Na+

Al3+

S2–

2Ca2+ PO43–

3Cl–

Net ionic equations and solubility rules

Review: forming ions• Ionic (i.e. salt) refers to +ve ion plus -ve ion• Usually this is a metal + non-metal or metal +

polyatomic ion (e.g. NaCl, NaClO3, Li2CO3)• Polyatomic ions are listed on page 71• (aq) means aqueous (dissolved in water)• For salts (aq) means the salt exists as ions• NaCl(aq) is the same as: Na+(aq) + Cl–(aq)• Acids form ions: HCl(aq) is H+(aq) + Cl–(aq),

Bases form ions: NaOH(aq) is Na+ + OH–

Q - how is charge determined (+1, -1, +2, etc.)?A - via valences (periodic table or see pg. 71)• F, Cl gain one electron, thus forming F–, Cl–

• Ca loses two electrons, thus forming Ca2+

• Charge can also be found via the compound• E.g. in NaNO3(aq) if you know Na forms Na+, then

NO3 must be NO3– (NaNO3 is neutral)

• By knowing the valence of one element you can often determine the other valences

Q - Write the ions that form from Al2(SO4)3(aq)?Step 1 - look at the formula: Al2(SO4)3(aq)Step 2 - determine valences: Al3 (SO4)2

Background: valences and formulas

(Al is 3+ according to the periodic table) Step 3 - write ions: 2Al3+(aq) + 3SO4

2–(aq)

• Note that there are 2 aluminums because Al has a subscript of 2 in the original formula

Practice with writing ionsQ - Write ions for Na2CO3(aq)

A - 2Na+(aq) + CO32–(aq) (from the PT Na is 1+.

There are 2, thus we have 2Na+. There is only one CO3. It must have a 2- charge)

• Notice that when ions form from molecules, charge can be separated, but the total charge (and number of each atom) stays constant.

Q - Write ions for Ca3(PO4)2(aq) & Cd(NO3)2(aq)

A - 3Ca2+(aq) + 2PO43–(aq)

A - Cd2+(aq) + 2NO3–(aq)

Q - Write ions for Na2S(aq) and Mg3(BO3)2(aq)

A - 2Na+(aq) + S2–(aq), 3Mg2+(aq)+ 2BO33–(aq)

Types of chemical equationsEquations can be divided into 3 types (pg. 396)

1) Molecular, 2) Ionic, 3) Net ionic• Here is a typical molecular equation:Cd(NO3)2(aq) + Na2S(aq) CdS(s) + 2NaNO3(aq)

• We can write this as an ionic equation (all compounds that are (aq) are written as ions):

Cd2+(aq) + 2NO3–(aq) + 2Na+(aq) + S2–(aq)

CdS(s) + 2Na+(aq) + 2NO3–

(aq)• To get the NET ionic equation we cancel out all terms

that appear on both sides:

Net: Cd2+(aq) + S2–(aq) CdS(s)

Equations must be balanced• There are two conditions for molecular, ionic,

and net ionic equationsMaterials balance

Both sides of an equation should have the same number of each type of atom

Electrical balanceBoth sides of a reaction should have the same net charge

Q- When NaOH(aq) and MgCl2(aq) are mixed, _______(s) and NaCl(aq) are produced. Write balanced molecular, ionic & net ionic equations

Mg(OH)2

NaOH(aq) + MgCl2(aq) Mg(OH)2(s) + NaCl(aq)

Next, balance the equation

First write the skeleton equation2

2

Ionic equation:2Na+(aq) + 2OH-(aq) + Mg2+(aq) + 2Cl-(aq)

Mg(OH)2(s) + 2Na+(aq) + 2Cl-(aq)

Net ionic equation:2OH-(aq) + Mg2+(aq) Mg(OH)2(s)

Write balanced ionic and net ionic equations:CuSO4(aq) + BaCl2(aq) CuCl2(aq) + BaSO4(s)

Fe(NO3)3(aq) + LiOH(aq) ______(aq) + Fe(OH)3(s)

Na3PO4(aq) + CaCl2(aq) _________(s) + NaCl(aq)

Na2S(aq) + AgC2H3O2(aq) ________(aq) + Ag2S(s)

