Take these two practice problems and their solutions from Ebbing (8th ed) that involve completing the molecular equation, then writing the net ionic equation:

4.42a: $\ce{Ca(OH)2(aq) + 2H2SO4(aq) -> }$

soln: $$\ce{Ca(OH)2(aq) + 2H2SO4(aq) -> 2H2O(l) + Ca(HSO4)2(aq)}$$ $$\ce{OH-(aq) +2H+(aq) +SO4^{2-}(aq) -> H2O(l) + HSO4-(aq)\\ \text{ (ionizes to } H+ + SO4^{2-})}$$

4.42c: $\ce{NaOH(aq) + H2SO4(aq) -> }$

soln: $$\ce{NaOH(aq) + H2SO4(aq) -> NaHSO4(aq) + H2O(l)}$$ $$\ce{OH-(aq) + H+(aq) -> H2O(l)}$$

In 4.42a the authors do not ionize the product, only note to the side that it does ionize. In 4.42c the authors do ionize the product and account for this in the net ionic equation.

In general, is one practice preferable or more correct than the other?

Please note that I did not intend to answer my own question. After reading some of the answers, I feel it is necessary to provide clarification for future readers. Hopefully, my answer is enlightening.

Let us begin with a point of interest, reading again from Ebbing:

Phosphoric acid is an example of a triprotic acid. By reacting this acid with different amounts of a base, you can obtain a series of salts:

$$\ce{H_3PO4(aq) + NaOH(aq) \rightarrow NaH_2PO_4(aq) + H_2O(l)}$$ $$\ce{H_3PO4(aq) + 2NaOH(aq) \rightarrow Na_2HPO_4(aq) + 2H_2O(l)}$$ $$\ce{H_3PO4(aq) + 3NaOH(aq) \rightarrow Na_3PO_4(aq) + 3H_2O(l)}$$

And now, let's consider the problem.

Problem 4.42 Complete the right side of each of the following molecular equations. Then write the net ionic equations. Assume all salts formed are soluble. Acid salts are possible.

a. $\ce{2H_2SO_4(aq) + Ca(OH)_2(aq) \rightarrow} $

To give us insight into the nature of this reaction, let us first consider a complete neutralization reaction for calcium hydroxide and sulfuric acid and see what that looks like.

$$\ce{1H_2SO_4(aq) + 1Ca(OH)_2(aq) \rightarrow CaSO_4(s) +2H_2O(l)}$$

Now, let us again consider our original problem in terms of the complete neutralization. We divide the coefficients by two.

$$\ce{1H_2SO_4(aq) + \frac{1}{2}Ca(OH)_2(aq) \rightarrow \text{ incomplete neutralization}} $$

This helps give us an idea of how to predict the products for the molar ratio provided. We know the acid salt formed is soluble as we are told this in the problem statement.

$$\ce{2H_2SO_4(aq) + Ca(OH)_2(aq) \rightarrow Ca(HSO_4)_2(aq) + 2H_2O(l)}$$

For the net ionic equation, we consider that $H_2SO_4$ is a strong acid and will ionize completely in solution and $Ca(OH)_2$ is a strong base and will dissociate in solution to be present entirely as ions.

$$\ce{4H^+(aq) + 2SO_4^{2-}(aq) +} Ca^{2+}(aq) \ce{+ 2OH^-(aq)} \rightarrow Ca^{2+}(aq) \ce{+2HSO_4^-(aq) + 2H_2O(l)}$$

Eliminate the spectator ions.

$$\ce{4H^+(aq) + 2SO_4^{2-}(aq) + 2OH^-(aq) \rightarrow 2HSO_4^-(aq) + 2H_2O(l)}$$

Reduce coefficients.

$$\ce{2H^+(aq) + SO_4^{2-}(aq) + OH^-(aq) \rightarrow HSO_4^-(aq) + H_2O(l)}$$

Note that $\ce{HSO_4^-}$ ionizes to $\ce{H^+}$ and $\ce{SO_4^{2-}}$. Substitute and reduce.

$$\ce{H^+(aq) + OH^-(aq) \rightarrow H_2O(l)}$$

We accept either form of the answer, though if to go with the former you must notate the species that ionizes, otherwise is incorrect.

$$\ce{2H^+(aq) + SO_4^{2-}(aq) + OH^-(aq) \rightarrow H_2O(l) + HSO_4^-(aq) \text{ (ionizes to } H^+ \text{ and } SO_4^{2-}\text{)}}$$

