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    DR. JIM BLOXTONS OPTIONAL CHEM 1021 NOMENCLATURE ANDMOLECULAR, FULL IONIC AND NET IONIC EQUATIONS

    BACKGROUND:

    Scientists use chemical symbols to represent the names of elements in chemical equationsand in chemical formulas. The names and formulas for elements that you will need for Chem1023 are given below:

    Group IA Group IIA Transition Elements Transition ElementsH hydrogen Mg magnesium Fe iron Ag silver Li lithium Ca calcium Co cobalt Au goldNa sodium Sr stronium Pt platinum Zn zincK potassium Ba barium Cu copper Cd cadmium

    Ra radium Mn manganese Hg mercuryCr chromium Mo molybdenium

    Group IIIA Group IVA Group VA Group VIAAl aluminum C carbon N nitrogen O oxygen

    Sn tin P phosphorus S sulfur Pb lead Se selenium

    Group VIIA Group VIIIAF fluorine Rn RadonCl chlorineBr bromineI iodine

    Most of these elements are represented as monatomic or one atom when they are writtenin chemical equations. An example is sodium which is written as Na in a chemical equation.The absence of a subscript after an elements symbol indicates that there is one atom of thatelement. Elements such as hydrogen, mercury, nitrogen, oxygen, fluorine, chlorine, bromineand iodine are more stable as diatomic molecules or two atom molecules. These elements arerepresented as follows in chemicals equations: H2, Hg2, N2, O2, F2, Cl2, Br2 and I2. The presenceof the subscript two after an elements symbol indicates that there are two atoms of that element.

    Ions play a very important role in maintaining life itself. An ion has either a positive ornegative charge and enables aqueous solutions to carry electrical charge. A cation is an ion witha positive charge. An anion is an ion with a negative charge. Ions are involved in many cellularprocess in our bodies. Some of the important ions in biological systems include K+, Na+, Li+,Mg2+, Fe3+, Cl-, I-, HCO3

    -, H2PO4-, and HPO4

    2-. The names and formulas of some of the ions thatyou will need to know for this course are given below.

    Group IA Cations Group IIA Cations Group IIIA CationH+ hydrogen ion Mg2+ magnesium ion Al3+ aluminum ionLi+ lithium ion Ca2+ calcium ion

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    Na+ sodium ion Ba2+ barium ionK+ potassium ion

    Transition Metal Cations Transition Metal Cations Group IVA CationsAg+ silver ion Fe2+ iron(II) ion or Sn2+ tin(II) ion orZn2+ zinc ion ferrous ion stannous ionCu+ copper(I) ion or Fe3+ iron(III) ion or Sn4+ tin(IV) ion or

    cuprous ion ferric ion stannic ionCu2+ copper(II) ion or Hg2

    2+ mercury(I) ion or Pb2+ lead(II) ioncupric ion mercurous ion Pb4+ lead(IV) ion

    Hg2+ mercury(II) ion orPolyatomic Cations mercuric ionNH4

    + ammonium ionH3O

    + hydronium ion

    The elements group number, Roman numeral for groups IA, IIA, and IIA corresponds tothe charge for the cations of these elements. Thus Group IA elements such as H, Li, Na and K

    will lose one electron to form a cation with a 1+ charge. The number of valence electrons thatan element has corresponds to the elements group number for the A group or representativeelements. Li, Na and K lose their 1 valence electron so that they can have the electronconfiguration of a noble gas.

    Group IIA elements such as Mg, Ca and Ba will lose two electrons to form a cation witha 2+ charge. Mg, Ca and Ba lose their 2 valence electrons so that they can have the electronconfiguration of a noble gas. Likewise Group IIIA elements such as Al will lose three electronsto form a cation with a 3+ charge. Al loses 3 valence electrons so that it can have the electronconfiguration of a noble gas.

    NOMENCLATURE RULE: The rules for naming cations are as follows:1. Write the name of the element followed by the word ion.Examples: What are the names of Li+, Ca2+ and Al3+?Answers: lithium ion, calcium ion, and aluminum ion.2. If an element has more than one oxidation state or charged form, the oxidation number orcharge must be specified as a Roman numeral that is enclosed in parenthesis after the name ofthe element. The word ion is also included in the name.Examples: What are the names of Cu2+, Fe3+ and Hg2

    2+?Answers: copper(II) ion, iron(III) ion, mercury(I) ion.Each mercury atom in Hg2

    2+ has a charge of 1+ since the charge on the Hg 22+ ion is 2+ and there

    are 2 mercury atoms in the ion. Each mercury atom in the ion has a charge of 1+.3. Polyatomic ions are ions that are composed of more than one atom. Polyatomic ions arenamed by writing the name of the ion followed by the word ion.Example: What is the name of NH4

    +? Answer: ammonium ion.Group IA Anion Group VIA Anions Group VIIA AnionsH- hydride ion O2- oxide ion F- fluoride ion

    S2- sulfide ion Cl- chloride ionSe2- selenide ion Br - bromide ion

    I- iodide ion

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    Polyatomic Anions Polyatomic AnionsHSO4

    - hydrogen sulfate ion HCO3- hydrogen carbonate ion or

    SO42- sulfate ion bicarbonate ion

    SO32- sulfite ion CO3

    2- carbonate ionNO2

    - nitrite ion OH- hydroxide ionNO3

    - nitrate ion CN- cyanide ionH2PO4

    - dihydrogen phosphate ion C2H3O2- acetate ion

    HPO42- hydrogen phosphate ion O2

    2- peroxide ionPO4

    3- phosphate ion

    The number of valence electrons that an element has corresponds to the elements groupnumber for the A group or representative elements. H is a group IA element can either gain orlose one electron. If H gains one electron it will have the electron configuration of the noble gasHe. When H gains one electron the H becomes an anion that has a charge of 1- and thus the H-

    or hydride ion is formed.

    Group VIA elements such as O, S and Se already have six valence electrons and will

    gain two electrons to form an anion with a 2- charge. O, S and Se gain 2 more electrons so thatthey can have the electron configuration of a noble gas.

    Likewise Group VIIA elements such as F, Cl, Br and I already have seven valenceelectrons and will gain one electron to form an anion with a 1- charge. F, Cl, Br and I gain 1more electron so that they can have the electron configuration of a noble gas.

    NOMENCLATURE RULE: The rules for naming anions are as follows:1. Write the name of the element and change the ending to end in ide.2. Add the word ion.Examples: What are the names of the following: A. H-, B. O2- and C. F-?

