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:

  1. 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

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

  3. 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.


2 Answers 2


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.

  • $\begingroup$ Thanks for your prompt answer. It's embarrassing to note that I made several mistakes in my calculations (concentration conversion and solution of the quadratic eqn with $K_a$). I'm aware that I should use activities and not concentrations ($c_i$) in the eqn for $K_a$ but I was simply unable to find proper activation coeffs ($f_i$) for alkyl phosphates. For such condensed solutions, Pitzer eqn for $f_i$ should be used. But its solution is not trivial so I'm wandering to what extent, the values of $n_{\ce{H2O}}$, $n_{\ce{A-}}$, etc. could change. $\endgroup$
    – norkbes
    Apr 21, 2013 at 10:37
  • $\begingroup$ I thought this question was asked as lower level chemistry homework, however, it seems you're more advanced than that. In that case, perhaps you should investigate how to solve the problem while taking into account the concentration of water. I've never actually encountered such a problem myself, so I'm not exactly sure how to go about it. The problem is in dilute solutions, $[H_2O]$ is implicitly approximated to $\ce 55.55\ mol.L$ in all equilibria. As you can see, this is very far from true. You may have to use $K'_{a}=\frac{[H_{3}O^{+}] [A^{-}]}{[H_2O][HA]}$ and the expression for $K_w$. $\endgroup$ Apr 21, 2013 at 14:17

While searching for a correct answer to my own question I performed a simple simulation using PHREEQC. It is critical to realize that the calculation procedure outlined by Nicolau in his answer to my question works properly only for dilute solutions. The highly concentrated solutions have high ionic strength ($IS$) which makes Debye–Hückel theory inadequate. Thus the activity coefficients need to be find in a different way.

It is claimed that Pitzer equations can handle solutions with $IS > 2$. This morning I got accustomed with basic functionality of PHREEQC program (it has amazing possibilities, though it seems to have extremely steep learning curve..). At the moment it seems that it is not possible to perform calculations for alkyl phosphates, which I had been originally interested in, so I decided to determine population of chemical moieties for phosphoric acid. In the calculations described below, I basically start with 14.616 mole of $\ce{H3PO4}$ dissolved in $1 kg$ of water. Then I trigger evaporation to finally get 85% wt solution.

The input file for PHREEQC is as follows (these are my first calculations in this program so I'd appreciate if someone more experienced could comment on the correctness of the input):

  pH 2.0 charge
  pe 0
  P 14616  # 85% wt solution has 14.62 moles of H3PO4
  water 1.0 # start with 1 kg of water
  H2O -1.0
  41.520 moles in 10 steps # end with 252g of water
  -macinnes   true
  -use_etheta true
  -redox      true
  -iterations 200

The results include calculated $pH$ ($0.197$), ionic strength ($1.363$), and molality ($58$). The really interesting, however, are final activities calculated for such a condensed solution of $\ce{H3PO4}$:

$\ce{H3O+}$: $0.635$
$\ce{H2O}$: $0.346$
$\ce{H3PO4}$: $56.64$
$\ce{H2PO4-}$: $0.635$

Using concentrations instead of activities in solving the equation for acid dissociation constant as shown by Nicolau, lead to $\ce{H3O+}$ of $0.324$ and $\ce{H3PO4}$ of $14.756$.

  • $\begingroup$ +1 for the effort. Unfortunately I can't exactly repeat your calculations and compare results. The more you play around with the program, the more likely you'll reach the right values. Also, perhaps it is worth searching for the theory behind acid/base dissociation in concentrated solutions, which I think is not necessarily equivalent to the theory of high ionic strength solutions. Honestly, I'm afraid there is not much I can do to help you. $\endgroup$ Apr 22, 2013 at 12:34

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