For the dissociation of a weak acid: $$\ce{HA(aq)<=>H+ + A-}$$ To calculate $\mathrm{pH}$ of a weak acid, we use the expression: $$K_\mathrm{a}=\ce{\frac{[H+] [A-]}{HA}}$$

The $\ce{[H+]}$ has two sources:

  1. Water
  2. Acid

Now, many problems that I've read on the calculation of $\mathrm{pH}$ of weak acids that doesn't take $\ce{H+}$ due to dissociation of water into account. I speculated that maybe in the expression of $K_\mathrm{a}$, the $\ce{[H+]}$ is the total hydrogen ion concentration (including hydrogen ions from water). Is that true? Or is that an approximation we have to make?

Which one of the following correctly represents acid dissociation in water? $$\ce{H2O + HA <=> H3O+ + A-},\tag{R1}$$

$$\ce{HA <=> H+ + A-}?\tag{R2}$$

Does the first reaction imply that water also dissociated and the $\ce{H+}/\ce{H3O+}$ concentration takes into account the contribution from water?

  • $\begingroup$ More exact expressions do, simplified ones leading to pH=0.5*(pKa - log c) do not. $\endgroup$ – Poutnik Oct 19 at 14:43
  • 2
    $\begingroup$ The only difference between those two reactions is the former explicitly mentions hydration of protons, as protons do not exist naked in water, being an extremely strong Lewis acid, reacting with the Lewis base H2O. $\endgroup$ – Poutnik Oct 19 at 15:05
  • $\begingroup$ It may help to have a look at this answer chemistry.stackexchange.com/questions/60068/…. and the answer listed in the comments. $\endgroup$ – porphyrin Oct 24 at 17:03

Your question and your formulas are really asking about two different things.

Let's look at your formulas first.

$$\ce{H2O + HA <=> H3O+ + A-},\tag{R1}$$

$$\ce{HA <=> H+ + A-}?\tag{R2}$$

For R1 the equilibrium equation should be:

$$\mathrm{K_a} = \dfrac{\ce{[H3O+][A-]}}{\ce{[H2O][HA]}}\tag{Eq-1}$$

but for R2 the equilibrium equation should be:

$$\mathrm{K_a} = \dfrac{\ce{[H+][A-]}}{\ce{[HA]}}\tag{Eq-2}$$

Thus the two equilibrium equations differ by a factor of $\ce{[H2O]}$ in the denominator. Since these equations are really only good for "dilute" aqueous solutions (say less than 0.1 molar ionic strength) the concentration of water can be taken as a constant. Thus equation Eq-2 is the accepted way to consider the equilibrium.

Not to leave the explanation hanging, let me point out that rather than concentrations the equilibrium expression should really be written using activities.

$$\mathrm{K_a} = \dfrac{a_\ce{H+}a_\ce{A-}}{a_\ce{HA}}\tag{Eq-3}$$

where the activity of a species does take into account the ionic strength of the solution.

Now let's consider water's contribution to the $\ce{H+}$ concentration.

Water disproportionates according to the chemical formula:

$$\ce{H2O <=> H+ + OH-}\tag{R3}$$

with the equilibrium expression

$$\mathrm{K_w} = \ce{[H+][OH-]}\tag{Eq-4}$$

Thus if an acid, HA, is added to water, then both equations Eq-2 and Eq-4 must be satisfied simultaneously for an exact solution. Thus for a monoprotic acid this leads to a cubic equation, for a diprotic acid a quartic equation, and so on. Fortunately in chemistry only 2-3 significant figures are usable, so we can usually make assumptions about the chemistry and simplify the mathematics to a quadratic equation for most situations.

| improve this answer | |

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.