    Answers: A. H-

    is derived from H, hydrogen. The ogen is dropped from hydrogen and replacedwith the suffix ide. This produces the word hydride. We need to follow hydride with the wordion to give the name of H- as hydride ion.B. O2- is derived from O, oxygen. The ygen is dropped from oxygen and replaced with thesuffix ide. This produces the word oxide. We need to follow oxide with the word ion to givethe name of O2- as oxide ion.C. F - is derived from F, fluorine. The ine is dropped from fluorine and replaced with the suffixide. This produces the word fluoride. We need to follow fluoride with the word ion to give thename of F- as fluoride ion.3. Polyatomic ions are named by writing the name of the ion followed by the word ion.Example: What is the name of OH-? Answer: hydroxide ion.

    The names and formulas of some of the acids that you will need to know for this courseare given below. An acid is a source of hydrogen ion or H+.

    Monoprotic Acids Diprotic AcidsHF hydrogen fluoride or H2SO4 sulfuric acidHF(aq) hydrofluoric acid H2CO3 carbonic acidHCl hydrogen chloride or HCl(aq) hydrochloric acid Triprotic AcidHBr hydrogen bromide or H3PO4 phosphoric acid

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    HBr(aq) hydrobromic acidHI hydrogen iodide or Other Important AcidsHI(aq) hydroiodic acid HC2H3O2 acetic acidHNO3 nitric acid NH4Cl ammonium chloride

    Acids can also be defined as hydrogen containing molecular compounds that whendissolved in water loses one or more hydrogen ions, H+. This process is called ionization ordissociation. Acids can be further classified as monoprotic, diprotic, or triprotic acids.Polyprotic acid is a general term for acids that release more than one H+ ion upon ionization.

    A monoprotic acid releases one H+ ion upon ionization. Examples of monoprotic acidsare HF, HCl, HBr, HI and HNO3.

    Ionization: Ex. HF(aq) H+

    (aq) + F-(aq)

    Hydrofluoric acid is separating into a hydrogen ion and a fluoride ion.Hydrofluoric acid: The acid formed when hydrogen fluoride (HF) is dissolved in water.

    H+: Called a proton and is also known as a hydrogen ion. Note in an aqueous solution, the H+

    ion or hydrogen ion bonds with H2O to form H3O+ which is called a hydronium ion.

    Ionization: Ex. HCl(aq) H+

    (aq) + Cl-(aq)

    Hydrochloric acid is separating into a hydrogen ion and a chloride ion.Hydrochloric acid: The acid formed when hydrogen chloride (HCl) is dissolved in water.

    Ionization: Ex. HBr(aq) H+

    (aq) + Br-(aq)

    Hydrobromic acid is separating into a hydrogen ion and a bromide ion.Hydrobromic acid: The acid formed when hydrogen bromide (HBr) is dissolved in water.

    Ionization: Ex. HI(aq) H+

    (aq) + I-

    (aq)

    Hydroiodic acid is separating into a hydrogen ion and an iodide ion.Hydroiodic acid: The acid formed when hydrogen iodide (HI) is dissolved in water.

    Ionization: Ex. HNO3(aq) H+

    (aq) + NO3-(aq)

    Nitric acid is separating into a hydrogen ion and a nitrate ion.

    Acetic acid is usually treated as a monoprotic acid. The equation for the ionization ofacetic acid is as follows:

    Ionization: Ex. HC2H3O2(aq) H+

    (aq) + C2H3O2-(aq)

    Acetic acid is separating into a hydrogen ion and an acetate ion.

    Ammonium chloride, NH4Cl, is a source of ammonium cation, NH4+. The ammonium

    cation is a weak acid that is also a monoprotic acid. The equation for the formation of theammonium cation in water or an aqueous solution is as follows:

    NH4Cl (aq) NH4+

    (aq) + Cl-(aq)

    Ammonium chloride dissolves in water to form ammonium ion and chloride ion.

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    Ionization: Ex. NH4+

    (aq) H+

    (aq) + NH3(aq)Ammonium ion is separating into a hydrogen ion and ammonia.

    A diprotic acid releases two H+ ions upon complete ionization. However these two H+

    ions are released in two separate steps. Examples of diprotic acids are H2SO4 and H2CO3.

    Example:Step 1 Ionization: Ex. H2SO4(aq) H

    +(aq) + HSO4

    -(aq)

    Sulfuric acid is separating into a hydrogen ion and a hydrogen sulfate ion.

    Step 2 Ionization: Ex. HSO4-

    (aq) H+

    (aq) + SO42-

    (aq)

    Hydrogen sulfate ion is separating into a hydrogen ion and a sulfate ion.

    A triprotic acid releases three H+ ions upon complete ionization. However these threeH+ ions are released in three separate steps. An example of a triprotic acid is H3PO4.

    Example:

    Step 1 Ionization: Ex. H3PO4(aq) H+(aq) + H2PO4-(aq)Phosphoric acid is separating into a hydrogen ion and a dihydrogen phosphate ion.

    Step 2 Ionization: Ex. H2PO4-(aq) H

    +(aq) + HPO4

    2-(aq)

    Dihydrogen phosphate ion is separating into a hydrogen ion and a hydrogen phosphate ion.

    Step 3 Ionization: Ex. HPO42-

    (aq) H+

    (aq) + PO43-

    (aq)

    hydrogen phosphate ion is separating into a hydrogen ion and a phosphate ion.

    The names and formulas of some of the bases that you will need to know for this courseare given below. A base is a source of hydroxide ion, OH-, or an acceptor of H+.

    Metal Hydroxide Bases Metal Oxide Bases NaOH sodium hydroxide Na2O sodium oxideKOH potassium hydroxide K 2O potassium oxideMg(OH)2 magnesium hydroxideCa(OH)2 calcium hydroxide

    Metal Bicarbonate Bases Metal Carbonate BasesNaHCO3 sodium bicarbonate or Li2CO3 lithium carbonate

    sodium hydrogen carbonate Na2CO3 sodium carbonateKHCO3 potassium bicarbonate or K 2CO3 potassium carbonate

    potassium hydrogen carbonate

    Other Bases

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    NH3 ammoniaNH4OH ammonium hydroxide

    NOMENCLATURE RULE: All of these bases except for ammonia are named as if they wereionic compounds. Actually all of these bases except ammonia are ionic compounds. The rulesfor naming these types of ionic compounds are as follows:1. Write the name of the cation first omitting the word ion.2. Write the name of the anion next and omit the word ion.

