Explain the self-ionisation of water, define and use the ionic product Kw, and relate pH, pOH and temperature

What this dot point is asking

Pure water conducts electricity very slightly, which shows it contains a small concentration of ions. This is because water undergoes self-ionisation (also called autoionisation):

often written more simply as $\text{H}_2\text{O}(l) \rightleftharpoons \text{H}^+(aq) + \text{OH}^-(aq)$. One water molecule acts as a Bronsted-Lowry acid and another as a base, transferring a proton.

Relating pH and pOH

Taking the negative logarithm of the $K_w$ expression gives a relationship you use constantly:

In pure water $[\text{H}^+] = [\text{OH}^-] = 1.0 \times 10^{-7}$ mol L$^{-1}$, so pH $= 7$ and the water is neutral. Adding an acid raises $[\text{H}^+]$ above $10^{-7}$; because the product must stay equal to $K_w$, $[\text{OH}^-]$ falls below $10^{-7}$. The two concentrations are inversely linked through $K_w$ at all times.

The temperature dependence of Kw

Self-ionisation is an endothermic process (it requires energy to break the O-H bond). By Le Chatelier's principle, raising the temperature shifts the equilibrium to the right, increasing both $[\text{H}^+]$ and $[\text{OH}^-]$, so $K_w$ increases. For example, at 50 degrees Celsius $K_w$ is larger than $10^{-14}$.

This has an important consequence: neutral water at a higher temperature still has $[\text{H}^+] = [\text{OH}^-]$, but both are greater than $10^{-7}$, so its pH is below 7. The water is still neutral; only at exactly 25 degrees does neutral equal pH 7. This is a common point of confusion that examiners like to test.

pOH, and finding the pH of bases

Many WACE questions hand you a base, not an acid, and the cleanest route uses pOH. Define $\text{pOH} = -\log_{10}[\text{OH}^-]$, the hydroxide analogue of pH. For a strong base such as sodium hydroxide, the hydroxide concentration equals the base concentration (full dissociation), so you can find pOH directly and then use $\text{pH} = 14 - \text{pOH}$ at $25\ ^\circ\text{C}$. For a base supplying two hydroxide ions per formula unit, such as $\text{Ca(OH)}_2$, remember that $[\text{OH}^-]$ is twice the formula concentration. Whichever route you take, the answer must be consistent: $[\text{H}^+][\text{OH}^-]$ should multiply back to $K_w$.

The link to Ka and Kb

The ionic product also ties together the strength of a conjugate acid-base pair. For any conjugate pair, $K_a \times K_b = K_w$, which is why a strong acid (large $K_a$) must have a negligibly weak conjugate base (tiny $K_b$). This single equation, derived straight from the water equilibrium, lets you calculate $K_b$ of a base from the $K_a$ of its conjugate acid, and it explains quantitatively why salts of weak acids give basic solutions.

Why this matters

$K_w$ is the bridge between the acid and base sides of any aqueous system. It lets you calculate the pH of bases (which directly supply $\text{OH}^-$), underpins the relationship between $K_a$ and $K_b$ for a conjugate pair, and explains why pH measurements must specify temperature.