writing chemical equations sound - black hills state university

Writing Chemical Equations

Chemical equations for solution reactions can be written in th diff t f l l ti l t i ithree different forms; molecular equations, complete ionic equations, and net ionic equations. In class, so far, we have been writing molecular equations.

This tutorial will show you the differences between these types of equations, and give you some practice in writing equations in the different forms.

The three types of equations are:yp q

I. The molecular equation that gives the overall reaction stoichiometry but not necessarily the actual forms of thestoichiometry, but not necessarily the actual forms of the reactants and products in solution.

II. The complete ionic equation that represents all strong electrolyte products and reactants in their ionic form.

III. The net ionic equation that gives only those reactant and product components that undergo change in the p p g greaction, and does not mention other components, called spectator ions, that do not undergo change.

The Molecular EquationLet’s look at the reaction where aqueous silver nitrate and aqueous sodium chloride react together to form aaqueous sodium chloride react together to form a silver chloride precipitate and sodium nitrate that remains in solution.

First let’s connect names with molecular formulas:aqueous silver nitrate = AgNO3(aq)aqueous silver nitrate AgNO3(aq)aqueous sodium chloride = NaCl(aq)silver chloride precipitate = solid silver chloride = AgCl(s)sodium nitrate in solution = aqueous sodium nitratesodium nitrate in solution = aqueous sodium nitrate

= NaNO3(aq)

The Molecular Equation

If we combine these molecular formulas into a reaction we get the molecular equation:

AgNO3(aq) + NaCl(aq)=AgCl(s) + NaNO3(aq)AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)

Fortunately for you this simple example is already balanced y y p p yas written, so you don’t have to put any extra effort intobalancing the equation.

The Complete Ionic EquationThe molecular equation contained three different ionic compounds in aqueous solution. All three of these ionic p qcompounds are strong electrolytes and would ionize,so in reality the molecules AgNO3(aq), NaCl(aq), and NaNO (aq) don’t exist!and NaNO3(aq) don t exist!

The better way to represent the true chemistry in this ti t b k ll t l t l t i t th i treaction to break all strong electrolytes into their component

ions. Thus, AgNO3(aq) = Ag+ (aq) + NO3

- (aq) NaCl(aq)= Na+ (aq) + Cl-(aq)NaNO3(aq) = Na+ (aq) + NO3

- (aq)

The Complete Ionic Equation

When we break all strong electrolyes into their componentg y pions we get the complete ionic equation. For this reaction the compete ionic equation is:

Ag+ (aq) + NO3- (aq) + Na+ (aq) + Cl-(aq)=

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

The Complete Ionic EquationNotice the only thing you can separate into ions are aqueousionic compounds, emphasis on aqueous, emphasis on ionic.

Never try to separate covalent compounds into ions.Never try to separate a solid(s) , liquid (l) or a gas(g)

into ions.

In this equation AgCl(s) is ionic, but it doesn’t get changed into ions because it is a solid

Don’t forget this. It will trip you up if you don’t remember,d h ill i ff

into ions because it is a solid.

and that will mean points off.

The Net Ionic EquationThe complete ionic equation,

Ag+ (aq) + NO3- (aq) + Na+ (aq) + Cl-(aq)=

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

contains two ions that appear on both sides of the chemical equation, Na+(aq) and NO3

-(aq). Things that appear unchanged on both sides of a chemical reaction are called spectators. Essentially the showed up and watched the reaction, but they didn’t actually take part in it. , y y p

The Net Ionic Equation

In the Net Ionic reaction we remove the spectator ions. Thi h l h i h i l l d iThis helps us to emphasize the important molecules and ions that are undergoing the reaction, and generally makes the equation shorter and easier to write down.

The net ionic equation for this reaction is:

Ag+ (aq) + Cl-(aq)=AgCl(s)

Example problem

Write the balanced molecular, complete ionic and net ionic pequations for the reaction of barium nitrate and potassium chromate in solution, where they react to form a barium chromate precipitate and aqueous potassium nitratechromate precipitate and aqueous potassium nitrate.

W it th ti f th ti f b i it t dWrite the equations for the reaction of barium nitrate and potassium chromate to form solid barium chromate and aqueous potassium nitrate.

First, what are the molecules?

