The pH of Solutions of Weak Bases
The pHA logarithmic measure of the concentration of hydrogen (hydronium) ion; pH = -log10([H+]) or pH = -log10([H3O+]). of a solutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture. of a weak baseAn base that ionizes only partially in a given solvent. can be calculated in a way which is very similar to that used for a weak acidAn acid that ionizes only partially in a given solvent.. Instead of an acid constant Ka, a base constant Kb must be used. If a weak base B accepts protons from water according to the equation
B + H2O BH+ + OH– (1)
then the base constant is defined by the expression
A list of Kb values for selected bases arranged in order of strength is given in the table below. This table is part of our larger collection of acid-base resources.
The Base Constants for Some Bases at 25°C.
|Base||Formula and Ionization Equation||Kb|
|Amide ion||NH2– + H2O NH3 + OH–||large|
|Ammonia||NH3 + H2O NH+4 + OH–||1.77 × 10–5|
|Aniline||C6H5NH2 + H2O C6H5NH+3 + OH–||3.9 × 10–10|
|Carbonate ion||CO32– + H2O HCO–3 + OH–||2.1 × 10–4|
|Dimethylamine||(CH3)2NH + H2O (CH3)2NH+2 + OH–||5.8 × 10–4|
|Ethylenediamine|| (CH2)2(NH2)2 + H2O (CH2)2(NH2)2H+ + OH–
(CH2)2(NH2)2H+ + H2O (CH2)2(NH2)2H2+2 + OH–
| K1 = 7.8 × 10–5
K2 = 2.1 × 10–8
|Hydrazine|| N2H4 + H2O N2H+5 + OH–
N2H+5 + H2O N2H2+6 + OH–
| K1 = 1.2 × 10–6
K2 = 1.3 × 10–15
|Hydride ion||H– + H2O H2 + OH–||large|
|Hydroxylamine||NH2OH + H2O NH3OH + + OH–||9.3 × 10–9|
|Methylamine||CH3NH2 + H2O CH3NH+3 + OH–||5.0 × 10–4|
|Phosphate ion||PO43– + H2O HPO2-4 + OH–||5.9 × 10–3|
|Pyridine||C5H5N + H2O C5H5NH+ + OH–||1.6 × 10–9|
|Trimethylamine||(CH3)3N + H2O (CH3)3NH+ + OH–||6.2 × 10–5|
- Taken from Hogfelt, E. Perrin, D. D. Stability Constants of MetalAn element characterized by a glossy surface, high thermal and electrical conductivity, malleability, and ductility. Ion Complexes, 1st ed. Oxford; New Pergamon, 1979-1982. International Union of Pure and Applied Chemistry, Commission on EquilibriumA state in which no net change is occurring, that is, in which the concentrations of reactants and products remain constant; chemical equilibrium is characterized by forward and reverse reactions occurring at the same rate.. ISBN: 0080209580
To find the pH we follow the same general procedure as in the case of a weak acid. If the stoichiometric concentrationA measure of the ratio of the quantity of a substance to the quantity of solvent, solution, or ore. Also, the process of making something more concentrated. of the base is indicated by cb, the result is entirely analogous to equation 4 in the section on the pH of weak acids; namely,
Under most circumstances we can make the approximation
cb – [OH–] ≈ cb
in which case Eq. (3) reduces to the approximation
[OH–] ≈ (4)
which is identical to the expression obtained in the acid case (approximation shown in equation 6 in the section on the pH of weak acids) except that OH– replaces H3O+ and b replaces a. Once we have found the hydroxide-ion concentration from this approximation, we can then easily find the pOHA logarithmic measure of the concentration of hydroxide ion expressed as -log10([OH-])., and from it the pH.
EXAMPLE Using the value for Kb listed in the table, find the pH of 0.100 M NH3.
Solution It is not a bad idea to guess an approximate pH before embarking on the calculation. Since we have a dilute solution of a weak base, we expect the solution to be only mildly basic. A pH of 13 or 14 would be too basic, while a pH of 8 or 9 is too close to neutral. A pH of 10 or 11 seems reasonable. Using Eq. (4) we have
Checking the accuracyThe extent to which an experimental value agrees with true value for a quantity. of the approximation, we find
The approximation is valid, and we thus proceed to find the pOH.
pH = 14.00 – pOH = 14.00 – 2.87 = 11.13
This calculated value checks well with our initial guess.
Occasionally we will find that the approximation
cb – [OH–] ≈ cb
is not valid, in which case we must use a series of successive approximations similar to that outlined above for acids. The appropriate formula can be derived from Eq. (3) and reads
[OH–] ≈ (5)