    Example. What is the name of Mg(OH)2?Answer: Mg(OH)2 is composed of the magnesium ion, Mg

    2+, and two hydroxide ions OH-.Therefore the name of Mg(OH)2 is magnesium hydroxide.

    Example. What is the name of K2O?Answer: K2O is composed of the two potassium ions, K

    +, and one hydroxide ion OH-.Therefore the name of K2O is potassium hydroxide.

    Example. What is the name of NaHCO3?Answer: NaHCO3 is composed of the sodium ion, Na

    +, and the bicarbonate ion HCO3- which is

    also called hydrogen carbonate ion. Therefore the name of NaHCO3 is sodium bicarbonate orsodium hydrogen carbonate.

    Example. What is the name of Li2CO3?Answer: Li2CO3 is composed of the two lithium ions, Li

    +, and one carbonate ion CO32-.

    Therefore the name of Li2CO3 is lithium carbonate.

    Example. What is the name of Fe(OH)3?Answer: Fe(OH)3 is composed of the iron(III) ion, Fe

    3+, and three hydroxide ions OH-.

    Therefore the name of Fe(OH)3 is iron(III) hydroxide.The next two nomenclature examples involve ionic compounds that are not bases.

    Example. What is the name of BaI2?Answer: BaI2 is composed of one barium ion, Ba

    2+, and two iodide ions I-. Therefore the nameof BaI2 is barium iodide.

    Example. What is the name of KCl?Answer: KCl is composed of the one potassium ion, K+, and one chloride ion Cl-. Therefore thename of KCl is potassium chloride.

    The next two examples involve naming ionic compounds that contain more than one polyatomicion in their formula. We have to enclose the polyatomic ion in parenthesis and use a subscriptafter the closing parenthesis to indicate the how many polyatomic ions are in the formula.

    Example. What is the name of Al2(SO4)3?Answer: Al2(SO4)3 is composed of two aluminum ions, Al

    3+, and three sulfate ions SO42-.

    Therefore the name of Al2(SO4)3 is aluminum sulfate.

    Example. What is the name of Ca(NO3)2?

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    Answer: Ca(NO3)2 is composed of one calcium ion, Ca2+, and two nitrate ions NO3

    -. Thereforethe name of Ca(NO3)2 is calcium nitrate.

    METHOD FOR WRITING THE FORMULA OF AN IONIC COMPOUNDFROM THE NAME OF THE IONIC COMPOUND

    This section involves writing the formula of an ionic compound from the name of theionic compound. The steps involved are in some ways the reverse of naming an ionic compoundfrom the formula of the ionic compound.

    Example: What is the formula of magnesium bromide?Answer: The ionic compound is composed of the Mg2+ cation and the Br-(aq) anion. In ioniccompounds, the overall charge has to be neutral or zero. This means that there must be an equalnumber of positive charges as well as an equal number of negative charges. Thus we need tofigure out how many Mg2+ cations and how many Br- anions we need. In figuring this out it isconvenient to use the following general format:

    CationxAniony where x is the number of cations in the formula and y is the number ofanions in the formula. We figure out the value of x by letting x equal the absolute value of thecharge on the anion. In a similar manner, the value of y is equal to the charge of the cation.

    For this example we will now solve for the values of x and y in MgxBry. Since Br- has a

    charge of 1- we take the absolute value of 1- which is 1 and this is our value of x. Since x = 1this means there is one atom of Mg in the formula or our ionic compound. Likewise since Mg2+

    has a charge of 2+, the value of y = 2. This means that we have two atoms of bromine in ourionic compound. The formula for MgxBry is MgBr2. The formula of magnesium bromide isMgBr2.Example: What is the formula of ammonium oxide?

    Answer: The ionic compound is composed of the NH4+

    cation and the O2-

    (aq) anion. In ioniccompounds, the overall charge has to be neutral or zero. This means that there must be an equalnumber of positive charges as well as an equal number of negative charges. Thus we need tofigure out how many NH4

    + cations and how many O2- anions we need. In figuring this out it isconvenient to use the following general format:

    CationxAniony where x is the number of cations in the formula and y is the number ofanions in the formula. We figure out the value of x by letting x equal the absolute value of thecharge on the anion. In a similar manner, the value of y is equal to the charge of the cation.

    For this example we will now solve for the values of x and y in (NH 4)xOy. Since O2- has

    a charge of 2- we take the absolute value of 2- which is 2 and this is our value of x. Since x = 2this means there are two formula units of NH4 in the formula or our ionic compound. Likewisesince NH4

    + has a charge of 1+, the value of y = 1. This means that we have one atom of oxygenin our ionic compound. The formula for (NH4)xOy is (NH4)2O. The formula of ammoniumoxide is (NH4)2O.

    Example: What is the formula of barium sulfate?

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    Answer: The ionic compound is composed of the Ba2+ cation and the SO42- anion. In ionic

    compounds, the overall charge has to be neutral or zero. This means that there must be an equalnumber of positive charges as well as an equal number of negative charges. Thus we need tofigure out how many Ba2+ cations and how many SO4

    2- anions we need. In figuring this out it isconvenient to use the following general format:

    CationxAniony where x is the number of cations in the formula and y is the number ofanions in the formula. We figure out the value of x by letting x equal the absolute value of thecharge on the anion. In a similar manner, the value of y is equal to the charge of the cation.

    For this example we will now solve for the values of x and y in Bax(SO4)y. Since SO42-

    has a charge of 2- we take the absolute value of 2- which is 2 and this is our value of x. Since x= 2 this means there are two atoms of Ba in the formula or our ionic compound. Likewise sinceBa2+ has a charge of 2+, the value of y = 2. This means that we have two formula units of thesulfate ion in our ionic compound. The formula for Bax(SO4)y appears to be Ba2(SO4)2.However we need to reduce the values of x and y from x = 2 and y = 2 to x = 1 and y = 1. Thenthe correct formula for Bax(SO4)y becomes BaSO4. The formula of barium sulfate is BaSO4.

    METHODS FOR WRITING MOLECULAR EQUATIONS, FULLIONIC EQUATIONS AND NET IONIC EQUATIONS

    Notes:

    1. The following terms and definitions are useful:Solute: a solid or gas that is dissolved in a solvent.Solvent: a liquid that the solute is dissolved in.Solution: a combination of solute and solvent.Saturated solution: a solution that cannot have more solute dissolved in it at a giventemperature.Unsaturated solution: A solution that can have more solute dissolved in it.Soluble: dissolves in a particular solvent.

