Writing Chemical Equations

(1)

Writing Chemical Equations

Chemical equations for solution reactions can be written in

th diff t f l l ti l t i i

three 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.

(2)

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 the stoichiometry, 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 g

reaction, and does not mention other components, called spectator ions, that do not undergo change.

(3)

The Molecular Equation

Let’s look at the reaction where aqueous silver nitrate and aqueous sodium chloride react together to form a

aqueous 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 nitrate

sodium nitrate in solution = aqueous sodium nitrate

(4)

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 y as written, so you don’t have to put any extra effort into

(5)

The Complete Ionic Equation

The molecular equation contained three different ionic compounds in aqueous solution. All three of these ionic p q compounds 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 t

reaction to break all strong electrolytes into their component ions. Thus,

AgNO₃(aq) = Ag+ _{(aq) + NO}

3- (aq) NaCl(aq)= Na+ _{(aq) + Cl}-_(aq)

(6)

The Complete Ionic Equation

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

Ag+ _{(aq) + NO}

3- (aq) + Na+ (aq) + Cl-(aq)=

AgCl(s) + Na+ _{(aq) + NO}

(7)

The Complete Ionic Equation

Notice the only thing you can separate into ions are aqueous ionic 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.

(8)

The Net Ionic Equation

The complete ionic equation,

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

AgCl(s) + Na+ _{(aq) + NO}

3- (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

(9)

The Net Ionic Equation

In the Net Ionic reaction we remove the spectator ions.

Thi h l h i h i l l d i

This 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:

(10)

Example problem

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

(11)

W it th ti f th ti f b i it t d Write 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)₂ Barium nitrate Ba(NO₃)₂

Potassium chromate = K₂CrO₄ Barium chromate = BaCrO₄ Potassium nitrate KNO Potassium nitrate = KNO₃

(12)

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 KNO

make it 2 KNO₃

Balanced molecular equation:

(13)

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) + 2NO}

3- (aq) 2 KNO₃(aq) 2 K (aq) 2NO₃ (aq) and our complete ionic equation becomes:

Ba2+ (aq) + 2 NO₃- (aq) + 2 K+(aq) + CrO₄2- (aq) =

BaCrO₄(s) +2 K+_{(aq) + 2NO}

(14)

In making the complete ionic equation:

B 2+ ( ) 2 NO ( ) 2 K+( ) C O 2 ( ) Ba2+ (aq) + 2 NO₃- (aq) + 2 K+(aq) + CrO₄2- (aq) =

BaCrO₄(s) +2 K+_{(aq) + 2NO}

3- (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 CrO}

4- . Covalent ions cannot be broken into

ll i i i h t b h t

smaller ionic pieces, so you have to remember what your

(15)

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

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

(16)

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.

(17)

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

Associating molecular formula’s with names

R t t

Reactants:

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

Products: ??

(18)

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.

(19)

But BaNO₃doesn’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 O

(20)

Balancing we get:

Ba(OH)₂+ 2 HNO₃ = Ba(NO₃)₂ + 2 H₂O

And adding physical forms to get our molecular equation h

we have:

(21)

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 NO}

3-(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)

(22)

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) = H}

2O(l)

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

(23)

Practice Problems

1. Write the molecular, complete ionic and net ionic equations that describe the solution reaction of sodium equations 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)

(24)

W it th ti th t d ib th l ti ti f Write 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 = Na₂SO₄ lead nitrate = Pb(NO ) lead nitrate = Pb(NO₃)₂ lead sulfate = PbSO₄

Some other unnamed product? 1st try at molecular equation:

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

C bi i h h h l f h

Combining the other halves of the reactants to make the other aqueous product

Na₂₂SO₄₄(aq) + Pb(NO( q) ( ₃₃))₂₂ ( q)(aq) = PbSO₄₄ ( )(s) + NaNO₃₃(aq)( q) Balancing:

(25)

Starting guess:

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

Combining the halves of the reactants that weren’t used in Making PbSO₄ you predict that missing product is NaNO₃. Next guess

Na₂₂SO₄₄(aq) + Pb(NO( q) ( ₃₃))₂₂ ( q)(aq) = PbSO₄₄ ( )(s) + NaNO₃₃(aq)( 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)

(26)

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 NO}

3-(aq)

Complete ionic equation:

2Na+(aq) + SO₄2-(aq) + Pb2+(aq) + 2 NO₃-(aq) =

PbSO₄₄(s) + 2 Na( ) +(aq) + 2 NO( q) ₃₃-(aq)( q) Net ionic equation:

SO 2-(aq) + Pb2+(aq) = PbSO (s) SO₄2 (aq) + Pb2 (aq) = PbSO₄(s)

(27)

Write the equations that describe the solution reaction

of sulfuric acid with the aluminum hydroxide.

Reactants:

Sulfuric acid = H SO Sulfuric acid = H₂SO₄

Aluminum hydroxide = Al(OH)₃ Products:

?

Acid and base, so H₂O

Combing the other halves of the molecules Al3+ + SO₄

2-But the charges don’t work for a 1:1 complex But the charges don t work for a 1:1 complex

(28)

Write the equations that describe the solution reaction

of sulfuric acid with the aluminum hydroxide.

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 q 3 H₂SO₄ (aq) + 2 Al(OH)₃ (aq) = 6 H₂O(l) + Al₂(SO₄)₃(aq)

(29)

Write the equations that describe the solution reaction

of sulfuric acid with the aluminum hydroxide.

Separating aqueous ions:

3 H SO (aq) = 6H+(aq) + 3 SO 2-(aq) 3 H₂SO₄ (aq) = 6H (aq) + 3 SO₄2 (aq) 2 Al(OH)₃ (aq) = 2Al3+(aq) + 6 OH-(aq) Al₂(SO₄)₃(aq) = 2Al3+(aq) + 3 SO₄2-(aq) Putting into complete ionic equation:

6H+(aq) + 3 SO( q) ₄₄2-(aq) + 2Al( q) 3+(aq) + 6 OH( q) -(aq) = ( q)

(30)

Write the equations that describe the solution reaction

of sulfuric acid with the aluminum hydroxide.

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 H}