and Products aqueous ammonium hydroxide  aqueous sodium nitrate → aqueous ammonium nitrate  aqueous sodium hydroxide Step 4: Write Chemical Equation for Reaction Using information from the solubility rules, write a balanced chemical equation. Indicate the states of the reactants and products in the reaction. NH4OH(aq)  NaNO3(aq) → NH4NO3(aq)  NaOH(aq) Step 5: Balance Equation Since the products of this reaction are aqueous, you can conclude that there is no reaction. Step 6: Write Total Ionic Equation Since there is no reaction, all entities in the equation dissociate into ions. Therefore, the total ionic equation is    NH4 (aq)  OH (aq)  Na (aq)  NO3 (aq) →    NH4 (aq)  NO3 (aq)  Na (aq)  OH (aq)

Step 7: Write Net Ionic Equation Cancel identical amounts of identical ions that appear on both sides of the equation:    NH4 (aq)  OH (aq)  Na (aq)  NO3 (aq) →    NH4 (aq)  NO3 (aq)  Na (aq)  OH (aq)

Since there are no ions remaining (no reaction took place), there is no net ionic equation. Example Write the total ionic equation and net ionic equation for the reaction between aqueous sodium chloride and aqueous lead(II) nitrate. Solution Double displacement reaction: sodium chloride  lead(II) nitrate → lead(II) chloride  sodium nitrate

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lead(II) chloride: low solubility sodium nitrate: soluble aqueous sodium chloride  aqueous lead(II) nitrate → solid lead(II) chloride  aqueous sodium nitrate NaCl(aq)  Pb(NO3 )2(aq) → PbCl2(s)  NaNO3(aq) (unbalanced) → PbCl2(s)  2 NaNO3(aq) (balanced) 2 NaCl(aq)  Pb(NO3 ) 2(aq) 2(  2    2 Na (aq)  2 Cl (aq)  Pb(aq)  2 NO3 (aq) → PbCl2(s)  2 Na (aq)  2 NO3 (aq) (total ionic equation)  2    2 Na (aq)  2 Cl (aq)  Pb(aq)  2 NO3 (aq) → PbCl2(s)  2 Na (aq)  2 NO3 (aq) 2  2 Cl → PbCl Pb(aq) 2(s) (net ionic equation) (aq)

SUMMARY

Writing Net Ionic Equations

1. Write the double displacement reaction as a word equation. 2. Using the solubility rules, determine whether the products of the reaction have high or low solubility in water. 3. Using this information, indicate the states of the reactants and products. 4. Write a chemical equation for the reaction. 5. Balance the equation. 6. Rewrite the equation, with all the ionic compounds that are soluble in water separated into their respective ions. This equation is the total ionic equation. Cancel identical amounts of identical ions that appear on both the reactant and product sides of the equation. 7. Write the net ionic equation, reducing coefficients if necessary.

Practice Understanding Concepts 2. Write the total ionic equation and the net ionic equation for each of the following reactions: (a) the reaction between aqueous barium chloride and aqueous silver nitrate (b) the reaction between aqueous zinc chloride and aqueous lead(II) nitrate

Qualitative Chemical Analysis The information in the solubility rules can be applied to create diagnostic tests for the presence of specific ions in aqueous solution. Chemists can use the solubility rules to determine the presence of certain ions in a solution by conducting double displacement reactions: they use solutions that contain ions that form a precipitate with the ions they wish to detect. For example, a

