Self-ionisation of water and Kw: WACE Year 12 Chemistry

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.