3. Ionic product of water. pH and pOH

As a weak electrolyte, water dissociates in accordance with the equation:

2Н₂О ↔ Н₃О⁺ + ОН^–

or in simplification Н₂О ↔ Н⁺ + ОН^–.

Dissociation constant for the last equation looks like:

К_d = [Н⁺][ОН^–] / [Н₂О], (3.1)

where the concentrations in square brackets mean molarity at equilibrium state.

The concentration of water molecules is accepted as constant, therefore the numerator of (3.1) is also constant, which is called ionic product of water, К_w:

[Н⁺][ОН^–] = К_d [Н₂О] = К_w = 1 · 10^–14 (at 22 °С). (3.2)

Ionic product of water (К_w) slightly depends on temperature and allows to predict the hydrogen ion and hydroxyl ion concentrations.

Thus, in pure water: [Н⁺] = [ОН^–] = = 1 · 10^–7 mol/L. In acidic medium: [Н⁺] >[ОН^–], in basic medium: [ОН^–] >[Н⁺].

It is more convenient to use «–lg» (always denoted as “p”) instead of such small values with negative powers, that is why acidity (or basicity) is usually characterized with the help of pH and pOH:

рН = –lg[Н⁺] (3.3) and [Н⁺] = 10^–pH (mol/L); (3.4)

рОН = –lg[ОН^–] (3.5) and [ОН^–] = 10^–pOH (mol/L). (3.6)

Finding the negative logarithm of the left and right parts of equation (3.2), we have:

–lg([Н⁺]·[ОН^–]) = –lg10^–14 => –lg[Н⁺] + (–lg [ОН^–]) = 14 => рН + рОН = 14

Consequently, the following formula gives a relationship between pH and pOH:

рН + рОН = 14 (3.7)

In acidic medium рН <7, in neutral medium рН = 7, in basic medium рН >7.

To predict the value of pH in solutions of strong acids and bases, it is enough to know their concentrations as we usually consider that they dissociate completely (degree of dissociation 100%, or α = 1) and the concentrations of Н⁺ or ОН^– may be found very easily [read examples 6 (a–c), 7 below]. But in order to calculate the value of pH in weak acids and bases, it is necessary to know their dissociation constants or, at least, the degree of dissociation (read examples 6d, 8).

We use pH-meters to measure the pH value precisely, and indicators (or indicator paper) to determine it approximately. Any pH-meter consist of two electrodes (sometimes hidden in a tube, which looks like one electrode), and a set, which measures the potential difference. The indicator contains specific organic substances (or a mixture of several compounds), which are able to change the colour at a certain pH. The same phenomena you can observe when some wildflowers change their color after the acid rain.

All the biological liquids are complicated solutions, which contain lots of organic and inorganic compounds. The pH range in such systems depends on the composition, the concentration of components and their constants (e.g., hydrolysis or dissociation constants).

pH range for some biological fluids:

Control Questions

1. What is the pH?
2. Give a definition of the pOH.
3. What is the relationship between pH and pOH?
4. Why do we use 7 (and not 1, 0, 10,…) as the point of a neutral medium?
5. How to define the ionic product of water?
6. Which factors have an influence on the ionic product of water?
7. Write the formulae, which allow to find [H⁺] and [OH^–] knowing pH or pOH.
8. Is this possible to find the value of pH in weak acids and bases knowing only the concentrations of these electrolytes?
9. Name the biological fluid, which may have either acidic or basic medium.
10. Write the normal pH range of human blood.

Examples of calculations

Example 1.

[H⁺]_saliva = 1 · 10^–6 M. Find the value of pH, compare the result with the normal range, write a conclusion. What kind of medium is this: acidic, basic, neutral?

Answer.

In accordance with definition pH = –lg[H⁺] = –lg10^–6 = 6 (pH <7, acidic medium).

This value is less than the normal range, and such a decrease leads to caries.

Example 2.

pH_bile = 8. Find the value of [H⁺], [OH^–], pOH. What kind of medium is this: acidic, basic, or neutral?

Answer.

pH >7 means basic medium;

pOH = 14 – pH = 14 – 8 = 6;

[Н⁺] = 10^–pH = 10^–8 (mol/L);

[ОН^–] = 10^–pOH = 10^–6 (mol/L).

Example 3.

c (HCl)_{gastric juices} = 0.1 M. Find the value of pH, pOH, and [OH^–]. Compare the result with normal range.

Answer.

As a strong electrolyte, HCl dissociates completely in water medium:

1HCl ↔ 1 H⁺ + 1Cl^–.