Transcript 3.4 Writing Formula, Complete and Net Ionic Equations

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Objective:
write a formula equation, complete ionic equation,
and net ionic equation that represent a reaction
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We must be able to represent a reaction in three
different ways:
Formula Equation
 Total or Complete Ionic Equation
 Net Ionic Equation
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A balanced chemical equation in which all the reactants
and products are given by their chemical formula.
For Example:
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
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Soluble species are indicated by placing “(aq)” after the
formula.
Precipitates are indicated by placing “(s)” after the formula.
This way of expressing the reaction is used whenever we
want to indicate the chemical formula of the reactants and
products.
Shows all the soluble ionic species broken up into their respective ions.
For Example:
Ag+(aq)+ NO3-(aq)+ Na+(aq)+ Cl-(aq) → AgCl(s)+ Na+(aq)+ NO3-(aq)
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This method is not frequently used because it is cumbersome to write out.
It is used to emphasize the situation which exists before and after the
reaction.
Spectator ions:
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Ions that remain unchanged during a reaction and do not participate in the
reaction.
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The spectator ions in the above equation are NO3-(aq) and Na+(aq)
NEVER write a total ionic equation until you have first written a balanced
formula equation.
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If you proceed directly to a total ionic equation the result may be balanced
but still may not represent the actual situation.
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Shows only the reacting species (species which are actively involved
in the reaction…the one making the PPT) in the equation.
The net ionic equation is formed by omitting the spectator
ions from the total ionic equation.
For Example:
We had:
then:
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
Ag+(aq)+ NO3-(aq)+ Na+(aq)+ Cl-(aq) → AgCl(s)+ Na+(aq)+ NO3-(aq)
Canceling out spectator ions, we get:
Ag+(aq)+ Cl-(aq) → AgCl(s)
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Example:
Write a formula equation, a total ionic
equation, and a net ionic equation for the
reaction which occurs when 0.2 M solutions of
Al(NO3)3(aq) and MgS(aq) are mixed.
1.
Write the complete formula equation. Determine if there is a
precipitate using“(s)” to identify the precipitate. The balanced
formula equation is:
2 Al(NO3)3(aq) + 3 MgS(aq) → Al2S3(s) + 3 Mg(NO3)2(aq)
2. Write the complete ionic equation by breaking up the
balanced formula equation into ions.
Remember:
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The precipitate remains in molecular form.
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The number of ions is found by multiplying the subscript
after the ion, by the coefficient in front of the molecule.
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The final equation must still be balanced.
We get:
2Al3+ (aq)+6NO3-(aq)+3Mg2+ (aq)+3S2-(aq) 
Al2S3(s)+3Mg2+ (aq)+6NO3-(aq)
3. Write the net ionic equation by deleting the spectator ions
and simplifying the coefficients if necessary.
2 Al3+ (aq) + 3 S2-(aq) → Al2S3(s)
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1)
2)
3)
Write the formula, complete ionic, and net
ionic equation for each of the following
reactions:
Aqueous nickel (II) chloride reacts with
aqueous sodium hydroxide to give nickel (II)
hydroxide and sodium chloride.
Solid potassium metal reacts with water to
give potassium hydroxide and hydrogen gas.
Aqueous sodium hydroxide reacts with
phosphoric acid to give water and sodium
phosphate.