1
$\begingroup$

Let's say you are doing a standard titration problem, for example:

Find the pH after adding $10~\mathrm{mL}$ of $0.3~\mathrm{M}$ $\ce{NaOH}$ in the titration of $25~\mathrm{mL}$ of $0.3~\mathrm{M}$ $\ce{HF}$ with $0.3~\mathrm{M}$ $\ce{NaOH}$. The $K_a$ of $\ce{HF}$ value is $6.6 \cdot 10^{-4}$

Most sources start by considering the reaction between $\ce{HF}$ and $\ce{OH^-}$, finding the pre-equilibrium concentrations of $\ce{HF}$ and $\ce{F^-}$ and then using Henderson-Hasselbalch or the equilibrium expression directly to find $\ce{[H^+]}$ (as seen here).

However, don't you have to consider the dissociation of $\ce{HF}$ first,

$$\ce{HF <=> H^+ + F^-}$$

and then consider the reaction between those produced $\ce{H^+}$ and the added $\ce{OH^-}$? Because whenever an acid reacts with a base isn't it truly the hydrogen ion formed from the deprotonated acid reacting with the hydroxide ion from the base? How can you consider reactions of unprotonated acids/bases directly?

$\endgroup$
  • 1
    $\begingroup$ I'm not entirely sure on this but according to me the hydroxide contribution from the strong base (NaOH) is enough that we can consider HF to be completely dissociated. $\endgroup$ – Binary Geek Mar 10 '15 at 10:11
3
$\begingroup$

There are numerous approximations made when pH calculations like these are made.

In a problem like this the approximations that would usually be made are

  1. pH = -log[H+] (as opposed to pH being -log (H+ activity) which is the true definition).

  2. NaOH is an infinitely strong base, with NaOH completely dissociated. (really, [Na+][OH-]/[NaOH] is about 4)

  3. The dissciation of water can be ignored (really (Kw =[H+][OH-]= 10^-14).

  4. There exists a constant Ka for HF that is [H+][F-]/[HF], (whereas a true equilibrium constant would be a function of activities, rather than concentrations, of these species, and of water. The Ka expressed as concentrations is dependent upon ionic strength for example in accordance with the Davies equation approximation).

  5. Particular to this example, it is being assumed that the only fluorine species are HF and F- (really there is also HF2- and another equilbrium constant for it).

So you can take away one or more of these approximating assumptions to try to more accurately calculate pH, but it makes the problem very difficult.

However, don't you have to consider the dissociation of HF first, and then consider the reaction between those produced H+ and the added OH-? Because whenever an acid reacts with a base isn't it truly the hydrogen ion formed from the deprotonated acid reacting with the hydroxide ion from the base? How can you consider reactions of unprotonated acids/bases directly?

At the end of the titration (or at any other point during the titration along as the solution is homogeneously mixed) the system will be in equilibrium. The correct approach is to consider all the species that are present and all the equilbrium constants amoung the species.

Here, the species would be H2O, H+, OH-, HF, F-, Na+, and NaOH (and if I want to be picky HF2-).

The equilbrium constants are Kw, Ka of HF, Kb(NaOH), (and Ka of HF2-).

You need the same number of independent equations as unknowns to solve the problem.

[F-] + [HF] = (0.1/0.35)0.3M

[Na+] + [NaOH] = (0.1/0.35)0.3M

[H+] + [Na+] = [F-] + [OH-]

[H20] = 55M (or, if that is too much of assumption [H2O] + [OH-] + [NaOH] = constant by conservation of oxygen atoms)

plus the 3 equilibrium constant equations gives you 7 equations and 7 unknowns. That would be a comprehensive consideration of the problem dropping assumptions number 2 and 3.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.