Chapter 10: Chemical Tests and Chemical Reaction Types

Section 10-1: Laboratory Tests to Identify Chemicals

Section 10-3: Combustion of Organic Compounds

Section 10-4: Reactions Involving Metals or Metal Compounds

Section 10-5: Aqueous Ionic Compounds and Molarity of Ions

Section 10-6: Solubility Rules, Precipitation Reactions, and Net Ionic Equations

Section 10-7: Acid + Base Neutralization Reactions and Molecular Equations

Section 10-8: Experiment - Acid + Base Titration

Chapter 10 Practice Exercises and Review Quizzes

Section 10-1: Laboratory Tests to Identify Chemicals

The following lab tests have been used to indicate the likely presence of the common chemicals listed:

Chemical	Test
carbon dioxide gas, CO₂ (g)	flaming wooden splint extinguished
hydrogen gas, H₂ (g)	flaming wooden splint "pops"
oxygen gas, O₂ (g)	glowing wooden splint relights
significant concentration of aqueous hydrogen ion, H⁺ (aq)	solution is acidic, so: 1. low pH meter reading 2. litmus paper will be red 3. phenolphthalein indicator will be colorless
significant concentration of aqueous hydroxide ion, OH^-(aq)	solution is basic, so: 1. high pH meter reading 2. litmus paper will be blue 3. phenolphthalein indicator will be pink

Section 10-2: Diatomic Nonmetals

There are 7 nonmetals that exist in covalently-bonded pairs at room temperature and pressure. Each of these is listed below along with its state of matter at room temperature and pressure:

hydrogen gas, H₂ (g)

nitrogen gas, N₂ (g)

oxygen gas, O₂ (g)

fluorine gas, F₂ (g)

chlorine gas, Cl₂ (g)

liquid bromine, Br₂ (l)

solid iodine, I₂ (s)

When the elements above are reactants or products in a chemical reaction, they should be written as diatomic.

Section 10-3: Combustion of Organic Compounds

The complete combustion or burning of organic compounds with the condensed formula C_xH_yO_z requires sufficient oxygen gas and will produce carbon dioxide and water:

C_xH_yO_z+ O₂→ CO₂ + H₂O

The products carbon dioxide and water will be the same regardless of whether z = 0 or z > 0 or, in other words, regardless of whether or not the compound burned contains oxygen.

Sample Exercise 10A:

Write a balanced equation for the combustion of each compound below using the smallest possible whole-number coefficients:

(a) hexane

(b) propanol

Solution:

(a) 2 C₆H₁₄ + 19 O₂ → 12 CO₂ + 14 H₂O

(b) 2 C₃H₇OH + 9 O₂ → 6 CO₂ + 8 H₂O

Section 10-4: Reactions Involving Metals or Metal Compounds

Neutral metals or compounds containing metal ions are reactants in the following reactions:

(a) metal + nonmetal → ionic compound (also known as ionic salt)

For example, lithium metal reacts with nitrogen gas to produce lithium nitride according to the following balanced equation:

(lithium nitride = combine 3 Li⁺ and N^3- = Li₃N)

6 Li + N₂ → 2 Li₃N

(b) metal + water → hydrogen gas + metal cation + hydroxide ion

For example, sodium metal reacts with water according to the following balanced equation:

(be sure that the metal cation has the correct charge)

2 Na + 2 H₂O → H₂ + 2 Na⁺ + 2 OH^-

(note that the coefficient 2 in front of the Na⁺ is added to ensure that the total positive charge equals the total negative charge on the right side of the equation)

(c) metal oxide + water → metal cation + hydroxide ion

For example, potassium oxide reacts with water as follows:

(potassium oxide = combine 2 K⁺ and O^2- = K₂O)

K₂O + H₂O → 2 K⁺ + 2 OH^-

(d) metal carbonate → metal oxide + carbon dioxide

For example, magnesium carbonate decomposes when heated to produce magnesium oxide and carbon dioxide according to the following balanced equation:

(magnesium carbonate = combine Mg²⁺ and CO₃^2- = MgCO₃,

magnesium oxide = combine Mg²⁺ and O^2- = MgO)

MgCO₃ → MgO + CO₂

Sample Exercise 10B:

Write a balanced equation for each reaction described below using the smallest possible whole-number coefficients:

(a) aluminum metal reacts with solid iodine

(b) calcium metal reacts with water

(d) sodium carbonate decomposes upon heating

Solution:

