2.6 IONIC EQUATIONS

ionic equations provide a shorthand way to show the essential chemistry involving the ions in a reaction. They enable you to make generalisations about a reaction and to pick out the species that have lost or gained electrons.

Working from a full equation to an ionic equation

Consider how to work out the ionic equation for the reaction between hydrochloric acid and sodium hydroxide that produces sodium chloride and water.

First write out the full equation:

HClaqNaOHaqNaClaqH Ol ( ) + ( ) → ( ) + ( ) 2

Now write the equation as ions and cancel the ions that appear on both sides:

Calculations from ionic equations

You can also calculate amounts from ionic equations.

One mole of sodium chloride has a mass of 58.5 g. This is calculated as the mass of a mole of sodium atoms plus the mass of a mole of chlorine atoms 23 35 5 + ( ) .g.

But sodium chloride consists of sodium ions, Na+ , and chloride ions, Cl

Since the mass of the electrons lost or gained is negligible, the mass of 1 mol Na+ is taken as the same as the mass of 1 mol sodium atoms. Similarly, the mass of 1 mol Cl is taken as the same as the mass of 1 mol of chlorine atoms.

Example

Water molecules are covalently bonded, so they forms no ions.

The ionic equation is: HaqOHaqH Ol +− ( ) + ( ) → ( ) 2

The ions that have cancelled out are called spectator ions

They are not written in the final ionic equation.

Example

Write the ionic equation for the reaction between zinc and hydrochloric acid.

First write out the full equation:

Zn sHCl aq ZnClaqH g ( ) + ( ) → ( ) + ( ) 2 22

Then write the equation as ions:

Zn sH aq Cl aq Zn aq Cl aq +H ( ) + ( ) + ( ) → ( ) + ( ) +− +− 22 2 2 2 g ( )

Hydrogen molecules are covalently bonded, so they form no ions.

Finally, cancel out ions that appear on both sides to give the ionic equation:

Zn sH aq Zn aq Hg + ( ) + ( ) → ( ) + ( ) + 2 2 2

In this reaction the zinc has been oxidised (it has lost electrons) and the hydrochloric acid has been reduced (the hydrogen has gained electrons).

You can use an ionic equation to calculate an amount. Zinc metal reacts with copper sulfate solution to deposit copper metal. Calculate the mass of copper that can be obtained from 130 g of zinc, using excess copper sulfate.

The chemical equation is:

Zn sCuSOaq ZnSOaqCus ( ) + ( ) → ( ) + ( ) 44

Write the equation as separate ions:

Zn sCuaqSOaqZnaq+SO aq ( ) + ( ) + ( )

Write the ionic equation:

Zn sCuaqZnaqCus( ) + ( ) → ( ) + ( ) ++ 22

From the ionic equation:

1 mol zinc atoms → 1 mol copper atoms so 65 g Zn gCu → 63 5 and, 130 128 gZng Cu →

You will meet many ionic equations in this course.

15. Iron displaces silver from silver nitrate solution. The ionic equation is:

Fe sAgaqFeaqAgs( ) + ( ) → ( ) + ( ) ++ 22 2

What is the maximum mass of silver that can be displaced using 9.52 g iron and excess silver nitrate solution? (Ar: Fe 55.8; Ag 107.9)