    2. In writing molecular, full ionic and net ionic equations we must include the states with eachreactant and product. The states are indicated as abbreviations that are placed as subscripts aftereach reactant and product. The states and their abbreviations are as follows:gas (g)liquid (l)solid (g)aqueous (aq)The term aqueous is used when the solvent is water.

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    3. If a compound is water insoluble this means that the compound does not dissolve in waterand remains as a solid which is indicated as a subscript (s) after the compound. If a compound issoluble in water, this is indicated by the use of the subscript (aq) after the compound.

    4. In molecular equations, water soluble acids, bases and ionic compounds are not broken upinto ions. In molecular equations gases, liquids and solids are not broken up into ions.

    5. In full ionic equations, strong acids, strong bases and ionic compounds that are soluble inwater are broken up into anions and cations. In full ionic equations weak acids that are not ioniccompounds, weak bases, gases, liquids and solids are not broken up into ions.

    6. In net ionic equations, strong acids, strong bases and ionic compounds that are soluble inwater are broken up into anions and cations. In net ionic equations weak acids that are not ioniccompounds, weak bases, gases, liquids and solids are not broken up into ions.

    7. The net ionic equation is derived from the full ionic equation by identifying and eliminatingspectator ions.

    8. A species is a spectator if the following three items remain exactly the same when a species isa reactant and a product: formula, charge and state.

    Example 1: Write a balanced net ionic equation for the reaction between water insolublecalcium hydroxide Ca(OH)2 and an aqueous solution of hydrobromic acid HBr.

    We first need to realize that the reactants are Ca(OH)2 and HBr. Ca(OH)2 is a metalhydroxide base that supplies a hydroxide ion OH- that will react with H+ ion that is supplied bythe acid HBr. In these equations we must include the states with each reactant and product. The

    states are indicated as subscripts after each reactant and product. We will start out by writing thereactants part of the molecular equation as follows in partial equation 1:

    Ca(OH)2(s) + HBr(aq) Partial Equation 1

    The state for Ca(OH)2 is indicated as a solid since Ca(OH)2 is insoluble in water.Aqueous is the state for HBr since we are dealing with an aqueous solution of HBr. The acidHBr dissolves or is soluble in water.

    Before we can complete the molecular equation, we need to realize what is taking placewith the reactants and then write the full ionic equation. When we write full ionic equations, thereactants and products have to be broken apart into their ions if they are water soluble ioniccompounds, water soluble strong acids or water soluble strong bases. Since Ca(OH)2 isinsoluble in water the base Ca(OH)2 will largely remain in the solid form of Ca(OH)2(s). HBr is astrong acid that is soluble in water. Strong acids will ionize or break apart almost completely toform ions called anions and cations. The equation for the ionization or dissociation of HBr toform the cation H+ or hydrogen ion and the anion Br- or bromide ion is as follows in equation 2:

    HBr(aq) H+

    (aq) + Br-(aq) Equation 2

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    This means that the reactants Ca(OH)2(s) and HBr(aq) are represented as follows in thepartial full ionic equation in partial equation 3:

    Ca(OH)2(s) + H+

    (aq) + Br-(aq) Partial Equation 3

    Next we need to look at the interactions between Ca(OH)2(s) and H+

    (aq) and Br-(aq). The

    reaction of interest will be an acid base reaction that takes place between H+(aq) or hydrogen ionand the OH- or hydroxide ion that is part of Ca(OH)2(s). This reaction will produce liquid wateror H2O(l) as a product. The calcium ion that is part of Ca(OH)2(s) will now be in the form of Ca

    2+

    (aq) as a product. The Br-(aq) or bromide ion remains the same during this reaction and will be a

    product. We can now write an unbalanced full ionic equation as follows in equation 4:

    Ca(OH)2(s) + H+

    (aq) + Br-(aq) H2O(l) + Ca

    2+(aq) + Br

    -(aq) Equation 4

    The next step involves completing the molecular equation and then balancing the

    molecular equation. In molecular equations, anion and cation components of ionic compoundshave to be combined together. The ionic compound in this example consists of Ca2+(aq) cation

    and Br-(aq) anion. In ionic compounds, the overall charge has to be neutral or zero. This meansthat there must be an equal number of positive charges as well as an equal number of negativecharges. Thus we need to figure out how many Ca2+ cations and how many Br- anions we need.In figuring this out it is convenient to use the following general format:

    CationxAniony where x is the number of cations in the formula and y is the number ofanions in the formula. We figure out the value of x by letting x equal the absolute value of thecharge on the anion. In a similar manner, the value of y is equal to the charge of the cation.

    For this example we will now solve for the values of x and y in CaxBry. Since Br- has a

    charge of 1- we take the absolute value of 1- which is 1 and this is our value of x. Since x = 1

    this means there is one atom of Ca in the formula or our ionic compound. Likewise since Ca2+

    has a charge of 2+, the value of y = 2. This means that we have two atoms of bromine in ourionic compound. The formula for CaxBry is thus CaBr2. The name of CaBr2 is calcium bromide.

    Since H2O is our other product we can now write our unbalanced molecular equation asfollows in equation 5:

    Ca(OH)2(s) + HBr(aq) H2O(l) + CaBr2(aq) Equation 5

    Now we need to balance our molecular equation. In balanced equations, there must beequal numbers of each atom on the reactants and products side of the equation. We balanceequations by the use of coefficients that are placed in front of formulas. We never change theformula of a reactant or product when we balance an equation.

    In our unbalanced molecular equation we have 1 Ca atom on both sides of the equation.However on the reactants side we have 2 atoms of O, 3 atoms of H and 1 atom of Br. Each OHin Ca(OH)2 contributes 1 O atom and since there are two OH in Ca(OH)2 this gives us 2 atoms ofO on the reactants side. Likewise each OH in Ca(OH)2 contributes 1 H atom and since there aretwo OH in Ca(OH)2 this gives us 2 atoms of H from Ca(OH)2. Since HBr also supplies 1 Hatom the total number of H atoms from the reactants Ca(OH) 2 and HBr is 3. On the products

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    side we have 1 atom of O, 2 atoms of H and 2 atoms of Br. We can adjust the number of Bratoms by placing a coefficient of 2 in front of HBr on the reactants side. This gives us thefollowing unbalanced molecular equation in equation 6:

    Ca(OH)2(s) + 2HBr(aq) H2O(l) + CaBr2(aq) Equation 6

    In this unbalanced molecular equation we still have 1 Ca atom on both sides of theequation. However on the reactants side we now have 2 atoms of O, 4 atoms of H and 2 atomsof Br. On the products side we still have 1 atom of O, 2 atoms of H and 2 atoms of Br. We canadjust the number of O atoms by placing a coefficient of 2 in front of H 2O on the products side.This gives us the following balanced molecular equation in equation 7:

    Ca(OH)2(s) + 2HBr(aq) 2H2O(l) + CaBr2(aq) Balanced Molecular Equation 7

    In this balanced molecular equation we have 1 atom of Ca, 2 atoms of O, 4 atoms of Hand 2 atoms of Br on the reactants and products side of the equation.

