How To Write Ionic Equations

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)


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:

  • $\text{MnO}_{4}^{-} + \text{H}^{+} + \text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + \text{Fe}^{3+} + \text{H}_{2}\text{O}$
  • $\text{Cl}_{2} + \text{I}^{-} \rightarrow \text{Cl}^{-} + \text{I}_{2}$
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  • $\text{MnO}_{4}^{-}(aq) + 8\text{H}^{+}(aq) + 5\text{Fe}^{2+}(aq) \rightarrow \text{Mn}^{2+}(aq) + 5\text{Fe}^{3+}(aq) + 4\text{H}_{2}\text{O}(l)$
  • $\text{Cl}_{2}(aq) + 2\text{I}^{-}(aq) \rightarrow 2\text{Cl}^{-}(aq) + \text{I}_{2}(aq)$


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2 thoughts on “How To Write Ionic Equations”

    • You can get this information from the equation. For instance, refer to the equation below:

      $$\text{Na}^{+} + \text{OH}^{-} + \text{H}^{+} + \text{SO}_{4}^{2-} \rightarrow \text{Na}^{+} + \text{SO}_{4}^{2-} + \text{H}_{2}\text{O}$$

      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.

      Hope this helps!


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