Ionic Equations

Following Sundays lesson on ionic equations here are the notes promised regarding technique for dealing with them.

When you write an equation down of an acid reaction it’s obviously going to be in a solution as acids HAVE to be in solution right. And therefore it follows that there will be ions roaming about waiting to react.

Lets take for our example NaOH reacting with H2SO4, a common reaction

Sodium Hydroxide + Sulphuric Acid à Water + Sodium Sulphate

To create an ionic equation and understand how it works we need to follow some simple steps, as follows

1. Work out the compounds involved (They will be ionic obviously so we may need to use our ionic cross trick)
2. Balance the equation (once you’ve flipped the numbers from top to bottom it usually unbalances the equation so you need to make sure that the equation is balanced – otherwise the charges will later come back and bite you!)
3. Break the equation into individual ions and bracket them off
4. See which ones come out of the solution and hence are no longer reactive
5. Write out the ions that are left still IN solution

Worked Example 1
Step 1 – work out compounds involved
NaOH + H2SO4 --> H2O + Na2SO4

The 2 after the Na comes from the fact that SO4 has a valency of 2-

Step 2 – balance equation
2NaOH + H2SO4 --> 2H2O + Na2SO4
Notice again that it takes 2 OH- to neutralise that H2SO4 – this was Question 5c earlier

Step 3 – Break into ions
(2Na+)(2OH-) + (2H+)(SO4-) --> (2H2O) + (2Na+)(SO4-)
At this point note – there are 2 reasons that H2O is not written as ionsIt is NOT an ionic bond that links itIt is the solvent – it’s not actually IN solution and hence we can’t say it breaks up in solution because it IS the solution!

Step 4 – Which ones are no longer in solution?
We can tell which ones are still in solution because they are still ionic and dissociated. Because they started off dissociated and in solution then they appear the same on both sides so you can cross them off on either side because they are identical. After all, all these reactions can only take place IN water so if they are not soluble in water no further reactions can take place. What is left over will be either covalent, water or insoluble.

There are 2 ways to spot the ones that are no longer in solution

1. They will tell you that the product is insoluble The list on page 98 under the heading formation of salts gives you rules for which salts are and are not soluble.
2. They ARE the solvent - water.

Step 5 – Write out the ones left
This is done by crossing out the ions that are identical on both sides (for the reasons described above) (2Na+)(2OH-) + (2H+)(SO4-) --> (2H2O) + (2Na+)(SO4-) This gives us the final equation 2OH- + 2H+ --> 2H2O Hope this makes more sense now.