    Now we can write a balanced full ionic equation from our balanced molecular equation

    7. Equation 8 is the balanced full ionic equation.

    Ca(OH)2(s) + 2H+

    (aq) + 2Br-(aq) 2H2O(l) + Ca2+(aq) + 2Br-(aq) Equation 8

    We derive our balanced net ionic equation from our balanced full ionic equation byidentifying and eliminating spectator ions. Remember a species is a spectator if the followingthree items remain exactly the same when a species is a reactant and a product: formula, chargeand state. Br-(aq) ion meets this three point criteria. All of the other species in the full ionicequation had some type of change. Thus Br-(aq) is our spectator ion. Eliminating Br

    -(aq) ion from

    the reactant and product sides of balanced full ionic equation 8 gives us our balanced net ionicequation. Equation 9 is a balanced net ionic equation.

    Ca(OH)2(s) + 2H+

    (aq) 2H2O(l) + Ca2+(aq) Equation 9

    In checking to see if a net ionic equation is correct, the following items should bechecked: 1. Is the equation balanced? 2. Are the states included? 3. Are the appropriatecharges present? 4. Were spectator ions eliminated?

    Summary:Balanced Molecular Equation:Ca(OH)2(s) + 2HBr(aq) 2H2O(l) + CaBr2(aq) Equation 7

    Balanced Full Ionic Equation:Ca(OH)2(s) + 2H

    +(aq) + 2Br

    -(aq) 2H2O(l) + Ca2+(aq) + 2Br-(aq) Equation 8

    Balanced Net Ionic Equation:Ca(OH)2(s) + 2H

    +(aq) 2H2O(l) + Ca2+(aq) Equation 9

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    Example 2: Write a balanced net ionic equation for the reaction between an aqueous solution ofpotassium hydroxide KOH and an aqueous solution of hydroiodic acid HI.

    We first need to realize that the reactants are KOH and HI. KOH is a metal hydroxidebase that supplies hydroxide ion OH- that will react with H+ ion that is supplied by the acid HI.

    In these equations we must include the states with each reactant and product. The states areindicated as subscripts after each reactant and product. The states for KOH and HI are indicatedas aqueous since we are dealing with aqueous solutions of KOH and HI. We will start out bywriting the reactants part of the molecular equation as follows in partial molecular equation 10:

    KOH(aq) + HI(aq) Partial Equation 10

    Before we can complete the molecular equation, we need to realize what is taking placewith the reactants and then write the full ionic equation. When we write full ionic equations, thereactants and products have to be broken apart into their ions if they are water soluble ioniccompounds, water soluble strong acids or water soluble strong bases. KOH is a strong base that

    is soluble in water. Strong bases will ionize or break apart almost completely to form ions calledanions and cations. The equation for the ionization or dissociation of KOH to form the cation K+

    or potassium ion and the anion OH- or hydroxide ion is as follows in equation 11:

    KOH(aq) K+

    (aq) + OH-(aq) Equation 11

    HI is a strong acid that is soluble in water. Strong acids will ionize or break apart almostcompletely to form ions called anions and cations. The equation for the ionization ordissociation of HI to form the cation H+ or hydrogen ion and the anion I- or iodide ion is asfollows in equation 12:

    HI(aq) H+

    (aq) + I-(aq) Equation 12

    Using the results of equations 11 and 12 we realize that the reactants KOH (aq) and HI(aq)are represented as follows in the partial full ionic equation in equation 13:

    K+(aq) + OH-(aq) + H

    +(aq) + I

    -(aq) Partial Equation 13

    Next we need to look at the interactions between K+(aq) and OH-(aq) and H

    +(aq) and I

    -(aq). In

    reactions between strong acids and strong bases, it is useful to pair the anion from the base, OH -,

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    with the cation from the acid, H+, to form liquid water H2O(l) as one of the products. This is anacid base reaction that takes place between H+ or hydrogen ion and the OH- or hydroxide ion.The K+ cation or potassium ion will also be paired with the I- anion or iodide ion to form anionic compound or salt that is soluble in water. We will represent K+ cation and the I- anion asK+(aq) and I

    -(aq) on the products side of the full ionic equation. We can now write a full ionic

    equation as follows in equation 14:

    K+(aq) + OH-(aq) + H

    +(aq) + I

    -(aq) H2O(l) + K

    +(aq) + I

    -(aq) Equation 14

    The next step involves completing the molecular equation and then balancing the

    molecular equation. In molecular equations, anion and cation components of ionic compoundshave to be combined together. The ionic compound in this example consists of K+(aq) cation andI-(aq) anion. In ionic compounds, the overall charge has to be neutral or zero. This means thatthere must be an equal number of positive charges as well as an equal number of negativecharges. Thus we need to figure out how many K+ cations and how many I- anions we need. Infiguring this out it is convenient to use the following general format: CationxAniony where x isthe number of cations in the formula and y is the number of anions in the formula. We figure

    out the value of x by letting x equal the absolute value of the charge on the anion. In a similarmanner, the value of y is equal to the charge of the cation.For this example we will now solve for the values of x and y in KxIy. Since I

    - has a charge of 1-we take the absolute value of 1- which is 1 and this is our value of x. Since x = 1 this meansthere is one atom of K in the formula or our ionic compound. Likewise since K+ has a charge of1+, the value of y = 1. This means that we have one atom of iodine in our ionic compound. Theformula for KxIy is thus KI. The name of KI is potassium iodide. Since H2O is our otherproduct we can write our molecular equation as follows in equation 15:

    KOH(aq) + HI(aq) H2O(l) + KI(aq) Equation 15

    Now we need to balance our molecular equation. In balanced equations, there must beequal numbers of each atom on the reactants and products side of the equation. We balanceequations by the use of coefficients that are placed in front of formulas. We never change theformula of a reactant or product when we balance an equation. In our molecular equation wehave 1 K atom, 1 O atom, 2 H atoms, and 1 I atom on both sides of the equation. We have equalnumbers of each type of atom on both sides of our molecular equation and thus we have abalanced molecular equation. Equation 15 is a balanced molecular equation.

