6
$\begingroup$

I know the ionic product of water is $[\ce{H+}][\ce{OH-}]$, by following the equation:

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

But actually it's just another way of writing:

$$\ce{2H2O <=> H3O+ + OH-}$$

Since, $K_\mathrm{w}=K_\mathrm{c}[\ce{H2O}]$ and

$$K_\mathrm{c}[\ce{H2O}]= \frac{[\ce{H3O+}][\ce{OH-}]}{[\ce{H2O}]}$$

Why can't the following equation be used? $$K_\mathrm{w}= \frac{[\ce{H3O+}][\ce{OH-}]}{[\ce{H2O}]}$$

$\endgroup$
  • $\begingroup$ This website and references therein (particularly ref 1) may be useful to people attempting to answer. $\endgroup$ – R.M. Dec 20 '16 at 18:49
  • $\begingroup$ I am certain this is a duplicate but I cannot seem to locate it... $\endgroup$ – Todd Minehardt Dec 20 '16 at 19:19
  • 1
    $\begingroup$ You yourself state that $\mathrm{K_c [H_2O]}$ is a constant, and then arbitrarily rebrand it as $\mathrm{K_w}$, so why don't you multiply that by the remaining factor of $\mathrm{[H_2O]}$ and rebrand $\mathrm{K_w [H_2O]}$ as another constant, say $\mathrm{k_w}$? If you want to use $\mathrm{K_w}= \frac{[\ce{H3O+}][\ce{OH-}]}{[\ce{H2O}]}$, then you need to use $\mathrm{K_w=1.81 \times 10^{-16}\ mol\ L^{-1}}$ in ambient conditions. We generally just use $\mathrm{k_w= [\ce{H3O+}][\ce{OH-}]=1.01 \times 10^{-14}\ mol^2\ L^{-2}}$ because it's simpler. Either way is right. $\endgroup$ – Nicolau Saker Neto Dec 20 '16 at 22:10
3
$\begingroup$

The short answer is: substances in their standard form are not included in equilibria expressions, and most commonly this includes pure liquids and solids.

The long answer is: the equilibrium expression relates all species to a standard activity (this is why they are all unitless). This question and answer detail that well.

$\endgroup$
  • 1
    $\begingroup$ Just to hammer this home, notice that for autoionization, we do not write $\mathrm{K}_{eq}$. We write $\mathrm{K}_{w}$ because it's a different value. $\endgroup$ – Zhe Dec 20 '16 at 22:26
  • $\begingroup$ One might want to add, that it is a (usually good) approximation to assume [H2O] as constant. If we go to high acid concentrations or other unusual environments this might not be applicable anymore. $\endgroup$ – mcocdawc Dec 21 '16 at 8:35
  • $\begingroup$ @mcocdawe, I would say that it is usually a good approximation to assume the activity of H2O is a constant. Then, not only do you have to be cautious of having high concentrations of other species, but how much other species interact with and order water, decreasing the "free water" in the system. That might be significant even if the concentration of water stays constant within significant figures. $\endgroup$ – NMJD Dec 22 '16 at 3:28
2
$\begingroup$

It is because water is in excess.

At pH = 7 which is assumed to be normal pH of water, though it is generally more acidic due to auto absorption of carbon dioxide gas making acidic HCO3-, we can write find the concentration of water molecules expected to auto-ionize.

Water reacting with itself: H2O + H2O ---> H3O+ + HO-, remember that pH = -log[H3O+], so then:

7 = -log[H3O+] Rearrange:

10^(-7) M or moles / liter = [H3O+] which is actually the same about of HO- that has formed from water reacting with itself at pH 7, so appreciably not much H2O has auto ionized.

Some calculations: Density of water at room temperature is about 1.000 grams / mL Molecular weight of water is about 18 grams / mole.

Use that to convert to moles / liter by cancellation of units.

(1 grams / mL) * (1000 mL / 1 L) * (1 mole / 18 grams) = 55.56 M H2O

Which H2O is much in excess of the concentration of concentration of hydroxide and hydronium (across the whole pH scale like pH 1 to pH 14), so when you have a reagent in unreacted excess, you ignore it in your equilibrium calculations with respect to chemical kinetics.

$\endgroup$
  • $\begingroup$ I would argue that it's not fundamentally because water is in excess, it's the fact that water is in excess that the activity of water in the system does not change. The fact that the activity does not change, and is thus the same as the standard activity, means that it doesn't appear in the expression. The fundamental issue is the activity of the species. $\endgroup$ – NMJD Dec 22 '16 at 3:32
0
$\begingroup$

Because $K_w$ is defined in terms of activity, and then the approximation is made that "activity of water" = 1.

$$\ce{2H2O <=> H3O+ + OH-}$$

$$K_w = \frac{a(\ce{H3O+})a(\ce{OH-})}{[a(\ce{H2O})]^2}$$

$\endgroup$

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.