(a) (product is aluminum iodide = combine Al³⁺ and 3 I^- = AlI₃)

2 Al + 3 I₂ → 2 AlI₃

(b) (metal cation produced is Ca²⁺)

Ca + 2 H₂O → H₂ + Ca²⁺ + 2 OH^-

SrO + H₂O → Sr²⁺ + 2 OH^-

(d) (sodium carbonate = combine 2 Na⁺ and CO₃^2- = Na₂CO₃,

product is sodium oxide = combine 2 Na⁺ and O^2- = Na₂O)

Na₂CO₃ → Na₂O + CO₂

Section 10-5: Aqueous Ionic Compounds and Molarity of Ions

When a solid ionic compound is dissolved in water, the ions separate from each other and become surrounded by water molecules in the aqueous solution. For example, KI separates in water into one K⁺ cation and one I^- anion per formula unit, while (NH₄)₂SO₄ separates in water into two NH₄⁺ cations and one SO₄^2- anion per formula unit. A molarity written on the bottle of an aqueous ionic compound will be accompanied by the formula of the neutral ionic compound but will not specify the molarity of individual ions in the solution. As such, if we wish to know the molarity of each type of ion in the solution, we must translate the information on the bottle using the number of cations and anions per formula unit of the ionic compound. For example, in a bottle labeled 0.026 M Na₃PO₄, the molarity of Na⁺ will be 3 x 0.026 M = 0.078 M Na⁺ because there are 3 Na⁺ cations per formula unit. The molarity of PO₄^3- will be 1 x 0.026 M = 0.026 M PO₄^3- since there is one PO₄^3- anion per formula unit.

Sample Exercise 10C:

What is the molarity of each ion in the following solutions?

(a) 1.2 M CaCl₂

(b) 0.31 M Al(NO₃)₃

Solution:

(a) 1 x 1.2 M = 1.2 M Ca²⁺ and 2 x 1.2 M = 2.4 M Cl^-

(b) 1 x 0.31 M = 0.31 M Al³⁺ and 3 x 0.31 M = 0.93 M NO₃^-

Section 10-6: Solubility Rules, Precipitation Reactions, and Net Ionic Equations

A solid ionic compound that dissolves significantly in water is said to be soluble. A solid ionic compound that does not dissolve to a significant extent in water is said to be insoluble. The following general solubility rule indicates several groups of ionic compounds that will be soluble:

Ionic compounds containing a Group 1 metal cation, the ammonium cation (NH₄⁺), or the nitrate anion (NO₃^-) are soluble and, therefore, will exist as separate ions and not as a solid in the presence of water.

Note that there are other ionic compounds which do not include any of the ions in the solubility rule above but will also still be soluble in some cases. For example, ionic compounds containing the anions Cl^-, Br^-, I^-, or SO₄^2- combined with cations other than NH₄⁺ or Group 1 cations may be soluble in some cases and insoluble in others.

When two different ionic compound solutions are mixed, a reaction will occur if cations from one solution can combine with anions from the other solution to form a solid precipitate that is insoluble. Since a solid precipitate is observed when the solutions Mg(NO₃)₂ (aq) and K₃PO₄ (aq) are mixed, we can deduce the formula of the precipitate by eliminating the combination of cation + anion that cannot possibly form an insoluble solid precipitate in water according to the solubility rule in bold print above:

K⁺ (aq) (Group 1 alkali metal cation) + NO₃^- (aq) → no reaction

Mg²⁺ (aq) + PO₄^3- (aq) → Mg₃(PO₄)₂ (s)

The balanced equation for the reaction of magnesium cations and phosphate anions is known as a net ionic equation because it only includes the ions that are truly involved in the precipitation reaction:

net ionic equation: 3 Mg²⁺ (aq) + 2 PO₄^3- (aq) → Mg₃(PO₄)₂ (s)

The K⁺ cations and NO₃^- anions are known as spectator ions because, although they are present in the solution, they do not actually take part in the reaction and, thus, are omitted from the net ionic equation.

If neither combination of cation + anion can form an insoluble solid precipitate, all ions will continue to exist separately in the mixture of solutions and, thus, no overall reaction occurs. For example, when a solution of lithium sulfate is mixed with a solution of ammonium nitrate, neither combination of cation from one solution + anion from the second solution can form an insoluble solid precipitate, so no overall reaction occurs and no solid will be observed:

Li⁺ (aq) + NO₃^- (aq) → no reaction

NH₄⁺ (aq) + SO₄^2- (aq) → no reaction

Note that it is possible for both combinations of cation + anion to yield a precipitate, in which case a mixture of two different solid precipitates will be produced in the same reaction.

