The pH Scale

At a given temperature pH of all solutions will be falling within a given range. This range depends on the extent to which water is ionized at that given temperature.

Water, being a weak electrolyte, will be partially ionized. The partial ionization of water can be represented as below;

\[H_2O \rightleftharpoons H^+ + HO^-\]

At \(25^o C\), the degree of dissociation (\(\alpha\)) of water is \(1.8 \times 10^{-9}\). This means that for every mole of water taken \(1.8 \times 10^{-9}\) moles of water is ionized.

The acid dissociation constant of water (\(K_a\)) can be written as,

\(K_a = \frac {[H^+][HO^-]}{[H_2O]} = \frac{C \alpha \times C \alpha}{C(1-\alpha)}\)

where C is the concentration of water and \(\alpha\) is the degree of dissociation of water.

The concentration of water is \(\frac {1000 g/L}{18 g/mol} \approx 55.5 mol/L \)

As \(\alpha \ll 1\), we can neglect it with respect to 1. Neglecting it with respect to 1 and substituting the values of C and \(\alpha\) in the above equation for \(K_a\) we get,

\(K_a = C\alpha^2 = \frac{1000}{18} \times 1.8 \times 1.8 \times 10^{-18} = 1.8 \times 10^{-16}\)

We can now define another constant \(K_w\), the ionic product of water as,

\(K_w = K_a \times [H_2O] = [H^+][HO^-]\)

\(K_w = 1.8 \times 10^{-16} \times \frac{1000}{18} = 10^{-14}\)

\(K_w = 1.8 \times 10^{-14} = [H^+][HO^-] = [H^+]^2 (because [H^+] = [HO^-])\)

So, \([H^+]^2 = 10^{-7}\)

\(pH = -log[H^+] = -log 10^{-7} = 7\)

pH of water at \(25^oC\) is 7. This will be the neutral pH as the \([H^+] = [HO^-]\). If the \([H^+] > [HO^-]\), then the value of pH will become less than 7 and if the \([H^+] < [HO^-]\), then the value of pH will become more than 7. To determine the lower and upper limit of the pH value at this temperature and thus the pH scale at this temperature we have to down by 7 units and up by 7 units because 7 will be the mid point of the scale. Thus, the pH scale at \(25^oC\) will be 0 to 14.

Let us say that, at \(125^oC\), \(K_w\) of water is \(1.0 \times 10^{-12}\). What would be the pH of pure water at this temperature and if a solution with pH = 7, is available at this temperature, will it be acidic, basic or neutral?

\(K_w = 1.0 \times 10^{-12} = [H^+][HO^-] = [H^+]^2 (because [H^+] = [HO^-])\)

So, \([H^+]^2 = 10^{-6}\)

\(pH = -log[H^+] = -log 10^{-6} = 6\)

pH of water at \(125^oC\) is 6. This will be the neutral pH as the \([H^+] = [HO^-]\). If the \([H^+] > [HO^-]\), then the value of pH will become less than 6 and if the \([H^+] < [HO^-]\), then the value of pH will become more than 6. The pH scale at this temperature will be 0 to 12. Also, the solution whose pH is 7 at this temperature would be basic.