Building and balancing chemical equations (3) Ionic precipitation

So you want to be able to build an equation for an ionic precipitation reaction.

They are sooo (I could go on but I won’t!!) difficult so you had better be prepared to persevere with them, big time.

If you are an English GCSE student, this is definitely an A* learning objective more usually found in Advanced level/college courses.

There is a beautiful demonstration of this type of reaction on YouTube here

The video comes complete with Pachabel’s Canon sound track—cool!!

But enough of the soft stuff, what about the hard formulae and equations for the reaction between lead nitrate and potassium iodide?

1) What particles are involved in these reactions?

2) Why are these reactions so quick?

3) How are they connected with salts?

Let’s take these questions in reverse order.

A precipitate is a substance insoluble in water because it does not dissolve in water.

So precipitation reactions can be used to prepare insoluble salts. 

The salt solubility rules tell you which salts are and are not soluble.

I came across a typical table of salt solubility rules and an original diagram that I’d never seen before in years of teaching chemistry. 

Here they are:

The point is these rules tell you the salts that can be formed using precipitation.

Now I’ve used the word precipitation and if you are also a student of geography you may get a tad confused at this point.

You are probably saying to yourself “precipitation that’s another word for rainfall isn’t it?”

And you would be correct.

Why is “precipitation” used for both formation of insoluble salts and rainfall?

Well, both rainfall and insoluble salts “come down” or fall: rain to earth, salts to the bottom of the test tube or beaker.

The word is derived from the word for a vertical cliff face or “precipice”

Push something (like your worst enemy) over a precipice, it he or she will precipitate or fall down to the bottom of the cliff and precipitation of it he or she will have taken place. 

What a lovely word and thought!!!

So question 2 why are these precipitation reactions so quick?

To make the insoluble salt we use two solutions of soluble salts.

In the example above it was lead nitrate and potassium iodide both soluble in water.

But more importantly both are solutions of IONS.

As in this diagram below:

As you can see lead nitrate consists of two ions: lead(II) ions: Pb²⁺  and nitrate(V) ions: NO₃^–  ions

Potassium iodide consists of K⁺   and I^—  ions

So why is the reaction between these two solutions so quick?

Answer: on mixing the two solutions the ions of opposite charge are electrostatically attracted to each other in aqueous solution and immediately precipitate.

It takes about a pico second that is 10^–12 seconds—pretty quick, if you ask me. 

The solid precipitate of lead iodide crystals (sometimes called golden snow!!) contain oppositely charged ions are held together by electrostatic attraction.

So how can we build and balance the equation for an ionic precipitation?

Let’s, as always, lay down the word equation reactants first:

Lead nitrate   +  potassium iodide

The easy way to see how the products form is to realise that the highlighted metals change places so:

Lead nitrate   +  potassium iodide    =    potassium nitrate   +  lead iodide

Now for the formulae:

First you need a table of ion formulae and to just memorise it.

There are really no short cuts to this no matter what you have been told or read or heard elsewhere.

Just do the traditional thing and commit this data to memory like learning your tables or how to ride a bike or play the piano. 

You’ve got to be at NIKE level so you can “Just do it!!!”

Here’s a table of ions and their formulae but you can find any number of these, like I did, on t’internet.

The other thing of course is to read your bottled water as you drink it because there are several common ions and their names usually listed there on the bottled water label.

Like so:

So let’s add the ions

Lead nitrate   +  potassium iodide    =    potassium nitrate   +  lead iodide

Pb²⁺  NO₃^–          K⁺   I^— 

And create the correct formulae ensuring that the number of positive charges and negative charges are equal since compounds are electrically neutral. 

In this example we double up the number of nitrate ions so that there are two negative charges on two nitrates for every positive charge on the lead ion.

Lead nitrate   +  potassium iodide    =    potassium nitrate   +  lead iodide

Pb²⁺  (NO₃^–)₂         K⁺   I^— 

Because nitrate ions are composed of two or more particles brackets are  needed to go round the whole thing.

Lets now tackle toe product formulae

Lead nitrate   +  potassium iodide    =    potassium nitrate   +  lead iodide

Pb²⁺  (NO₃^–)₂         K⁺   I^—                        K⁺   NO₃^–                   Pb²⁺  I^— 

Again lets sort out the formulas and double up on the iodide ions in lead iodide so

Lead nitrate   +  potassium iodide    =    potassium nitrate   +  lead iodide

Pb²⁺  (NO₃^–)₂         K⁺   I^—                        K⁺   NO₃^–                   Pb²⁺  I^—  ₂

Lets remove all charges and see what the thing looks like

Lead nitrate   +  potassium iodide    =    potassium nitrate   +  lead iodide

Pb(NO₃)₂              K  I                                       K  NO₃                   Pb I₂

All that’s left to do now is to balance up the numbers of particles left and right

In this case the iodides need doubling so

Lead nitrate   +  potassium iodide    =    potassium nitrate   +  lead iodide

Pb(NO₃)₂             2 K  I                                       K  NO₃                   Pb I₂

But that leaves too few potassium’s on the right so double up that too.

Lead nitrate   +  potassium iodide    =    potassium nitrate   +  lead iodide

Pb(NO₃)₂             2 K  I                                      2 K NO₃                   Pb I₂

And you are done!!

Tedious maybe difficult yes but we got there in the end.

Now let’s do the spectator ion bit.

You say, what’s that mean?

Well, in a precipitation reaction only two ions form the precipitate the other two remain in the solution unaffected

These two ions that remain in solution are just looking on as the reaction to form the precipitate takes place

We’ll call them the spectators of the chemistry or spectator ions. 

We can extract from our full equation above the two ions that form the precipitate by working back from the precipitate itself.

If lead(II) iodide is the precipitate then the lead and iodide ions are the ions involved in the reaction

The potassium and the nitrate ions are the spectator ions

So the equation was this

 Lead nitrate   +  potassium iodide    =    potassium nitrate   +  lead iodide

Pb(NO₃)₂             2 K  I                                      2 K NO₃                   Pb I₂

But removing the spectator ions it becomes this

 Lead(II)   +   iodide    =     lead(II) iodide

Pb²⁺            2 I^—                           Pb I₂

And to go the whole hog we can add state symbols to our equations like this:

Full equation:

Lead nitrate   +  potassium iodide    =    potassium nitrate   +  lead iodide

Pb(NO₃)₂ (aq)            2 K I  (aq)                2 K NO₃ (aq)                 Pb I₂ (s)

Ionic equation:

 Lead(II)   +      iodide                =     lead(II) iodide

Pb²⁺ (aq)         2 I^— (aq)                     Pb I₂ (s)

In this post, I have tried to give you all the ideas you need to crack any precipitation equation and reduce it down to what is called its ionic equation.

You need to practice these by the dozen so here are a few to start you off

Copper sulphate   +   sodium hydroxide

Iron(II)sulphate   +  potassium hydroxide

Zinc nitrate   +  sodium hydroxide

Nickel sulphate  +  potassium hydroxide

Iron (III) sulfate   +   sodium hydroxide

Aluminium nitrate   + potassium hydroxide

Etc…………………..