# Why add H2O when calculating the pH of an acid?

My question is very easy.

Case 1: I have to calculate the $\mathrm{pH}$ of $0.01\ \mathrm{M}$ of $\ce{HNO3}$ solution.

This is my solution:

Case 2:I have to calculate the $\mathrm{pH}$ of a $0.01\ \mathrm{M}$ $\ce{H3PO4}$ solution. This is the provided solution:

My question is why do we add the watermolecule in our calculations in the second case and not in the first? (or vice-verca)

Based on what I can make out, you did not use the concentration of $\ce{H2O}$ in either of your calculations.

In the second case, you wrote it as a part of your reaction equation. Now, the reason for doing this becomes clear when you realise that when you are talking about a $\ce{H+}$ ion, you are in fact referring to a "proton"-- a bare positive charge. Moreover, a proton is extremely tiny, and thus consequently the charge density is quite high.

This makes it extremely reactive (in a sense), and thus in a chemical system of any sort would immediately seek out and associate with the electron clouds of a surrounding molecule.

In an aqueous solution an obvious, readily available target is the water molecule, and you end up with hydronium ($\ce{H3O+}$) ions.

In these particular problems, the solvent i.e water doesn't have a significant effect on the equilibria, however, this may not always be the case; which is why I recommend writing your reaction equations as you did in the second case, as it allows you to identify the acid/base, and conjugate acid/conjugate base in the reaction.

The reason appears to be that in the nitric acid case you are assuming that it is a strong acid and dissociates completely. You could write $\ce{HNO3 + H2O ->H3O+ + NO3-}$ if you wanted to, as its technically more correct, and we can define pH as $\ce{pH = log10([H3O+])}$.
In the second example the phosphoric acid is not such a strong acid as nitric acid. It can also loose up to three protons, one at a time, so an equilibrium is necessary to calculate how much of the various phosphates are present. For the first step you could just write $\ce{H3PO4<=>H+ + H2PO4-}$ if you wanted. If dissociation was complete to $\ce{3H+ + PO4^3-}$ then you would treat it just as you did for a strong acid.

I don't think you did add h2o to that calculation...you just added it to the reaction equation. The concentration of water doesn't change and so is not included in the calculation.

• ok, but why is it there in the second example and not in the first? – privetDruzia Aug 15 '16 at 20:21
• Because acids donate H ions to water,they don't just ping them off. Strictly speaking h2o should be in the first equation also but we sometimes just leave the water out for simplicity. ..maybe we shouldn't – user33810 Aug 15 '16 at 20:24