When writing the net ionic equation, if one of the products ionizes, what is the most appropriate way to account for this in the answer?
We consider mixing of $\ce{H2SO4(aq)}$ and $\ce{Ca(OH)2}$ in molar ratio 2:1.
Molecular equation version:
$$\ce{2 H2SO4 + Ca(OH)2 -&gt; Ca(HSO4)2(aq) + 2 H2O}$$
Eventually if $[\ce{Ca^2+}][\ce{SO4^2-}] = [\ce{Ca^2+}] \cdot K_\mathrm{a2} \cdot \frac {[\ce{HSO4-}]}{[\ce{H3O+}]}&gt; K_{\mathrm{sp,}\ce{ CaSO4}}$ ($\mathrm{p}K_\mathrm{a2}=1.99$):
$$\ce{Ca(HSO4)2(aq)  &lt;=&gt; CaSO4(s) + H2SO4(aq)}$$
Ionic equation version:
Separated solutions:
\begin{align}
  \ce{H2SO4(aq) + H2O &amp;-&gt; H3O+(aq) + HSO4-(aq)}\\
  \ce{HSO4-(aq) + H2O &amp;&lt;&lt;=&gt; H3O+(aq) + SO4^2-(aq)}\\
  \ce{Ca(OH)2(aq) &amp;-&gt; Ca^2+(aq) + 2 OH-(aq)}
\end{align}
More exactly:
\begin{align}
  \ce{Ca(OH)2(aq) &amp;&lt;=&gt;&gt; Ca(OH)+(aq) + OH-(aq)}\\
  \ce{Ca(OH)+(aq) &lt;=&gt;&gt; Ca^2+(aq) + OH-(aq)}\\
\end{align}
Mixing solutions with 2:1 molar ratio:
$$\ce{2 H3O+(aq) + 2 HSO4-(aq) + Ca^2+(aq) + 2 OH-(aq)-&gt; 4 H2O + 2 HSO4-(aq) + Ca^2+(aq) }$$
And after elimination of spectator ions:
$$\ce{2 H3O+(aq) +  2 OH-(aq)-&gt; 4 H2O }$$
Then, further ionizations of the bisulphate, as $\ce{[H3O+]}$ decreased:
$$\ce{ HSO4-(aq) &lt;=&gt; H3O+(aq) + SO4^2-(aq)}$$
Again, eventually if $[\ce{Ca^2+}][\ce{SO4^2-}] = [\ce{Ca^2+}] \cdot K_\mathrm{a2} \cdot \frac {[\ce{HSO4-}]}{[\ce{H3O+}]}&gt; K_{\mathrm{sp, }\ce{CaSO4}}$ ($\mathrm{p}K_\mathrm{a2}=1.99$):
$$\ce{Ca^2+(aq) + SO4^2-(aq)  &lt;=&gt; CaSO4(s)}$$