The pH of Solutions of Weak Bases

Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 14:53

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 \rightleftharpoons BH+ + OH      (1)

then the base constant is defined by the expression

K_{b}=\frac{\text{ }\!\![\!\!\text{ BH}^{\text{+}}\text{ }\!\!]\!\!\text{  }\!\![\!\!\text{ OH}^{-}\text{ }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ B }\!\!]\!\!\text{ }}      (2)

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 \rightarrow NH3 + OH large
Ammonia NH3 + H2O \rightleftharpoons NH+4 + OH 1.77 × 10–5
Aniline C6H5NH2 + H2O \rightleftharpoons C6H5NH+3 + OH 3.9 × 10–10
Carbonate ion CO32– + H2O \rightleftharpoons HCO3 + OH 2.1 × 10–4
Dimethylamine (CH3)2NH + H2O \rightleftharpoons (CH3)2NH+2 + OH 5.8 × 10–4
Ethylenediamine (CH2)2(NH2)2 + H2O \rightleftharpoons (CH2)2(NH2)2H+ + OH

(CH2)2(NH2)2H+ + H2O \rightleftharpoons (CH2)2(NH2)2H2+2 + OH

K1 = 7.8 × 10–5

K2 = 2.1 × 10–8

Hydrazine N2H4 + H2O \rightleftharpoons N2H+5 + OH

N2H+5 + H2O \rightleftharpoons N2H2+6 + OH

K1 = 1.2 × 10–6

K2 = 1.3 × 10–15

Hydride ion H + H2O \rightarrow H2 + OH large
Hydroxylamine NH2OH + H2O \rightleftharpoons NH3OH + + OH 9.3 × 10–9
Methylamine CH3NH2 + H2O \rightleftharpoons CH3NH+3 + OH 5.0 × 10–4
Phosphate ion PO43– + H2O \rightleftharpoons HPO2-4 + OH 5.9 × 10–3
Pyridine C5H5N + H2O \rightleftharpoons C5H5NH+ + OH 1.6 × 10–9
Trimethylamine (CH3)3N + H2O \rightleftharpoons (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,

K_{b}=\frac{\text{ }\!\![\!\!\text{ OH}^{-}\text{ }\!\!]\!\!\text{ }^{\text{2}}}{c_{b}-\text{ }\!\![\!\!\text{ OH}^{-}\text{ }\!\!]\!\!\text{ }}      (3)

Under most circumstances we can make the approximation

cb – [OH] ≈ cb

in which case Eq. (3) reduces to the approximation

[OH] ≈ \sqrt{K_{b}c_{b}}      (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

  & \text{ }\!\![\!\!\text{ OH}^{-}\text{ }\!\!]\!\!\text{ }=\sqrt{K_{b}c_{b}} \ 
 & \text{           }=\sqrt{\text{1}\text{.8 }\times \text{ 10}^{-\text{5}}\text{ mol dm}^{-\text{3}}\text{ }\times \text{ 0}\text{.100 mol dm}^{-\text{3}}} \ 
 & \text{           }=\sqrt{\text{1}\text{.8 }\times \text{ 10}^{-\text{6}}\text{ mol}^{\text{2}}\text{ dm}^{-6}}=\text{1}\text{.34 }\times \text{ 10}^{-\text{3}}\text{ mol dm}^{-\text{3}} \ 

Checking the accuracyThe extent to which an experimental value agrees with true value for a quantity. of the approximation, we find

\frac{\text{ }\!\![\!\!\text{ OH}^{-}\text{ }\!\!]\!\!\text{ }}{c_{\text{b}}}=\frac{\text{1}\text{.34 }\times \text{ 10}^{-\text{3}}}{\text{0}\text{.1}}\approx \text{1 percent}

The approximation is valid, and we thus proceed to find the pOH.

\text{pOH}=-\text{log}\frac{\text{ }\!\![\!\!\text{ OH}^{-}\text{ }\!\!]\!\!\text{ }}{\text{mol dm}^{-\text{3}}}=-\text{log(1}\text{.34 }\times \text{ 10}^{-\text{3}}\text{)}=\text{2}\text{.87}

From which

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] ≈ \sqrt{K_{b}\text{(}c_{b}-\text{ }\!\![\!\!\text{ OH}^{-}\text{ }\!\!]\!\!\text{ )}}      (5)