# How To Write Ionic Equations

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Writing ionic equation is extremely similar to writing chemical equations. Recall that ionic compounds that dissolved in water will dissociate completely into ions (have charge).

In an ionic equation:

• Number of atoms of each elements must be balanced
• Total charges carried by the ions must be balanced (e.g. +3 on left must have +3 on right as well)
• Spectator ions are omitted from the ionic equation. (Spectator ions are ions that do not take part in the reaction)
• State of ions is always aqueous (aq)

### Examples:

Reaction between sodium hydroxide and sulphuric acid to form sodium sulphate and water.

• sodium hydroxide + sulphuric acid $\rightarrow$ sodium sulphate + water
• Unbalanced chemical equation: $\text{NaOH} \, + \text{H}_{2}\text{SO}_{4} \, \rightarrow \, \text{Na}_{2}\text{SO}_{4} \, + \, \text{H}_{2}\text{O}$
• Change the chemical equation to ionic equation: $\text{Na}^{+} + \text{OH}^{-} + \text{H}^{+} + \text{SO}_{4}^{2-} \rightarrow \text{Na}^{+} + \text{SO}_{4}^{2-} + \text{H}_{2}\text{O}$
• Omit the spectator ions ($\text{Na}^{+}$ and $\text{SO}_{4}^{2-}$): $\text{OH}^{-} + \text{H}^{+} \rightarrow \text{H}_{2}\text{O}$
• Notice that the number of H and O are balanced so we do not have to multiply the chemical symbols
• Notice that the charges are balanced as well. (Neutral on the left and neutral on the right)
• Include the state: $\text{OH}^{-} (aq) + \text{H}^{+}(aq) \rightarrow \text{H}_{2}\text{O} (l)$

Reaction between aqueous sodium hydroxide and iron(III) nitrate solution to form iron(III) hydroxide precipitate and sodium nitrate.

• iron(III) nitrate + sodium hydroxide $\rightarrow$ iron(III) hydroxide + sodium nitrate
• Unbalanced chemical equation: $\text{Fe}(\text{NO}_{3})_{3} + \text{NaOH} \rightarrow \text{Fe}(\text{OH})_{3} + \text{NaNO}_{3}$
• Change chemical equation to ionic equation: $\text{Fe}^{3+} + \text{NO}_{3}^{-} + \text{Na}^{+} + \text{OH}^{-} \rightarrow \text{Fe}(\text{OH})_{3} + \text{Na}^{+} + \text{NO}_{3}^{-}$
• Omit the spectator ions ($\text{Na}^{+}$ and $\text{NO}_{3}^{-}$): $\text{Fe}^{3+} + \text{OH}^{-} \rightarrow \text{Fe}(\text{OH})_{3}$
• Balance number of $\text{OH}^{-}$ ions: $\text{Fe}^{3+} + 3\text{OH}^{-} \rightarrow \text{Fe}(\text{OH})_{3}$
• Notice that Fe and number of charges are balanced.
• Include the state: $\text{Fe}^{3+}(aq) + 3\text{OH}^{-}(aq) \rightarrow \text{Fe}(\text{OH})_{3}(s)$

Write the balanced ionic equation for aluminium + copper(II) sulphate solution $\rightarrow$ copper + aluminium sulphate solution

• Unbalanced chemical equation: $\text{Al} + \text{CuSO}_{4} \rightarrow \text{Cu} + \text{Al}_{2}(\text{SO}_{4})_{3}$
• Change to ionic equation: $\text{Al} + \text{Cu}^{2+} + \text{SO}_{4}^{2-} \rightarrow \text{Cu} + \text{Al}^{3+} + \text{SO}_{4}^{2-}$
• Omit spectator ions ($\text{SO}_{4}^{2-}$): $\text{Al} + \text{Cu}^{2+} \rightarrow \text{Cu} + \text{Al}^{3+}$
• Balance total charges: $\text{Al} + 3\text{Cu}^{2+} \rightarrow \text{Cu} + 2\text{Al}^{3+}$
• Balance Al and Cu atoms: $2\text{Al} + 3\text{Cu}^{2+} \rightarrow 3\text{Cu} + 2\text{Al}^{3+}$
• Include state: $2\text{Al}(s) + 3\text{Cu}^{2+}(aq) \rightarrow 3\text{Cu}(s) + 2\text{Al}^{3+}(aq)$

Practice: Balance these ionic equations:

• $\text{MnO}_{4}^{-} + \text{H}^{+} + \text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + \text{Fe}^{3+} + \text{H}_{2}\text{O}$
• $\text{Cl}_{2} + \text{I}^{-} \rightarrow \text{Cl}^{-} + \text{I}_{2}$

• $\text{MnO}_{4}^{-}(aq) + 8\text{H}^{+}(aq) + 5\text{Fe}^{2+}(aq) \rightarrow \text{Mn}^{2+}(aq) + 5\text{Fe}^{3+}(aq) + 4\text{H}_{2}\text{O}(l)$
• $\text{Cl}_{2}(aq) + 2\text{I}^{-}(aq) \rightarrow 2\text{Cl}^{-}(aq) + \text{I}_{2}(aq)$ ### 2 thoughts on “How To Write Ionic Equations”

1. how do we know that which one is the spectator ions?////////

• $$\text{Na}^{+} + \text{OH}^{-} + \text{H}^{+} + \text{SO}_{4}^{2-} \rightarrow \text{Na}^{+} + \text{SO}_{4}^{2-} + \text{H}_{2}\text{O}$$
Spectator ions are the ones that do not get “changed” during a reaction. Notice that $\text{Na}^{+}$ and $\text{SO}_{4}^{2-}$ are present on both sides of the equation. They are the spectator ions.