Auto-ionization equilibrium of water shifted with acid-base addition?

Question

Adding acids or bases to water, so that either pH or pOH decreases independently of the other, will that affect the auto-ionization of water? For example, pure water + auto-ionized state, with some base added to remove some protons, will it auto-ionize a bit more (create more H+ and OH-) or a bit less (remove some H+ and OH-)?

Answer 1

Adding acids or bases to water, so that either pH or pOH decreases independently of the other, will that affect the auto-ionization of water?

Yes, the auto-ionisation of water is affected by the addition of acid or base (alkali). The auto-ionisation reaction is:

$$2\text{ H}_2 \text{O}(l) \to \text{H}_3\text{O}^+(aq) + \text{OH}^-(aq)$$

Note that the arrow should point both ways, because it is an equilibrium reaction.

The equilibrium constant $K_w$, aka the water constant, is given by:

$$K_w=[\text{H}_3\text{O}^+]\times [\text{OH}^-] \approx 10^{-14}\tag{1}$$

The concentration of e.g. $\text{H}_3\text{O}^+$ (the 'oxonium' ion) in neutral conditions is thus:

$$[\text{H}_3\text{O}^+]= [\text{OH}^-]\approx \sqrt{10^{-14}}=10^{-7}$$

By definition the $\text{pH}$ is:

$$\text{pH}=-\log[\text{H}_3\text{O}^+]=7.0$$

Now lets say we introduce a small amount of a strong acid $\text{HA}$, so that the acid concentration becomes $c_A$. Strong acids by definition dissociate completely dissociate:

$$\text{HA}(aq) + \text{ H}_2 \text{O}(l) \to \text{H}_3\text{O}^+(aq) + \text{A}^-(aq)\tag{2}$$

Assuming $c_A$ is much larger than $10^{-7}$, that is the initial $[\text{H}_3\text{O}^+]$ value, then:

$$[\text{H}_3\text{O}^+]\approx c_A$$

And:

$$[\text{OH}^-]\approx \frac{10^{-14}}{c_A}$$

This means the auto-ioninisation according to $(1)$ is strongly suppressed.

This is also true when $\text{HA}$ is a weak acid, albeit to a lesser extent.

Note that the water constant $K_w$ is not affected by the addition of acid or base.