    KOH (aq) + HI (aq) H2O (l) + KI (aq) Balanced Molecular Equation 15

    Now we can write a balanced full ionic equation from our balanced molecular equation15. Equation 16 is a balanced full ionic equation.

    K+ (aq) + OH-(aq) + H

    +(aq) + I

    -(aq) H2O (l) + K+ (aq) + I-(aq) Equation 16

    We derive our balanced net ionic equation from our balanced full ionic equation byidentifying and eliminating spectator ions. Remember a species is a spectator if the followingthree items remain exactly the same when a species is a reactant and a product: formula, chargeand state. K+(aq) ion and I

    -(aq) ion meets this three point criteria. All of the other species in the

    full ionic equation had some type of change. Thus K+(aq) and I-(aq) are our spectator ions.

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    Eliminating K+(aq) and I-(aq) ions from the reactant and product sides of the balanced full ionic

    equation 16 gives us our balanced net ionic equation. Equation 17 is a balanced net ionicequation.

    OH-(aq) + H+

    (aq) H2O(l) Equation 17In checking to see if a net ionic equation is correct, the following items should be

    checked: 1. Is the equation balanced? 2. Are the states included? 3. Are the appropriatecharges present? 4. Were spectator ions eliminated?Summary:Balanced Molecular Equation:KOH (aq) + HI(aq) H2O(l) + KI(aq) Equation 15

    Balanced Full Ionic Equation:

    K+(aq) + OH-(aq) + H

    +(aq) + I

    -(aq) H2O(l) + K+(aq) + I-(aq) Equation 16

    Balanced Net Ionic Equation:OH-(aq) + H

    +(aq) H2O(l) Equation 17

    Example 3: Write a balanced net ionic equation for the reaction between an aqueous solution of potassium hydrogen carbonate KHCO3 and an aqueous solution of hydroiodic acid HI.Potassium hydrogen carbonate KHCO3 is also known as potassium bicarbonate.

    We first need to realize that the reactants are KHCO3 and HI. KHCO3 is a metalbicarbonate base that supplies a hydrogen carbonate or bicarbonate ion HCO3

    - that will reactwith H+ ion that is supplied by the acid HI. In these equations we must include the states witheach reactant and product. The states are indicated as subscripts after each reactant and product.The states for KHCO3 and HI are indicated as aqueous since we are dealing with aqueous

    solutions of both KHCO3 and HI. We will start out by writing the reactants part of themolecular equation as follows in partial molecular equation 18:

    KHCO3(aq) + HI(aq) Partial Equation 18

    Before we can complete the molecular equation, we need to realize what is taking placewith the reactants and then write the full ionic equation. When we write full ionic equations, thereactants and products have to be broken apart into their ions if they are water soluble ioniccompounds, water soluble strong acids or water soluble strong bases. KHCO3 is a bicarbonatebase that is soluble in water and is also an ionic compound. Water soluble bicarbonate bases thatare ionic compounds will ionize or break apart almost completely to form ions called anions andcations. The equation for the ionization or dissociation of KHCO3 to form the cation K

    + orpotassium ion and the anion HCO3

    - or bicarbonate ion is as follows in equation 19:

    KHCO3(aq) K+

    (aq) + HCO3-(aq) Equation 19

    HI is a strong acid that is soluble in water. Strong acids will ionize or break apart almostcompletely to form ions called anions and cations. The equation for the ionization ordissociation of HI to form the cation H+ or hydrogen ion and the anion I- or iodide ion is asfollows in equation 20:

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    HI (aq) H+

    (aq) + I-(aq) Equation 20

    Using the results of equations 19 and 20 we realize that the reactants KHCO3(aq) and HI(aq)are represented as follows in the partial full ionic equation in equation 21:

    K+(aq) + HCO3-(aq) + H

    +(aq) + I

    -(aq) Partial Equation 21

    Next we need to look at the interactions between K+ (aq) and HCO3-(aq) and H

    +(aq) and I

    -(aq).

    In reactions between strong acids and bicarbonate bases, it is useful to pair the anion from thebicarbonate base, HCO3

    -, with the cation from the acid, H+, to form the water soluble carbonicacid, H2CO3(aq), as one of the products. This is an acid base reaction that takes place between H

    +

    or hydrogen ion and the HCO3- or bicarbonate ion. The formation of carbonic acid or H2CO3 is

    given in equation 22.

    HCO3-(aq) + H

    +(aq) H2CO3(aq) Equation 22

    However carbonic acid, H2CO3, breaks apart to form the gas carbon dioxide, CO 2, and

    liquid water, H2O, as is illustrated in equation 23.

    H2CO3(aq) CO2(g) + H2O(l) Equation 23

    The K+ cation or potassium ion will also be paired with the I- anion or iodide ion to forman ionic compound or salt that is soluble in water. We will represent K+ cation and the I- anionas K+(aq) and I

    -(aq) on the products side of the full ionic equation. Using information from

    equations 21, 22 and 23, we can now write a full ionic equation as follows in equation 24.

    K+(aq) + HCO3-(aq) + H

    +(aq) + I

    -(aq) CO2(g) + H2O(l) + K

    +(aq) + I

    -(aq) Equation 24

    The next step involves completing the molecular equation and then balancing themolecular equation. In molecular equations, anion and cation components of ionic compoundshave to be combined together. The ionic compound in this example consists of K+(aq) cation andI-(aq) anion. In ionic compounds, the overall charge has to be neutral or zero. This means thatthere must be an equal number of positive charges as well as an equal number of negativecharges. Thus we need to figure out how many K+ cations and how many I- anions we need. Infiguring this out it is convenient to use the following general format:

    CationxAniony where x is the number of cations in the formula and y is the number ofanions in the formula. We figure out the value of x by letting x equal the absolute value of thecharge on the anion. In a similar manner, the value of y is equal to the charge of the cation.

    For this example we will now solve for the values of x and y in KxIy. Since I- has a

    charge of 1- we take the absolute value of 1- which is 1 and this is our value of x. Since x = 1this means there is one atom of K in the formula or our ionic compound. Likewise since K+ hasa charge of 1+, the value of y = 1. This means that we have one atom of iodine in our ioniccompound. The formula for KxIy is thus KI. The name of KI is potassium iodide.

