Transcript PowerPoint - Net Ionic Equations, Ions, and Solubility Rules

PO43–
S2–
3Cl–
2Ca2+
Na+
Al3+
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+
Background: valences and formulas
• 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
(Al is 3+ according to the periodic table)
Step 3 - write ions:
2Al3+(aq) + 3SO42–(aq)
• Note that there are 2 aluminums because Al
has a subscript of 2 in the original formula
Practice with writing ions
Q - 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 equations
Equations 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 equations
Materials balance
Both sides of an equation should have the
same number of each type of atom
Electrical balance
Both sides of a reaction should have the same
net charge
Q- When NaOH(aq) and MgCl2(aq) are mixed,
Mg(OH)2 and NaCl(aq) are produced. Write
_______(s)
balanced molecular, ionic & net ionic equations
First write the skeleton equation
2 NaOH(aq) + MgCl2(aq)
 Mg(OH)2(s) + 2 NaCl(aq)
Next, balance the equation
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)
LiNO3
Fe(NO3)3(aq) + LiOH(aq)  ______(aq)
+ Fe(OH)3(s)
Ca3(PO4)2
Na3PO4(aq) + CaCl2(aq)  _________(s)
+ NaCl(aq)
Na2S(aq) + AgC2H3O2(aq)  NaC
________(aq)
+ Ag2S(s)
2H3O2
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 = SO42• 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 - 11.26
a) 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. 400
Ionic:
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. 400
Ionic:
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 soluble
Ba2+(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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