This is a quote from my Chemistry textbook:

Nitric acid ionises according to the following equation: $$\ce{HNO3(l) + H2O(l) -> H3O+(aq) + NO3- (aq)}$$

I know that the formula for nitric acid is $\ce{HNO3}$. But I cannot understand what is the product of this equation. What is $\ce{H3O+(aq) + NO3- (aq)}$? Is it a diluted acid with less concentration?

This is another quote (from my textbook) that confuses me:

Nitric acid and sulfuric acid are not quite as strong as hydrochloric acid. Chemical and spectroscopic studies have shown that concentrated solutions of both these acids do contain some unionised molecules. For our purpose, however, their ionisation is effectively complete. This is particularly true in dilute solutions.

I cannot understand the part that is in the bold. What are these unionised molecules? I know that these acids are considered strong acids which completely ionise in water solution. If that is the case, then why does the text indicate that there are some unionised molecules?

  • $\begingroup$ If you there is less water than acid in solution it can't dissociate completely. $\endgroup$ – Mithoron Jan 22 '15 at 0:02

In your first reaction, $\ce{H2O}$ acts as a base to abstract an $\ce{H+}$ from the nitric acid. The resulting $\ce{H3O+}$, called hydronium, is the conjugate acid, while the $\ce{NO3-}$, called nitrate, is the conjugate base (this is the molecule of nitric acid, but stripped of its $\ce{H+}$). You could view this as a dilution, in the sense that the resulting solution is less acidic than highly concentrated nitric acid would be, but the process of dilution of acid by mixing with water inevitably involves an acid-base reaction.

As to your second question, the unionized molecules are the neutral water and nitric acid on the reactant side, while the ions are the products resulting from the acid-base reaction. Nitric acid is relatively strong, meaning it tends to ionize almost completely in aqueous solution. Additionally, the less acidic the solution already is prior to addition of the acid, the more extensively the added acid will dissociate into ions (to understand this, please read up on acid-base equilibria and Le Chatelier's principle, as the topics are too lengthy and complex to cover adequately here). This is why the author emphasizes that the dissociation is more complete in dilute solutions. Statistically, however, some proportion of the nitric acid molecules will inevitably not dissociate into ions. The stronger an acid, the greater the extent of dissociation, but there is always some statistical chance that a proportion of the molecules will remain unreacted. For strong acids, however, the overwhelming preponderance of the molecules will dissociate, while only a tiny fraction remain unreacted. Hence, for many (if not most) purposes, it's reasonable when dealing with strong acids to proceed as if the dissociation is truly complete, since the unreacted portion is typically negligible in quantity.


The product is simply the ionized form of $\text{HNO}_3$; if you prefer it is equally correct to write $$\text{HNO}_3(aq)\rightarrow \text{H}^+(aq)+\text{NO}_3^-(aq)$$

Note that technically nitric acid is $\text{HNO}_3(aq)$and not $\text{HNO}_3(l)$ (which would presumably be the salt hydrogen nitrate in liquid form) As hydrogen nitrate will simply decompose under standard conditions if not dissolved so $\text{HNO}_3$ is sufficient to refer to nitric acid; however, it is understood to refer to $\text{HNO}_3$ dissolved in water, not just $\text{HNO}_3$ alone.


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