Barium nitrate = Ba(NO3)2Barium nitrate Ba(NO3)2Potassium chromate = K2CrO4Barium chromate = BaCrO4Potassium nitrate KNOPotassium nitrate = KNO3


Combining the molecules and putting in the physical forms to get an unbalanced molecular equation:to get an unbalanced molecular equation:

Ba(NO3)2 (aq) + K2CrO4 (aq)=BaCrO4(s) +KNO3 (aq)

Ba and CrO4are balanced, K and NO3 will balance if we make it 2 KNOmake it 2 KNO3

Balanced molecular equation:

Ba(NO3)2 (aq) + K2CrO4 (aq) = BaCrO4(s) +2 KNO3 (aq)


Breaking all aqueous ionic compounds into their component ions to get the complete ionic equation we have:

Ba(NO3)2 (aq) = Ba2+(aq)+ 2 NO3-(aq)

K2CrO4 (aq) = 2 K+(aq) + CrO42- (aq)

2 KNO3(aq) = 2 K+(aq) + 2NO3- (aq)2 KNO3(aq) 2 K (aq) 2NO3 (aq)

and our complete ionic equation becomes:

Ba2+ (aq) + 2 NO3- (aq) + 2 K+(aq) + CrO4

2- (aq) =BaCrO4(s) +2 K+(aq) + 2NO3

- (aq)


In making the complete ionic equation:

B 2+ ( ) 2 NO ( ) 2 K+( ) C O 2 ( )Ba2+ (aq) + 2 NO3- (aq) + 2 K+(aq) + CrO4

2- (aq) = BaCrO4(s) +2 K+(aq) + 2NO3

- (aq)

Remember that you can’t break the solid into its component ions.

In addition, notice that this equation contained the covalent ions NO3

- and CrO4- . Covalent ions cannot be broken into

ll i i i h t b h tsmaller ionic pieces, so you have to remember what your covalent ions are, and never break them down into smaller pieces.


Finally we remove the spectator ions that occur on both sides y pof the complete ionic equation to get the net ionic equation:

Ba2+ (aq) + CrO42- (aq) = BaCrO4(s)Ba (aq) + CrO4 (aq) BaCrO4(s)

One last problem….Write the molecular, complete ionic and net ionic equations for the reaction of barium hydroxide and nitric acid.

I stepped it up a notch here. I told you the reactants, but I didn’t give you’re the products, that is up to you.

Actually this is not as bad as it sounds. The ‘hydroxide’ part of the name ‘barium hydroxide’ tells you that this compound

g y p , p y

of the name barium hydroxide tells you that this compound is a base. The name ‘nitric acid’ tells you the other reactant is an acid. So you have an acid base reaction, that means you should be looking for H+ and OH- to be reacting to make H2O somewhere in this reaction.

Write the molecular, complete ionic and net ionic equations for the reaction of barium hydroxide and nitric acid.

Associating molecular formula’s with namesR t tReactants:

Barium hydroxide = Ba(OH)2Nitric Acid = HNO33

Products:??H2O (product of an acid-base reaction)


Let’s write the reactants and the only known product to see if we can figure out the missing productg g p

Ba(OH)2 + HNO3 = ? + H2O

In acid base reactions like this, the OH of one molecule reacts with the H of the other molecule to make water. Let’s combine the other parts of each reactant (Ba and NO3) to make the other product.

BaNO3


But BaNO3 doesn’t look right.

Ba should be a +2 ion, and NO3 is a -1 ion, so the charges don’t balance. The proper product should be:

Ba(NO3)2

The unbalanced molecular equation then is :

B (OH) HNO B (NO ) H OBa(OH)2 + HNO3 = Ba(NO3)2 + H2O


Balancing we get:

Ba(OH)2 + 2 HNO3 = Ba(NO3)2 + 2 H2O

And adding physical forms to get our molecular equationhwe have:

Ba(OH)2 (aq) + 2 HNO3 (aq) = Ba(NO3)2 (aq) + 2 H2O(l)( )2 ( q) 3 ( q) ( 3)2 ( q) 2 ( )


Breaking the aqueous ionic compounds into their ions we get:Ba(OH)2 (aq) = Ba2+(aq) + 2OH-(aq)2 HNO ( ) 2H+( ) 2 NO ( )2 HNO3 (aq) = 2H+(aq) + 2 NO3

-(aq)Ba(NO3)2 (aq) = Ba2+(aq) + 2 NO3

-(aq)

(And I hope you didn’t try to break the H2O down. It is NOT ionic and it is NOT aqueous)

The complete ionic equation then is:

Ba2+(aq) + 2OH-(aq) + 2H+(aq) + 2 NO3-(aq) =

Ba2+(aq) + 2 NO3-(aq) + 2H2O(l)


Removing the spectators we have the net ionic equation:

2OH ( ) 2H+( ) 2H O(l)2OH-(aq) + 2H+(aq) = 2H2O(l)

Taking all terms to the lowest common denominator we have the final net ionic equation:

OH-(aq) + H+(aq) = H2O(l)

(Which is a true acid base reaction)(Which is a true acid-base reaction)

Practice Problems1. Write the molecular, complete ionic and net ionic equations that describe the solution reaction of sodiumequations that describe the solution reaction of sodium sulfate with lead nitrate. The solid product of this reaction is lead sulfate.

