Lakhasly

Water is weak electrolyte.In the similar manner to pH and pOH, pKa is equal to the negative decimal logarithm of the constant of dissociation for an acid, while pKb is equal to the decimal logarithm of the constant of dissociation for a base.To avoid the usage of low numbers, it was suggested to use negative decimal logarithm of the concentration of H+ ions to describe the level of acidity in water solution.Since the most of the constants of dissociation are expressed by very low numbers, it is better to use their negative decimal logarithms to compare them with each other.The level of pH in the solution of strong acid is calculated from the activity of H+ ions that is equal to their concentration multiplied by the coefficient of activity (fa).For alkali you need to calculate the activity of OH- ions, convert it to pOH, and then minus the resulting number from 14.We can calculate it for 1 L of water: the density of water is 1 g/ml, the mass of water is 1000 g, the molar mass is 18 g/mol, and the number of moles is 55.6 mol.pH = -log[H+] pOH = -log[OH-] pH + pOH = 14 In the same manner we can calculate pOH from the concentration of OH- ions.pKa = -log Kdiss for an acid ( Ka - acid ionization constant).

Water is weak electrolyte. Its constant of dissociation is rather low: 1.8∙10–16 at 25 °C. So, we
can write the equation about this process.
Kdiss = ([H+]∙[OH–]) / [H2O] = 1.8∙10–16
The above written equation includes the concentration of water in pure water. We can calculate
it for 1 L of water: the density of water is 1 g/ml, the mass of water is 1000 g, the molar mass is 18
g/mol, and the number of moles is 55.6 mol. So, the concentration of water in pure water is equal to
55.6 mol/L, and we can re-write the equation with this concrete number.
Kdiss = ([H+]∙[OH–]) / 55.6 = 1.8∙10–16
Kwater = Kdiss∙[H2O] = 55.6∙1.8∙10–16 = 1∙10–14
[H+]∙[OH–] = 1∙10–14
The multiple of concentrations of H+ and OH– in water solution is always equal to the
same number at the same temperature. At 25 °C that constant of water (water ion product) is
equal to 1∙10–14. It means that if we know the concentration of H+, we can calculate the concentration
of OH–, and vice versa.
To avoid the usage of low numbers, it was suggested to use negative decimal logarithm of the
concentration of H+ ions to describe the level of acidity in water solution. For example, if the
concentration of H+ is equal to 0.01 mol/L, pH is equal to 2, if the concentration of H+ is equal to
0.00001 mol/L, pH is equal to 5.
pH = –log[H+] pOH = –log[OH–]
pH + pOH = 14
In the same manner we can calculate pOH from the concentration of OH– ions. Since the ion
product of water is 10–14, the sum of pH and pOH is equal to 14. Using this simple equation we can get
pH from the concentration of alkali in water solution.
The level of pH in the solution of weak acid is calculated with the help of the degree of
dissociation (α). In each step of dissociation the concentration of H+ is equal to the multiple of
concentration of a weak acid and the degree of its dissociation. The level of pH in the solution of
strong acid is calculated from the activity of H+ ions that is equal to their concentration multiplied by
the coefficient of activity (fa). For alkali you need to calculate the activity of OH– ions, convert it to
pOH, and then minus the resulting number from 14.
Since the most of the constants of dissociation are expressed by very low numbers, it is better
to use their negative decimal logarithms to compare them with each other. In the similar manner to
pH and pOH, pKa is equal to the negative decimal logarithm of the constant of dissociation for an
acid, while pKb is equal to the decimal logarithm of the constant of dissociation for a base.
pKa = –log Kdiss for an acid ( Ka – acid ionization constant).