Sample Exercise 10D:

Only two of the following four solution mixtures will react to form a precipitate. Indicate which two combinations yield no reaction and also write a balanced net ionic equation, including states of matter, for the two combinations that do yield a precipitate:

(i) AgNO₃ (aq) + (NH₄)₂CO₃ (aq)

(ii) NaNO₃ (aq) + CuSO₄ (aq)

(iii) aqueous ammonium chloride + aqueous lithium nitrate

(iv) aqueous potassium hydroxide + aqueous nickel(II) nitrate

Solution:

No reaction will occur for (ii) and (iii) as all combinations of cation + anion will be soluble. One precipitate will form in each of (i) and (iv):

(i) NH₄⁺ (aq) + NO₃^- (aq) → no reaction

net ionic equation: 2 Ag⁺ (aq) + CO₃^2- (aq) → Ag₂CO₃ (s)

(ii) Na⁺ (aq) + SO₄^2- (aq) → no reaction

Cu²⁺ (aq) + NO₃^- (aq) → no reaction

(iii) NH₄⁺ (aq) + NO₃^- (aq) → no reaction

Li⁺ (aq) + Cl^- (aq) → no reaction

(iv) K⁺ (aq) + NO₃^- (aq) → no reaction

net ionic equation: Ni²⁺ (aq) + 2 OH^- (aq) → Ni(OH)₂ (s)

Section 10-7: Acid + Base Neutralization Reactions and Molecular Equations

The reaction of an aqueous acid with an aqueous metal hydroxide can be considered a neutralization reaction because the acidic H⁺ ions from the acid solution and the basic OH^- ions from the metal hydroxide solution combine to form neutral water. Although these reactions can be written as net ionic equations, we can also write a molecular equation with spectator ions included as follows:

acid (aq) + metal hydroxide (aq) → water (l) + ionic compound (or ionic salt) (aq)

The correct formula of the ionic compound (or ionic salt) produced is a combination of the metal cation from the metal hydroxide and the anion from the acid. For example, in the reaction of sulfuric acid with sodium hydroxide, the Na⁺ cations from the NaOH combine with the SO₄^2- anions from the H₂SO₄ to yield the formula Na₂SO₄ in the balanced molecular equation:

H₂SO₄ (aq) + 2 NaOH (aq) → 2 H₂O (l) + Na₂SO₄ (aq)

Note that it is possible in some cases for the ionic compound produced to be an insoluble solid rather than aqueous.

Sample Exercise 10E:

Write a balanced molecular equation, including states of matter, for the neutralization reaction between solutions of acetic acid and calcium hydroxide.

Solution:

For neutralization reactions, we will write acetic acid as HCH₃COO:

2 HCH₃COO (aq) + Ca(OH)₂ (aq) → 2 H₂O (l) + Ca(CH₃COO)₂ (aq)

Section 10-8: Experiment – Acid + Base Titration

In a titration experiment, a precisely measured volume of one reactant solution is added dropwise from a buret (or burette) to a second reactant solution until the stoichiometric equivalence point of the reaction has been reached. When the stoichiometric equivalence point has been reached, neither reactant is in excess and, therefore, we can use the mole ratio from the balanced chemical reaction in stoichiometry calculations to determine the unknown molarity of one of the reactants.

One common titration experiment utilizes an aqueous acid + aqueous metal hydroxide neutralization reaction. An acid of known molarity and volume is placed in an Erlenmeyer flask along with a few drops of phenophthalein indicator, which will be colorless in the acid solution. A metal hydroxide solution with unknown molarity is then added dropwise from a buret until the phenolphthalein indicator in the mixture obtains a faint pink color signifying that the endpoint has been reached. Although this pink endpoint is caused by a very slight excess of basic hydroxide ions from the final drop of metal hydroxide solution that was added, we can assume that the exact stoichiometric equivalence point has been reached and, therefore, that neither the acid nor the metal hydroxide is in excess at that point.

