Answer
Acid and base used to form this ion, respectively, : $H_2SO_4$ and $Ca(OH)_2$,
Complete ionic equation:
$H^+(aq) + HS{O_4}^{-}(aq) + Ca^{2+}(aq) + 2OH^-(aq) -- \gt CaSO_4(s) + 2H_2O(l) $
Net ionic equation:
$H^+(aq) + HS{O_4}^{-}(aq) + Ca^{2+}(aq) + 2OH^-(aq) -- \gt CaSO_4(s) + 2H_2O(l) $
Work Step by Step
1. Identify the ions of the salt: $(NaNO_2)$:
$Ca^{2+}$ and $S{O_4}^{2-}$:
To the cation, add hydroxide ions: $Ca(OH)_2$; this is the base.
To the anion, add hydrogen ions: $H_2SO_4$; this is the acid.
2. Now, write the balanced overall equation between, which is:
$Acid + Base -- \gt Salt + Water$
We already have the salt, so:
$H_2S{O_4}(aq) + Ca(OH)_2(aq) -- \gt CaSO_4(s) + H_2O(l) $
** Acording to table 3.1, $CaSO_4$ is not soluble in water.
3. Write the complete ionic equation.
- For the completely dissociated/ionized compounds, separate them by their ions:
$H^+(aq) + HS{O_4}^{-}(aq) + Ca^{2+}(aq) + OH^-(aq) -- \gt CaSO_4(s) + H_2O(l) $
* $H_2SO_4$ is a strong acid, but only for one of its hydrogens; therefore, it is totally dissociated in $H^+$ and $HS{O_4}^-$.
* $Ca(OH)_2$ is a strong base, so it is completely ionized in water.
Balance it:
$H^+(aq) + HS{O_4}^{-}(aq) + Ca^{2+}(aq) + 2OH^-(aq) -- \gt CaSO_4(s) + 2H_2O(l) $
4. Remove the repeated ions:
$H^+(aq) + HS{O_4}^{-}(aq) + Ca^{2+}(aq) + 2OH^-(aq) -- \gt CaSO_4(s) + 2H_2O(l) $
This is the net ionic equation.
