Chapter 10:  Chemical Tests and Chemical Reaction Types

 

Section 10-1:  Laboratory Tests to Identify Chemicals

Section 10-2:  Diatomic Nonmetals

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

singnificant 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 CwHxOySz requires sufficient oxygen gas and will generally produce carbon dioxide, water, and sulfur dioxide:

 

CwHxOySz + O2 CO2 + H2O + SO2

 

The products will be the same regardless of whether y = 0 or y > 0 or, in other words, regardless of whether or not the compound burned contains oxygen.  If the compound burned does not contain carbon (w = 0), then no carbon dioxide will be produced. If the compound burned does not contain hydrogen (x = 0), then no water will be produced. If the compound burned does not contain sulfur (z = 0), then no sulfur dioxide will be produced:

 

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

(c) carbon disulfide

 

Solution:

 

(a) 2 C6H14 + 19 O2 12 CO2 + 14 H2O

 

(b) 2 C3H7OH + 9 O2 6 CO2 + 8 H2O

 

(c) CS2 + 3 O2 CO2 + 2 SO2

 

 

 

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 N3- = 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 O2- = 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 CO32- = MgCO3,

magnesium oxide = combine Mg2+ and O2- = 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

(c) strontium oxide 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-

 

(c) (strontium oxide = combine Sr2+ and O2- = SrO)

 

SrO + H2O Sr2+ + 2 OH-

 

(d) (sodium carbonate = combine 2 Na+ and CO32- = Na2CO3,

product is sodium oxide = combine 2 Na+ and O2- = 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 SO42- 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 PO43- will be 1 x 0.026 M = 0.026 M PO43- since there is one PO43- 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.  We will use the following widely accepted general solubility rules to determine which common ionic compounds will be soluble versus insoluble in water (solubility rules from other sources may differ slightly):

 

Water Solubility of Ionic Compounds

 

1) always soluble:  compounds containing Group 1 alkali metal cations, NH4+, CH3COO-, or NO3-

 

2) usually soluble:  compounds containing Br-, Cl-, or I- (except when combined with Ag+, Hg22+, or Pb2+)

 

3) usually soluble:  compounds containing SO42- (except when combined with Ba2+, Ca2+, Pb2+, or Sr2+)

 

4) usually insoluble:  compounds containing CO32-, CrO42- (chromate), or PO43- (except when combined with Group 1 alkali metal cations or NH4+)

 

5) usually insoluble:  compounds containing OH- or S2- (except when combined with Group 1 alkali metal cations, NH4+, Ba2+, Ca2+, or Sr2+)

 

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.  When the solutions Mg(NO3)2 (aq) and K3PO4 (aq) are mixed, we can eliminate one combination of cation + anion that cannot possibly form an insoluble solid precipitate in water according to the solubility rules above, and we can also determine that the second combination of cation + anion will form an insoluble precipitate:

 

K+ (aq) (Group 1 alkali metal cation) + NO3- (aq) no reaction     [Rule 1]

 

Mg2+ (aq) + PO43- (aq) Mg3(PO4)2 (s)     [Rule 4]

 

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 PO43- (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) + SO42- (aq) no reaction

 

 

Sample Exercise 10D:

 

Indicate if any of the combinations below yield no reaction and also write a balanced net ionic equation, including states of matter, for any 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) + CO32- (aq) Ag2CO3 (s)

 

(ii) Na+ (aq) + SO42- (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)

 

 

Note that it is also 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.  For example, when CoSO4 (aq) is mixed with Ba(OH)2 (aq), two precipitates are formed simultaneously according to the following net ionic equation:

 

Co2+ (aq) + 2 OH- (aq) + Ba2+ (aq) + SO42- (aq) Co(OH)2 (s) + BaSO4 (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 SO42- anions from the H2SO4 to yield the formula Na2SO4 in the balanced molecular equation:

 

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

 

 

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)

 

 

Note that it is possible for the ionic compound produced to be an insoluble solid rather than aqueous based on the solubility rules above.  For example, when solutions of sulfuric acid and barium hydroxide are mixed, the balanced molecular equation is:

 

H2SO4 (aq) + Ba(OH)2 (aq) 2 H2O (l) + BaSO4 (s)

 

 

 

 

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

(c) C4H4S

Click for Solution

 

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

 

           (b) 2 C2H6 + 7 O2 4 CO2 + 6 H2O

 

           (c) C4H4S + 6 O2 4 CO2 + 2 H2O + SO2

 

 

 

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

(c) cesium oxide reacts with water

(d) calcium carbonate decomposes upon heating

Click for Solution

 

10-2)

(a) (product is barium oxide = combine Ba2+ and O2- = BaO)

 

2 Ba + O2 2 BaO

 

(b)

2 Rb + 2 H2O H2 + 2 Rb+ + 2 OH-

 

(c) (cesium oxide = combine Cs+ and O2- = Cs2O)

 

Cs2O + H2O 2 Cs+ + 2 OH-

 

(d) (calcium carbonate = combine Ca2+ and CO32- = CaCO3, product is calcium oxide = combine Ca2+ and O2- = 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

Click for Solution

 

10-3) (a) 2 x 0.18 M = 0.36 M NH4+ and 1 x 0.18 M = 0.18 M CO32-

 

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

 

 

 

10-4) Indicate if any of the combinations below yield no reaction and also write a balanced net ionic equation, including states of matter, for any combinations that do yield a precipitate:

 

(i) aqueous lithium nitrate + aqueous ammonium carbonate

(ii) aqueous sodium phosphate + aqueous calcium acetate

(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) + CO32- (aq) no reaction

 

(ii) Na+ (aq) + CH3COO- (aq) no reaction

 

net ionic equation:  3 Ca2+ (aq) + 2 PO43- (aq) Ca3(PO4)2 (s)

 

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

 

 net ionic equation:  Pb2+ (aq) + SO42- (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.

Click for Solution

 

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?

Click for Solution

 

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?

Click for Solution

 

10-7)

 

 

 

Click for Review Quiz 1

Click for Review Quiz 1 Answers