LiNO3

Ca3(PO4)2

NaC2H3O2

Cu2+(aq) + SO42–(aq) + Ba2+(aq) + 2Cl–(aq) Cu2+(aq) +

2Cl–(aq) + BaSO4(s)

Net: SO42–(aq) + Ba2+(aq) BaSO4(s)

Fe3+(aq) + 3NO3–(aq) + 3Li+(aq) + 3OH–(aq) 3Li+(aq) +

3NO3–(aq) + Fe(OH)3(s)

Net: Fe3+(aq) + 3OH–(aq) Fe(OH)3(s)

2Na3PO4(aq) + 3CaCl2(aq) Ca3(PO4)2(s)+ 6NaCl(aq)

6Na+(aq) + 2PO43–(aq) + 3Ca2+(aq) + 6Cl–(aq)

Ca3(PO4)2(s)+ 6Na+(aq) + 6Cl–(aq)

Net: 2PO43–(aq) + 3Ca2+(aq) Ca3(PO4)2(s)

2Na+(aq) + S2–(aq) + 2Ag+(aq) + 2C2H3O2–(aq) 2Na+(aq)

+ 2C2H3O2–(aq) + Ag2S(s)

Net: S2–(aq) + 2Ag+(aq) Ag2S(s)

Solubility• Precipitation refers to the formation of a solid

from ions. A precipitate is “insoluble”• Soluble and insoluble are general terms to

describe how much of a solid dissolves.• Solubility can be predicted from rules (pg.399)• These are general rules, based on observation• To determine solubility, follow the rules in order• Note: in rule 4 that sulfate = SO4

2-

• You will not have to memorize these rules, you will have to use the rules to predict solubility

• Read over example 11.2 (pg. 400)• Do 11.26 (435) (list the relevant rule for each)• Do PE 5 (pg. 400) and 11.28 (pg. 435)

Precipitation• Precipitation refers to the formation of a

solid from ions

• Metathesis refers to double displacement:

AB + CD AD + CB

• A metathesis involving ions going to one or more solids, is called precipitation:

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

• Recall: aq indicates the compound is aqueous (as ions)

Solubility - 11.26a) Ca(NO3)2 - Soluble

rule 2 (salts containing NO3- are soluble)

b) FeCl2 - Soluble

rule 3 (all chlorides are soluble)c) Ni(OH)2 - Insoluble

rule 5 (all hydroxides are insoluble)d) AgNO3 - Soluble

rule 2 (salts containing NO3- are soluble)

e) BaSO4 - Insoluble

rule 4 (Sulfates are soluble, except … Ba2+)f) CuCO3 - Insoluble

rule 6 (containing CO32- are insoluble)

Solubility PE 5 (a) - pg. 400Ionic:

Ag+(aq) + NO3-(aq) + NH4

+(aq) + Cl-(aq)

Note: combine, in your head, the positive and negative ions. If together a pair is insoluble, they will form a precipitate (s).

In this case AgCl is insoluble (rule 3)

Ag+(aq) + NO3-(aq) + NH4

+(aq) + Cl-(aq)

AgCl(s) + NO3-(aq) + NH4

+(aq)

Net ionic: Ag+(aq) + Cl-(aq) AgCl(s)If no solid is formed then write N.R.

Solubility PE 5 (b), (c) - pg. 400Ionic:2Na+(aq) + S2– (aq) + Pb2+(aq) + 2C2H3O2

–(aq)In this case PbS is insoluble (rule 6)2Na+(aq) + S2–(aq) + Pb2+(aq) + 2C2H3O2

–(aq) PbS(s) + 2Na+(aq) + 2C2H3O2

–(aq)Net ionic: Pb2+(aq) + S2–(aq) PbS(s)

Ionic:Ba2+(aq) + 2Cl–(aq) + NH4

+(aq) + NO3–(aq)

In this case all combinations are solubleBa2+(aq) + 2Cl–(aq) + NH4

+(aq) + NO3–(aq)

Ba2+(aq) + 2Cl–(aq) + NH4+(aq) + NO3

–(aq)Net ionic: N.R. (all ions cancel)

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