How many of different types of chemical species are in a certain volume of aqueous solution of an acid The acid in question has $\mathrm{p}K_\mathrm{a}$ about 1.29. Density of 98% w/w solution is $d=\pu{1.323 g/cm^3}$. I would like to find the number of undissociated acid molecules ($\ce{HX}$), water molecules ($\ce{H2O}$), dissociated acid molecules ($\ce{X-}$), and hydronium ions ($\ce{H3O+}$).
The molar mass of the acid ($\ce{HX}$) is $\pu{125.1 g/mol}$. Thus 98% w/w solution corresponds to $\pu{10.57 M (mol/L)}$. If I'm not missing sth, the solution contains 1.5 mole of water and 10.57 mole of acid. What I'm not sure is how these numbers convert into the number of ions/molecules which are, let's say in a volume of $1~\pu{\mu L}$ (one microliter).
Though my attempt would be:

Firstly I have to determine $\mathrm{pH}$ of the 98% (w/w) solution in question; I did some rough calculations and got $\mathrm{pH}$ = 0.509

Then, I need to establish the ratio $\dfrac{[\ce{X-}]}{[\ce{HX}]}$; this can be done using Henderson-Hasselbach equation; I got 0.166

Thus, for let's say 1000 $\ce{HX}$ molecules and 172 $\ce{H2O}$ molecules (this stems from 10.57:1.5 molar ratio), I have at equilibrium: 834 $\ce{HX}$, 166 $\ce{X-}$, 166 $\ce{H3O+}$ and 6 $\ce{H2O}$ molecules.


The solution, seems to be correct. 834 $\ce{HX}$, 166 $\ce{X-}$, 166 $\ce{H3O+}$ and 6 $\ce{H2O}$ molecules should have a total mass of $2.129\times10^{-19} \pu{g}$. Since the density is known, such a group of molecules should occupy a volume of $\pu{160.9 nm^3}$ which is perfectly reasonable.
The following is an approximate calculation, since among other things it makes the approximation that the activities of each substance are equal to their molarities, which is not true in such concentrated solutions. Also I take some shortcuts to make the calculation quicker.

Let's look at a volume of $1\ L$. The mass would be $1323\ g$, 98% of which ($1296.54\ g$) is due to the acid, and 2% ($26.46\ g$) due to water. This initially translates into $10.36\ mol$ of acid, and $1.47\ mol$ of water. However, some of the water and acid are consumed to create hydronium and conjugate base. 

To calculate the dissociation of the acid, you may want to apply $K_{a}=\frac{[H_{3}O^{+}] [A^{-}]}{[HA]}$ (the Henderson-Hasselbach equation is derived from this by taking the log on both sides, so they're equivalent), but this is actually a very bad approximation, since you need to take into account the concentration of water in non-dilute solutions. We'll do it anyway and I worked it out to be $[H_{3}O^{+}]=[A^-]=0.704\ mol.L^{-1}$. Since the volume we're working with is $1\ L$, that means $n_{H_{3}O^{+}} = n_{A^-} = 0.704\ mol$. Subtracting those amounts from the initial amounts of water and acid, we reach $$ n_{H_{2}O}=0.77\ mol$$ $$ n_{HA}= 9.66\ mol$$ $$n_{H_{3}O^{+}}=n_{A^-}=0.704\ mol$$
To find the actual number of each entity, just multiply by the Avogadro number $6.022\times 10^{23}$ and you have the amount of each in a litre of solution.