Identify the species involved in the reaction: H2CrO4 (chromic acid) and Ba(OH)2 (barium hydroxide) are both aqueous compounds that dissociate into ions in solution.
Write the dissociation of each compound into its ions: H2CrO4 dissociates into 2H^{+} and CrO4^{2-}, and Ba(OH)2 dissociates into Ba^{2+} and 2OH^{-}. Represent this as: \(\mathrm{H_2CrO_4 (aq) \rightarrow 2H^{+} (aq) + CrO_4^{2-} (aq)}\) and \(\mathrm{Ba(OH)_2 (aq) \rightarrow Ba^{2+} (aq) + 2OH^{-} (aq)}\).
Combine the ions from both reactants to write the complete ionic equation, showing all ions separately on the reactant side and the products formed: \(2H^{+} (aq) + CrO_4^{2-} (aq) + Ba^{2+} (aq) + 2OH^{-} (aq) \rightarrow\) products.
Determine the products formed: Ba^{2+} and CrO4^{2-} combine to form the insoluble precipitate BaCrO4 (s), and H^{+} and OH^{-} combine to form water, H2O (l). Write the products as \(\mathrm{BaCrO_4 (s) + 2H_2O (l)}\).
Write the complete ionic equation by putting it all together: \(2H^{+} (aq) + CrO_4^{2-} (aq) + Ba^{2+} (aq) + 2OH^{-} (aq) \rightarrow BaCrO_4 (s) + 2H_2O (l)\). This shows all ions and the formation of the precipitate and water.