After calculating the known moles of the acid, we can use the mole ratio from the balanced molecular equation to calculate the unknown moles of the metal hydroxide added from the buret. We then divide by the volume in liters of the metal hydroxide solution added from the buret to obtain the unknown molarity of the metal hydroxide solution. For example, if we start with 16.7 mL of 0.116 M hydrochloric acid in the Erlenmeyer flask and 14.9 mL of sodium hydroxide solution is required to titrate the acid (reach the equivalence point), the unknown molarity of the sodium hydroxide solution can be determined as follows:

Note that the calculated molarity of the NaOH solution may differ from the true value due to experimental error. For example, if extra drops of NaOH solution are inadvertantly added from the buret, the volume recorded for NaOH solution added will be too high. Since we would divide by too large a volume of NaOH solution in the final step, the calculated molarity of NaOH solution would be erroneously low.

If we know the volume and molarity of the acid in the Erlenmeyer flask as well as the molarity of the metal hydroxide solution being added from the buret, we can calculate the volume of metal hydroxide solution required for the titration, as demonstrated in the following problem:

Sample Exercise 10F:

How many milliliters of 0.019 M strontium hydroxide are required to titrate 24 mL of 0.026 M nitric acid?

Solution:

Chapter 10 Practice Exercises and Review Quizzes:

10-1) Write a balanced equation for the combustion of each compound below using the smallest possible whole-number coefficients:

(a) heptanol

(b) ethane

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10-1) (a) 2 C₇H₁₅OH + 21 O₂ → 14 CO₂ + 16 H₂O

(b) 2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O

10-2) Write a balanced equation for each reaction described below using the smallest possible whole-number coefficients:

(a) barium metal reacts with oxygen gas

(b) rubidium metal reacts with water

(d) calcium carbonate decomposes upon heating

Click for Solution

10-2)

(a) (product is barium oxide = combine Ba²⁺ and O^2- = BaO)

2 Ba + O₂ → 2 BaO

(b)

2 Rb + 2 H₂O → H₂ + 2 Rb⁺ + 2 OH^-

Cs₂O + H₂O → 2 Cs⁺ + 2 OH^-

(d) (calcium carbonate = combine Ca²⁺ and CO₃^2- = CaCO₃, product is calcium oxide = combine Ca²⁺ and O^2- = CaO)

CaCO₃ → CaO + CO₂

10-3) What is the molarity of each ion in the following solutions?

(a) 0.18 M (NH₄)₂CO₃

(b) 0.014 M FeBr₃

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10-3) (a) 2 x 0.18 M = 0.36 M NH₄⁺ and 1 x 0.18 M = 0.18 M CO₃^2-

(b) 1 x 0.014 M = 0.014 M Fe³⁺ and 3 x 0.014 M = 0.042 M Br^-

10-4) Only two of the following four solution mixtures will react to form a precipitate. Indicate which two combinations yield no reaction and also write a balanced net ionic equation, including states of matter, for the two combinations that do yield a precipitate:

(i) aqueous lithium nitrate + aqueous ammonium carbonate

(ii) aqueous sodium phosphate + aqueous calcium nitrate

(iii) Pb(NO₃)₂ (aq) + K₂SO₄ (aq)

(iv) NaI (aq) + NH₄Br (aq)

Click for Solution

10-4) No reaction will occur for (i) and (iv) as all combinations of cation + anion will be soluble. One precipitate will form in each of (ii) and (iii):

(i) NH₄⁺ (aq) + NO₃^- (aq) → no reaction

Li⁺ (aq) + CO₃^2- (aq) → no reaction

(ii) Na⁺ (aq) + NO₃^- (aq) → no reaction

net ionic equation: 3 Ca²⁺ (aq) + 2 PO₄^3- (aq) → Ca₃(PO₄)₂ (s)

(iii) K⁺ (aq) + NO₃^- (aq) → no reaction

net ionic equation: Pb²⁺ (aq) + SO₄^2- (aq) → PbSO₄ (s)

(iv) NH₄⁺ (aq) + I^- (aq) → no reaction

Na⁺ (aq) + Br^- (aq) → no reaction

10-5) Write a balanced molecular equation, including states of matter, for the neutralization reaction between solutions of sulfuric acid and lithium hydroxide.

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10-5) H₂SO₄ (aq) + 2 LiOH (aq) → 2 H₂O (l) + Li₂SO₄ (aq)

10-6) If 13 mL of 0.24 M barium hydroxide solution is required to titrate 28 mL of acetic acid solution, what is the molarity of the acetic acid solution?

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10-6)

10-7) How many milliliters of 0.0164 M hydrochloric acid can be titrated with 12.5 mL of 0.0214 M calcium hydroxide?

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10-7)

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