    Since CO2 and H2O are our other products we can now write our molecular equation asfollows in equation 25.

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    KHCO3(aq) + HI(aq) CO2(g) + H2O(l) + KI(aq) Equation 25

    Now we need to balance our molecular equation. In balanced equations, there must beequal numbers of each atom on the reactants and products side of the equation. We balanceequations by the use of coefficients that are placed in front of formulas. We never change theformula of a reactant or product when we balance an equation. In our molecular equation wehave 1 K atom, 2 H atoms, 1 C atom, 3 O atoms, and 1 I atom on both sides of the equation. Wehave equal numbers of each type of atom on both sides of our molecular equation and thus wehave a balanced molecular equation. Equation 25 is a balanced molecular equation.

    KHCO3(aq) + HI(aq) CO2(g) + H2O(l) + KI(aq) Balanced Molecular Equation 25

    Now we can write a balanced full ionic equation from our balanced molecular equation25. Equation 26 is a balanced full ionic equation.

    K+(aq) + HCO3-(aq) + H

    +(aq) + I

    -(aq) CO2(g) + H2O(l) + K+(aq) + I-(aq) Equation 26

    We derive our balanced net ionic equation from our balanced full ionic equation byidentifying and eliminating spectator ions. Remember a species is a spectator if the followingthree items remain exactly the same when a species is a reactant and a product: formula, chargeand state. K+(aq) ion and I

    -(aq) ion meets this three point criteria. All of the other species in the

    full ionic equation had some type of change. Thus K+(aq) and I-(aq) are spectator ions. Eliminating

    K+(aq) and I-(aq) ions from the reactant and product sides of the balanced full ionic equation gives

    us balanced net ionic equation 27. Equation 27 is a balanced net ionic equation.

    HCO3-(aq) + H

    +(aq) CO2(g) + H2O(l) Equation 27

    In checking to see if a net ionic equation is correct, the following items should bechecked: 1. Is the equation balanced? 2. Are the states included? 3. Are the appropriatecharges present? 4. Were spectator ions eliminated?

    Summary:

    Balanced Molecular Equation:KHCO3(aq) + HI(aq) CO2(g) + H2O(l) + KI(aq) Equation 25

    Balanced Full Ionic Equation:K+(aq) + HCO3

    -(aq) + H

    +(aq) + I

    -(aq) CO2(g) + H2O(l) + K+(aq) + I-(aq) Equation 26

    Balanced Net Ionic Equation:HCO3

    -(aq) + H

    +(aq) CO2(g) + H2O(l) Equation 27

    Example 4: Write a balanced net ionic equation for the reaction between an aqueous solution ofpotassium carbonate K2CO3 and an aqueous solution of hydroiodic acid HI.

    We first need to realize that the reactants are K2CO3 and HI. K2CO3 is a metal carbonatebase that supplies a carbonate ion CO3

    2- that will react with H+ ion that is supplied by the acidHI. In these equations we must include the states with each reactant and product. The states areindicated as subscripts after each reactant and product. The states for K2CO3 and HI areindicated as aqueous since we are dealing with aqueous solutions of both K2CO3 and HI. We

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    will start out by writing the reactants part of the molecular equation as follows in the partialmolecular equation 28:

    K2CO3(aq) + HI(aq) Partial Equation 28

    Before we can complete the molecular equation, we need to realize what is taking placewith the reactants and then write the full ionic equation. When we write full ionic equations, thereactants and products have to be broken apart into their ions if they are water soluble ioniccompounds, water soluble strong acids or water soluble strong bases. K2CO3 is a carbonate basethat is soluble in water and is also an ionic compound. Water soluble carbonate bases that areionic compounds will ionize or break apart almost completely to form ions called anions andcations. The equation for the ionization or dissociation of K2CO3 to form the cation K

    + orpotassium ion and the anion CO3

    2- or carbonate ion is as follows in equation 29:

    K2CO3(aq) 2K+

    (aq) + CO32-

    (aq) Equation 29

    HI is a strong acid that is soluble in water. Strong acids will ionize or break apart almost

    completely to form ions called anions and cations. The equation for the ionization ordissociation of HI to form the cation H+ or hydrogen ion and the anion I- or iodide ion is asfollows in equation 30:HI(aq) H

    +(aq) + I

    -(aq) Equation 30

    Using the results of equations 29 and 30 we realize that the reactants K2CO3(aq) and HI(aq)are represented as follows in the partial full ionic equation 31:

    2K+(aq) + CO32-

    (aq) + H+

    (aq) + I-(aq) Partial Equation 31

    Next we need to look at the interactions between K+(aq) and CO32-

    (aq) and H+

    (aq) and I-(aq). In

    reactions between strong acids and carbonate bases, it is useful to pair the anion from thecarbonate base, CO3

    2-, with the cation from the acid, H+, to initially form the water solublebicarbonate ion or HCO3

    -. The initially formed bicarbonate ion, HCO3-, will then react with a

    second H+ ion to form carbonic acid, H2CO3(aq), as one of the products. This is an acid basereaction that takes place between two H+ ions and the CO3

    2- or carbonate ion. The two stepformation of carbonic acid or H2CO3 from CO3

    2- is given in equations 32 and 33.

    CO32-

    (aq) + H+

    (aq) HCO3-(aq) Equation 32

    HCO3-(aq) + H

    +(aq) H2CO3(aq) Equation 33

    However carbonic acid, H2CO3, breaks apart to form the gas carbon dioxide, CO 2, andliquid water, H2O, as is illustrated in equation 34.

