The Ionic Product of Water, pH and pOH

1Why does the purest water still conduct?

Water purified as far as modern technology allows, the kind used to rinse computer chips, still conducts electricity: its conductivity at 25 °C is 0.055 microsiemens per centimetre. That is tiny, but it is not zero, and conduction needs ions. Since nothing is dissolved in the water, the ions must come from the water itself. A very small fraction of water molecules pass a proton to their neighbours:

$${\mathrm{H}}_2\mathrm{O}(\mathrm{l})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}(\mathrm{a}\mathrm{q})+\mathrm{O}{\mathrm{H}}^{-}(\mathrm{a}\mathrm{q})\backslash mathrm\{H\_2O(l)\; +\; H\_2O(l)\; \backslash rightleftharpoons\; H\_3O^\{+\}(aq)\; +\; OH^\{-\}(aq)\}$$

This equilibrium, present in every aqueous solution, is the key to the pH scale. In this lesson you will use it to calculate the pH and pOH of solutions of strong acids and strong bases.

2The ionic product of water

For the self-ionisation equilibrium,

$$K_c=\frac{[{\mathrm{H}}_3{\mathrm{O}}^{+}][\mathrm{O}{\mathrm{H}}^{-}]}{[{\mathrm{H}}_2\mathrm{O}{]}^2}K\_c\; =\; \backslash frac\{[\backslash mathrm\{H\_3O^\{+\}\}][\backslash mathrm\{OH^\{-\}\}]\}\{[\backslash mathrm\{H\_2O\}]^\{2\}\}$$

So few molecules ionise that $[{\mathrm{H}}_2\mathrm{O}][\backslash mathrm\{H\_2O\}]$ stays at about $55.5\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{d}{\mathrm{m}}^{-3}55.5\backslash ,\backslash mathrm\{mol\backslash ,dm^\{-3\}\}$ (1000 g of water is 55.5 mol) and can be taken into the constant:

$$K_w=K_c[{\mathrm{H}}_2\mathrm{O}{]}^2=[{\mathrm{H}}_3{\mathrm{O}}^{+}][\mathrm{O}{\mathrm{H}}^{-}]=1.0\times {10}^{-14}\mathrm{m}\mathrm{o}{\mathrm{l}}^2\mathrm{d}{\mathrm{m}}^{-6}\text{ at 25 °C}K\_w\; =\; K\_c[\backslash mathrm\{H\_2O\}]^\{2\}\; =\; [\backslash mathrm\{H\_3O^\{+\}\}][\backslash mathrm\{OH^\{-\}\}]\; =\; 1.0\; \backslash times\; 10^\{-14\}\backslash ,\backslash mathrm\{mol^\{2\}\backslash ,dm^\{-6\}\}\; \backslash text\{\; at\; 25\; °C\}$$

In pure water every ${\mathrm{H}}_3{\mathrm{O}}^{+}\backslash mathrm\{H\_3O^\{+\}\}$ is made together with one $\mathrm{O}{\mathrm{H}}^{-}\backslash mathrm\{OH^\{-\}\}$, so $[{\mathrm{H}}_3{\mathrm{O}}^{+}]=[\mathrm{O}{\mathrm{H}}^{-}]=\sqrt{1.0\times {10}^{-14}}=1.0\times {10}^{-7}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{d}{\mathrm{m}}^{-3}[\backslash mathrm\{H\_3O^\{+\}\}]\; =\; [\backslash mathrm\{OH^\{-\}\}]\; =\; \backslash sqrt\{1.0\; \backslash times\; 10^\{-14\}\}\; =\; 1.0\; \backslash times\; 10^\{-7\}\backslash ,\backslash mathrm\{mol\backslash ,dm^\{-3\}\}$, and pH = 7.00. In any aqueous solution, acid or alkali, the product $[{\mathrm{H}}_3{\mathrm{O}}^{+}][\mathrm{O}{\mathrm{H}}^{-}][\backslash mathrm\{H\_3O^\{+\}\}][\backslash mathrm\{OH^\{-\}\}]$ stays equal to $K_{w}K\_w$: adding acid raises $[{\mathrm{H}}_3{\mathrm{O}}^{+}][\backslash mathrm\{H\_3O^\{+\}\}]$ and pushes $[\mathrm{O}{\mathrm{H}}^{-}][\backslash mathrm\{OH^\{-\}\}]$ down.

Self-ionisation is endothermic, so $K_{w}K\_w$ rises with temperature (Table 2.1). Water is neutral when $[{\mathrm{H}}_3{\mathrm{O}}^{+}]=[\mathrm{O}{\mathrm{H}}^{-}][\backslash mathrm\{H\_3O^\{+\}\}]\; =\; [\backslash mathrm\{OH^\{-\}\}]$; its pH is 7.00 only at 25 °C.

Table 2.1: $K_{w}K\_w$ and the pH of neutral water at different temperatures.

3pH and pOH

Taking $-{\log }_{10}-\backslash log\_\{10\}$ of $K_w=[{\mathrm{H}}_3{\mathrm{O}}^{+}][\mathrm{O}{\mathrm{H}}^{-}]K\_w\; =\; [\backslash mathrm\{H\_3O^\{+\}\}][\backslash mathrm\{OH^\{-\}\}]$ gives

$$\mathrm{p}\mathrm{H}+\mathrm{p}\mathrm{O}\mathrm{H}=\mathrm{p}{K}_w=14.00\text{ at 25 °C}\backslash mathrm\{pH\}\; +\; \backslash mathrm\{pOH\}\; =\; \backslash mathrm\{p\}K\_w\; =\; 14.00\; \backslash text\{\; at\; 25\; °C\}$$

Because the scale is logarithmic, a change of one pH unit means a tenfold change in $[{\mathrm{H}}_3{\mathrm{O}}^{+}][\backslash mathrm\{H\_3O^\{+\}\}]$ (Figure 2.1). A pH can be negative (a 2 M strong acid has pH −0.30) or above 14.

Figure 2.1: The pH scale at 25 °C (M = mol dm⁻³).

Significant figures: the digits after the decimal point in a pH match the significant figures in the concentration. $[{\mathrm{H}}_3{\mathrm{O}}^{+}]=2.5\times {10}^{-3}[\backslash mathrm\{H\_3O^\{+\}\}]\; =\; 2.5\; \backslash times\; 10^\{-3\}$ (2 s.f.) gives pH 2.60 (2 decimal places).

4Strong acids and strong bases

A strong acid is completely ionised, so $[{\mathrm{H}}_3{\mathrm{O}}^{+}][\backslash mathrm\{H\_3O^\{+\}\}]$ equals the acid concentration times the number of protons released per formula. A strong base gives $[\mathrm{O}{\mathrm{H}}^{-}][\backslash mathrm\{OH^\{-\}\}]$ directly; then use $\mathrm{p}\mathrm{O}\mathrm{H}\backslash mathrm\{pOH\}$ and $\mathrm{p}\mathrm{H}=14.00-\mathrm{p}\mathrm{O}\mathrm{H}\backslash mathrm\{pH\}\; =\; 14.00\; -\; \backslash mathrm\{pOH\}$, or $[{\mathrm{H}}_3{\mathrm{O}}^{+}]=\frac{K_w}{[\mathrm{O}{\mathrm{H}}^{-}]}[\backslash mathrm\{H\_3O^\{+\}\}]\; =\; \backslash frac\{K\_w\}\{[\backslash mathrm\{OH^\{-\}\}]\}$.