(As in equations (3.14) and (3.16) the solvent term [H 2 O] is by convention omitted. Kh is simply related
to Kw and Ka by equation (3.26), and as such is a redundant constant whose use should be discouraged.)
Equation (3.24) shows that the amounts of AH and OH– generated in the hydrolysis are equal.
Furthermore, if it is assumed that only a small amount of the salt is hydrolysed, the concentration C of
the salt dissolved is approximately the same as the concentration of A–. Then from (3.26)
from equation (3.18), [H+] = Kw/[OH–] whence [H+] = (KwKa/C)1/2 and finally
The pH of the solution will be dependent upon pKa for the acid AH and on the concentration of the salt
dissolved in the solution. For example, the pH of solutions of sodium cyanide may be calculated as
follows:
Thus, when
(b)—
Strong Acid-weak Base
Similar reasoning shows hydrolysis leading to the production of hydrogen ions and a drop in pH,
and enables an analogous expression for pH to be derived, i.e.
(c)—
Weak Acid-weak Base
Using the same approach again, both hydrolytic processes seen above will be expected to occur and the
pH of the solution will depend on the relative values of pKa and pKb, but be independent of the