Uploaded by Sukhamrit Singh Babra

Ionic and Net Ionic Equations

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We are learning to write balanced net ionic
equations to represent precipitation and
neutralization reactions.
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PRECIPITATION REACTIONS
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In chemistry, you will likely see tons of
reactions that occur in water
When ions are involved in a reaction, the
chemical equation can be written with various
levels of detail.
Depending on which part of the reaction you
are interested in, you might write a chemical,
ionic, or net ionic equation.
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There are three basic types of chemical equations:
Chemical, Ionic, & Net ionic.
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Chemical EQUATIONS are written as if all
substances were molecular, even though some
substances may exist as ions.
HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)
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IONIC EQUATIONS have the substances which
exist as ions (i.e. dissociate in water) written in
Note: H O does not
ionic form.
break into ions
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H+ (aq) + Cl- (aq) + Na+ (aq) + OH-(aq)  Na+ (aq) + Cl- (aq) + H2O (l)
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Precipitation, Acid/base, and Redox reactions can all be
written depicting the appropriate substances as ions.
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NET IONIC EQUATIONS are ionic equations with
the Spectator ions removed.
H+ (aq) + OH-(aq)  H2O
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(l)
SPECTATOR IONS do not participate in a reaction (that is
they do not react to form a new substance). Common
Spectator ions are Group I, many Group II, and NO3(nitrate) and C2H3O2- (acetate) ions.
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Example #1:
No precipitate
NaCl in water
NaCl is soluble in
water. Solid NaCl
dissociates into
Na+ and Cl- ions in
aqueous solution:
NaCl(s)  Na+(aq) + Cl-(aq)
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Example #2:
Precipitate forms Pb(NO3)2 (aq) + 2NaI (aq)
Pb(NO3)2 + 2NaI  PbI2 + 2NaNO3
(aq)
(aq)
(s)
(aq)
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Solubility Rules
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Balanced Chemical Equation:
Pb(NO3)2(aq) + 2NaI(aq)  PbI2(s) + 2NaNO3(aq)
“Complete Ionic” Equation:
Pb2+(aq) + 2NO3-(aq) + 2Na+(aq)+ 2I-(aq)  PbI2(s) + 2Na+(aq) +
2NO3- (aq)
Cancel the “spectator ions” that appear on both sides of
the arrow
Pb2+(aq) + 2NO3-(aq) + 2Na+(aq)+ 2I-(aq)  PbI2(s) + 2Na+(aq) +
2NO3- (aq)
“Net Ionic” Equation:
Pb2+(aq) + 2I-(aq)  PbI2(s)
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Net Ionic Equations Revisited:
1) Write the (balanced!) molecular equation first
- Reaction products: swap cations and anions
- Predict solubility (using Solubility rules)
2) Write the complete ionic equation next
- (s) compounds don’t ionize
- (aq) compounds do ionize
ion subscripts in the molecular equation become
coefficients in the complete ionic equation!
3) Write the net ionic equation next
- cancel spectator ions
The net ionic equation is a “simplified” form
of the complete ionic equation
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Sample Problem
Balanced Molecular Equation:
Ba(NO3)2(aq)+ NiSO4(aq)
Ni(NO3)2 (aq)+ BaSO4(s)
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Sample Problem
Balanced Molecular Equation:
Ba(NO3)2(aq)+ NiSO4(aq)
Ni(NO3)2 (aq)+ BaSO4(s)
Complete Ionic Equation:
Ba2+(aq) + 2NO3-(aq) + Ni2+(aq) + SO42-(aq)
Ni2+(aq) + 2NO3-(aq) + BaSO4 (s)
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Sample Problem
Balanced Molecular Equation:
Ba(NO3)2(aq)+ NiSO4(aq)
Ni(NO3)2 (aq)+ BaSO4(s)
Complete Ionic Equation:
Ba2+(aq) + 2NO3-(aq) + Ni2+(aq) + SO42-(aq)
Ni2+(aq) + 2NO3-(aq) + BaSO4 (s)
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Sample Problem
Balanced Molecular Equation:
Ba(NO3)2(aq)+ NiSO4(aq)
Ni(NO3)2 (aq)+ BaSO4(s)
Complete Ionic Equation:
Ba2+(aq) + 2NO3-(aq) + Ni2+(aq) + SO42-(aq)
Ni2+(aq) + 2NO3-(aq) + BaSO4 (s)
Net Ionic Equation:
Ba2+(aq) + SO42-(aq)
BaSO4 (s)
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Balance the following molecular equations then write
both the ionic & net ionic equations:
1. __Na2CO3(aq) + __HNO3(aq)  __H2CO3(aq) + __NaNO3 (aq)
Note: carbonic acid decomposes into carbon dioxide and water.
2. __KBr(aq) + __I2(g)  __ KI(aq) + __Br2(l)
3. __AlCl3(aq) + __K3PO4(aq)  __AlPO4(s) + __KCl(aq)
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1. Na2CO3(aq) + 2 HNO3(aq)  H2CO3(aq) + 2 NaNO3(aq)
Note: carbonic acid decomposes into carbon dioxide and water.
Ionic: Na+(aq) + CO3 2-(aq) + 2H+(aq) + 2NO3-(aq) 
H2O(l) + CO2(g) + 2Na+(aq) + 2NO3- (aq)
Net ionic: CO32-(aq) + 2H+(aq)  H2O(l) + CO2(g)
2. 2 KBr(aq) + I2(g)  2 KI(aq) + Br2(l)
Ionic: 2K+(aq) + 2Br-(aq) + I2(g)  2K+(aq) + 2I-(aq) + Br2(l)
Net ionic: 2Br-(aq) + I2(g)  2I-(aq) + Br2(l)
3. AlCl3(aq) + __K3PO4(aq)  __AlPO4(s) + 3 KCl(aq)
Ionic: Al3+(aq) + Cl-(aq) + K+(aq) + PO43(aq) -  AlPO4(s) + K+(aq) + Cl-(aq)
Net ionic: Al3+(aq) + PO43(aq) -  AlPO4(s)
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By the end of this lesson..
I can write balanced net ionic equations to
represent precipitation and neutralization
reactions.
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