2. Write the molecular, complete ionic and net ionic equations that describe the solution reaction of sulfuric acid with the aluminum hydroxide.

(Try them first before you see my answers)(Try them first, before you see my answers)

W it th ti th t d ib th l ti ti fWrite the equations that describe the solution reaction of sodium sulfate with lead nitrate. The solid product of this reaction is lead sulfate.Molecules involved:

sodium sulfate = Na2SO4lead nitrate = Pb(NO )lead nitrate = Pb(NO3)2lead sulfate = PbSO4Some other unnamed product?

1st try at molecular equation: Na2SO4(aq) + Pb(NO3)2 (aq) = PbSO4 (s) + ?

C bi i h h h l f hCombining the other halves of the reactants to make the other aqueous productNa2SO4(aq) + Pb(NO3)2 (aq) = PbSO4 (s) + NaNO3(aq)2 4( q) ( 3)2 ( q) 4 ( ) 3( q)Balancing:Na2SO4(aq) + Pb(NO3)2 (aq) = PbSO4 (s) + 2 NaNO3(aq)

W it th ti th t d ib th l ti ti fWrite the equations that describe the solution reaction of sodium sulfate with lead nitrate. The solid product of this reaction is lead sulfate.Starting guess:

Na2SO4(aq) + Pb(NO3)2 (aq) = PbSO4 (s) + ?

Combining the halves of the reactants that weren’t used inMaking PbSO4 you predict that missing product is NaNO3.

Next guessNa2SO4(aq) + Pb(NO3)2 (aq) = PbSO4 (s) + NaNO3(aq)2 4( q) ( 3)2 ( q) 4 ( ) 3( q)

Balancing for final molecular equation:Na SO (aq) + Pb(NO ) (aq) = PbSO (s) + 2 NaNO (aq)Na2SO4(aq) + Pb(NO3)2 (aq) = PbSO4 (s) + 2 NaNO3(aq)

W it th ti th t d ib th l ti ti fWrite the equations that describe the solution reaction of sodium sulfate with lead nitrate. The solid product of this reaction is lead sulfate.Breaking up the aqueous ionic compounds:

Na2SO4(aq) = 2Na+(aq) + SO42-(aq)

Pb(NO ) (aq) = Pb2+(aq) + 2 NO -(aq)Pb(NO3)2 (aq) = Pb2+(aq) + 2 NO3 (aq)2 NaNO3(aq) = 2 Na+(aq) + 2 NO3

-(aq)

Complete ionic equation:2Na+(aq) + SO4

2-(aq) + Pb2+(aq) + 2 NO3-(aq) =

PbSO4 (s) + 2 Na+(aq) + 2 NO3-(aq)4 ( ) ( q) 3 ( q)

Net ionic equation:SO 2-(aq) + Pb2+(aq) = PbSO (s)SO4

2 (aq) + Pb2 (aq) = PbSO4 (s)

Write the equations that describe the solution reaction of sulfuric acid with the aluminum hydroxide.

Reactants:Sulfuric acid = H SOSulfuric acid = H2SO4Aluminum hydroxide = Al(OH)3

Products:?Acid and base, so H2O

Combing the other halves of the moleculesAl3+ + SO4

2-

But the charges don’t work for a 1:1 complexBut the charges don t work for a 1:1 complexso Al2(SO4)3 is better


Unbalanced molecular equation:H SO + Al(OH) = H O + Al (SO )H2SO4 + Al(OH)3 = H2O + Al2(SO4)3

Balancing:3 H2SO4 + 2 Al(OH)3 = 6 H2O + Al2(SO4)3

Adding physical forms to get our final molecular equation:g p y g q3 H2SO4 (aq) + 2 Al(OH)3 (aq) = 6 H2O(l) + Al2(SO4)3(aq)


Separating aqueous ions:3 H SO (aq) = 6H+(aq) + 3 SO 2-(aq)3 H2SO4 (aq) = 6H (aq) + 3 SO4

2 (aq)2 Al(OH)3 (aq) = 2Al3+(aq) + 6 OH-(aq)Al2(SO4)3(aq) = 2Al3+(aq) + 3 SO4

2-(aq)

Putting into complete ionic equation:6H+(aq) + 3 SO4

2-(aq) + 2Al3+(aq) + 6 OH-(aq) = ( q) 4 ( q) ( q) ( q)6H2O(aq) + 2Al3+(aq) + 3 SO4

2-(aq)


Removing spectators for net ionic:6H+(aq)+ 6 OH-(aq) = 6H O(aq)6H (aq)+ 6 OH (aq) = 6H2O(aq)

Taking to lowest common denominator to get our final net ionic equation:1 H+(aq)+ 1 OH-(aq) = 1 H2O(aq)

writing chemical equations sound - black hills state university

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