  • 2
    $\begingroup$ Preferable is the correct one. 4.42a is wrong. There is no way to have HSO4- after neutralisation, but SO4^2- before it. $\endgroup$
    – Poutnik
    Oct 4, 2020 at 4:59
  • 1
    $\begingroup$ Note that molecular equations are not generally correct, as ions in solution do not belong to particular counterions. E.g equimolar solution of KOH and NaCl Is equivalent to the equimolar solution of KCl and NaOH. $\endgroup$
    – Poutnik
    Oct 4, 2020 at 10:46
  • $\begingroup$ Note that for H2SO4 mass concentration > 1 g/L (about [H2SO4] >0.01 M, pH<2 ), the majority of H2SO4 exists as HSO4-. For pH=0 it is about 99%. So you cannot expect initial H2SO4 contains SO4^2-, but after partial neutralization by Ca(OH)2 there are suddenly HSO4-. The latter are there all the time. $\endgroup$
    – Poutnik
    Oct 6, 2020 at 6:41

2 Answers 2


We consider mixing of $\ce{H2SO4(aq)}$ and $\ce{Ca(OH)2}$ in molar ratio 2:1.

Molecular equation version:

$$\ce{2 H2SO4 + Ca(OH)2 -> Ca(HSO4)2(aq) + 2 H2O}$$

Eventually if $[\ce{Ca^2+}][\ce{SO4^2-}] = [\ce{Ca^2+}] \cdot K_\mathrm{a2} \cdot \frac {[\ce{HSO4-}]}{[\ce{H3O+}]}> K_{\mathrm{sp,}\ce{ CaSO4}}$ ($\mathrm{p}K_\mathrm{a2}=1.99$): $$\ce{Ca(HSO4)2(aq) <=> CaSO4(s) + H2SO4(aq)}$$

Ionic equation version:

Separated solutions: \begin{align} \ce{H2SO4(aq) + H2O &-> H3O+(aq) + HSO4-(aq)}\\ \ce{HSO4-(aq) + H2O &<<=> H3O+(aq) + SO4^2-(aq)}\\ \ce{Ca(OH)2(aq) &-> Ca^2+(aq) + 2 OH-(aq)} \end{align}

More exactly: \begin{align} \ce{Ca(OH)2(aq) &<=>> Ca(OH)+(aq) + OH-(aq)}\\ \ce{Ca(OH)+(aq) <=>> Ca^2+(aq) + OH-(aq)}\\ \end{align}

Mixing solutions with 2:1 molar ratio: $$\ce{2 H3O+(aq) + 2 HSO4-(aq) + Ca^2+(aq) + 2 OH-(aq)-> 4 H2O + 2 HSO4-(aq) + Ca^2+(aq) }$$

And after elimination of spectator ions: $$\ce{2 H3O+(aq) + 2 OH-(aq)-> 4 H2O }$$

Then, further ionizations of the bisulphate, as $\ce{[H3O+]}$ decreased: $$\ce{ HSO4-(aq) <=> H3O+(aq) + SO4^2-(aq)}$$

Again, eventually if $[\ce{Ca^2+}][\ce{SO4^2-}] = [\ce{Ca^2+}] \cdot K_\mathrm{a2} \cdot \frac {[\ce{HSO4-}]}{[\ce{H3O+}]}> K_{\mathrm{sp, }\ce{CaSO4}}$ ($\mathrm{p}K_\mathrm{a2}=1.99$): $$\ce{Ca^2+(aq) + SO4^2-(aq) <=> CaSO4(s)}$$


I am not sure where your solutions come from but 4.42a isn't correct. The products are calcium sulfate and water, and not, as Poutnik points out, $\ce{HSO4-}$. So the molecular formula should finish like this:

$\ce{Ca(OH)2(aq) + H2SO4(aq) -> CaSO4(aq) + 2H2O(l)}$

Calcium sulfate is actually pretty insoluble so would normally appear as a precipitate but the 5th edition of Ebbing that I looked at does say to treat all salts as soluble for this question. The ionic equation for the reaction should be this:

$\ce{2OH-(aq) + 2H+(aq) -> 2H2O(l)}$

Your solution to 4.42c is correct. So, that is what is intended.

  • $\begingroup$ In excess of acid, there will be hydrogen sulphate on the left and on the right, until CaSO4(s) starts precipitation, what is not much probable, considering Ca(OH)2(aq) at the beginning. pKa2=1.99 $\endgroup$
    – Poutnik
    Oct 4, 2020 at 7:16
  • $\begingroup$ $$\ce{Ca^2+ + 2 OH- + 2H3O+ + 2HSO4- -> Ca^2+ + 2 HSO4- + 3H2O}$$ $\endgroup$
    – Poutnik
    Oct 4, 2020 at 7:26
  • $\begingroup$ @Isobutane, the solutions come from General Chemistry: Solutions Manual, Ebbing, 8th ed. $\endgroup$ Oct 5, 2020 at 3:08

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