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chemist may suspect that a solution contains acetate ions, C2H3O2 (aq). The solubility rules indicate that silver acetate, AgC2H3O2(s), has low solubility. If the chemist adds a source of silver ions, Ag (aq), to the solution and a precipitate forms, then the chemist might infer that acetate ions are present. The chemist must be aware, however, that other ions, such as Cl (aq), may be present in solution, which may also cause the acetate ions to precipitate. To rid the solution completely of acetate ions, the chemist can continue adding silver ions until no more precipitate forms. Most solutions contain more than one type of ion. Therefore, chemists must design procedures to identify and remove any suspected ions one at a time. For example, a chemist may suspect that a solution contains both 2 , and iron(II) ions, Fe2 . In order to determine whether strontium ions, Sr(aq) (aq) one or both types of ions are present, the chemist must design and carry out a procedure that precipitates each type of suspected ion out of solution one at a time. The solubility rules indicate that if hydroxide ions, OH (aq), are added to a solution that contains both strontium and iron(II) ions, the hydroxide ions will react with only the iron(II) ions. Therefore, the chemist may add a solution of sodium hydroxide, NaOH(aq). If iron(II) ions are present, they will react with the hydroxide ions, producing a precipitate according to the following net ionic equation: 2  2 OH → Fe(OH) Fe(aq) 2(s) (aq)

If no precipitate forms, the chemist can infer that no iron(II) ions are present in the solution. To ensure that all the iron(II) ions are removed from the solution so that they do not interfere with any future reactions, the chemist may continue adding sodium hydroxide solution until no more precipitate forms. The chemist must then remove the iron(II) hydroxide precipitate, Fe(OH)2(s), from the solution. One method is to use a centrifuge. A centrifuge is a piece of laboratory equipment that separates different substances in a solution based on their densities. A centrifuge spins the solution at very high speeds, causing higher-density (heavier) particles to sink to the bottom of the test tube faster than lower-density particles. During centrifugation, the heavier particles of iron(II) hydroxide precipitate settle to the bottom of the container while the supernate, which contains all the other aqueous ions, is left at the top of the container. After centrifugation, the iron(II) hydroxide precipitate can be separated from the remaining supernate by pouring the supernate into another container. Once all the iron(II) ions have been removed from the solution as iron(II) hydroxide precipitate, the chemist can test for the presence of strontium ions. From the solubility rules, the chemist knows that sulfate ions, SO42 (aq), will react with strontium ions to form a precipitate. This reaction is represented by the following net ionic equation:

centrifuge a piece of laboratory equipment that spins solutions at very high speeds, to separate the different particles from each other based on their densities supernate the part of a centrifuged solution that does not settle to the bottom of the centrifuge tube

2  SO 2 → SrSO Sr(aq) 4 (aq) 4(s)

If a precipitate forms, the chemist can infer the presence of strontium ions. The strontium ions can be removed from the solution using aqueous sodium sulfate, Na2SO4(aq). The sulfate ions will react with the strontium ions. NEL

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Chemists use solubility to test for the presence of specific ions, and also to remove ions from solution (Figure 2). In the next section, you will use the solubility rules to design and perform an experiment that allows you to determine which ions are present in a solution.

solution known to contain 2+ Fe2+ (aq) and/or Sr(aq) ions add NaOH(aq) precipitate

no precipitate

solution contains Fe2+ ions, precipitated as Fe(OH)2(s)

no Fe2+ ions present

centrifuge

Fe(OH)2(s)

supernate add Na2SO4(aq)

Figure 2 This flow chart summarizes the steps taken to determine whether 2 and/or Fe2 ions are present Sr(aq) (aq) in solution.

precipitate

no precipitate

solution contains Sr2+ ions, precipitated as SrSO4(s)

no Sr2+ ions present

Section 1.15 Questions Understanding Concepts 1. Using the solubility rules, determine whether each of the following compounds is soluble in water: (a) lead(II) sulfate (b) ammonium sulfide (c) silver nitrate (d) silver chloride (e) calcium carbonate (f) ammonium hydroxide (g) barium hydroxide 2. Using the solubility rules, determine whether a precipitate will form if each of the following pairs of compounds is mixed. If a precipitate forms, identify the precipitate and write its chemical formula.

(a) strontium nitrate and sodium sulfate (b) sodium acetate and silver nitrate (c) barium nitrate and ammonium phosphate (d) sodium hydroxide and calcium nitrate 3. Write the total ionic equation and net ionic equation for each reaction in question 2. Making Connections 4. Some pollutants in natural waters, such as heavy metals and organic compounds, are classified as having low solubility. What are the origins of these pollutants in natural waters? If these pollutants are low-solubility compounds, why are they a problem? GO

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