    H2CO3(aq) CO2(g) + H2O(l) Equation 34

    The K+ cation or potassium ion will also be paired with the I- anion or iodide ion to forman ionic compound or salt that is soluble in water. We will represent K+ cation and the I- anionas K+(aq) and I

    -(aq) on the products side of the full ionic equation. Using the results of equations

    31, 32, 33 and 34, we can now write an unbalanced full ionic equation as follows in equation 35:

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    2K+(aq) + CO32-

    (aq) + H+

    (aq) + I-(aq) CO2(g) + H2O(l) + 2K

    +(aq) + I

    -(aq) Equation 35

    The next step involves completing the molecular equation and then balancing the

    molecular equation. In molecular equations, anion and cation components of ionic compoundshave to be combined together. The ionic compound in this example consists of K+(aq) cation andI-(aq) anion. In ionic compounds, the overall charge has to be neutral or zero. This means thatthere must be an equal number of positive charges as well as an equal number of negativecharges. Thus we need to figure out how many K+ cations and how many I- anions we need. Infiguring this out it is convenient to use the following general format:

    CationxAniony where x is the number of cations in the formula and y is the number ofanions in the formula. We figure out the value of x by letting x equal the absolute value of thecharge on the anion. In a similar manner, the value of y is equal to the charge of the cation. Forthis example we will now solve for the values of x and y in KxIy. Since I

    - has a charge of 1- wetake the absolute value of 1- which is 1 and this is our value of x. Since x = 1 this means there isone atom of K in the formula or our ionic compound. Likewise since K+ has a charge of 1+, thevalue of y = 1. This means that we have one atom of iodine in our ionic compound. The

    formula for KxIy is thus KI. The name of KI is potassium iodide. Since CO2 and H2O are ourother products we can now write our unbalanced molecular equation as follows in equation 36:

    K2CO3(aq) + HI(aq) CO2(g) + H2O(l) + KI(aq) Equation 36

    Now we need to balance our molecular equation. In balanced equations, there must beequal numbers of each atom on the reactants and products side of the equation. We balanceequations by the use of coefficients that are placed in front of formulas. We never change theformula of a reactant or product when we balance an equation. In our unbalanced molecularequation we have 2 K atoms, 1 C atom, 3 O atoms, 1 H atom, and 1 I atom on the reactants sideof the equation. We have 1 K atom, 1 C atom, 3 O atoms, 2 H atoms, and 1 I atom on the

    products side of the equation. We can balance out the number of K atoms by placing acoefficient of 2 in from of KI on the products side. This gives us the following unbalancedmolecular equation in equation 37:

    K2CO3(aq) + HI(aq) CO2(g) + H2O(l) + 2KI(aq) Equation 37

    In this unbalanced molecular equation we now have 2 K atoms, 1 C atom, 3 O atoms, 1H atom, and 1 I atom on the reactants side of the equation and there are now 2 K atoms, 1 Catom, 3 O atoms, 2 H atoms, and 2 I atoms on the products side of the equation. Next we canbalance out the number of H atoms and I atoms by placing a coefficient of 2 in from of HI onthe reactants side. We now have 2 K atoms, 1 C atom, 3 O atoms, 2 H atoms, and 2 I atoms onboth sides of the equation. This will give us the following balanced molecular equation 38:

    K2CO3(aq) + 2HI(aq) CO2(g) + H2O(l) + 2KI(aq) Balanced Molecular Equation 38

    Now we can write a balanced full ionic equation from our balanced molecular equation38. Equation 39 is a balanced full ionic equation

    2K+(aq) + CO32-

    (aq) + 2H+

    (aq) + 2I-(aq) CO2(g) + H2O(l) + 2K+(aq) + 2I-(aq) Equation 39

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    We derive our balanced net ionic equation from our balanced full ionic equation byidentifying and eliminating spectator ions. Remember a species is a spectator if the followingthree items remain exactly the same when a species is a reactant and a product: formula, chargeand state. K+(aq) ion and I

    -(aq) ion meets this three point criteria. All of the other species in the

    full ionic equation had some type of change. Thus K+(aq) and I-(aq) are spectator ions. Eliminating

    K+(aq) and I-(aq) ions from the reactant and product sides of the balanced full ionic equation 39

    gives us balanced net ionic equation 40. Equation 40 is a balanced net ionic equation.

    CO32-

    (aq) + 2H+

    (aq) CO2(g) + H2O(l) Equation 40In checking to see if a net ionic equation is correct, the following items should be

    checked: 1. Is the equation balanced? 2. Are the states included? 3. Are the appropriatecharges present? 4. Were spectator ions eliminated?

    Summary:

    Balanced Molecular Equation:K2CO3(aq) + 2HI(aq) CO2(g) + H2O(l) + 2KI(aq) Equation 38

    Balanced Full Ionic Equation:2K+(aq) + CO3

    2-(aq) + 2H

    +(aq) + 2I

    -(aq) CO2(g) + H2O(l) + 2K

    +(aq) + 2I

    -(aq) Equation 39

    Balanced Net Ionic Equation:CO3

    2-(aq) + 2H

    +(aq) CO2(g) + H2O(l) Equation 40

    PRELABRATORY QUESTIONS:

    1. What is the symbol for sodium?

    2. What is the symbol for the element oxygen as it is used in a chemical reaction?

    3. What is the symbol for copper(II) ion?

    4. What is the name of HCO3-?

    5. What is the name of HCl(aq)?

    6. What is the name of Na3PO4?

    7. What is the formula of calcium hydrogen sulfate?

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    8. Give an example of a metal hydroxide base.

    9. Write a balanced molecular equation for the reaction between an aqueous solution of thestrong base Ba(OH)2(aq) and an aqueous solution of the strong acid HBr(aq).

    10. Write a balanced full equation for the reaction between an aqueous solution of the strongbase Ba(OH)2(aq) and an aqueous solution of the strong acid HBr(aq).

    11. Write a balanced net equation for the reaction between an aqueous solution of the strongbase Ba(OH)2(aq) and an aqueous solution of the strong acid HBr(aq).

    12. What are the formulas and names of the spectator ions in question 10?

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    LABORATORY REPORT SHEET PAGE 1. IN ADDITION TO THIS SHEET PAGES149 AND 150 FROM EXPERIMENT 26B IN HOLUM ARE ALSO TO BE TURNED INFOR THE LAB REPORT.

    PRELABORATORY QUESTIONS:

    1. What is the symbol for sodium?

    2. What is the symbol for the element oxygen as it is used in a chemical reaction?

    3. What is the symbol for copper(II) ion?

    4. What is the name of HCO3-?

    5. What is the name of HCl(aq)?

    6. What is the name of Na3PO4?

    7. What is the formula of calcium hydrogen sulfate?

    8. Give an example of a metal hydroxide base.

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    9. Write a balanced molecular equation for the reaction between an aqueous solution of thestrong base Ba(OH)2(aq) and an aqueous solution of the strong acid HBr(aq).

    10. Write a balanced full equation for the reaction between an aqueous solution of the strongbase Ba(OH)2(aq) and an aqueous solution of the strong acid HBr(aq).

    11. Write a balanced net equation for the reaction between an aqueous solution of the strongbase Ba(OH)2(aq) and an aqueous solution of the strong acid HBr(aq).

    12. What are the formulas and names of the spectator ions in question 10?