Acidity of Solutions of Acid-Base Buffers

James Richard Fromm


Acids do not react with acids nor bases with bases because acid-base reactions are proton transfer reactions. In order to transfer a proton, one species (the acid) must donate it to another species (the base) which will accept it. An amphiprotic substance can serve either as an acid or a base, but it serves only one of these functions in any one equilibrium. For example, the amphiprotic dihydrogen phosphate ion acts as a base in the first ionization of phosphoric acid (K1 = 7.11 x 10-3) and as an acid in the second (K2 = 6.23 x 10-8). Acids react only with bases and bases react only with acids.

Acid-base reactions are always proton transfers between conjugate pairs. Under the classical Arrhenius definitions, reactions of acids with bases always form water because H+ + OH- rarrow.gif (63 bytes) H2O. Under the Bronsted definitions, no water need be directly involved in proton transfer reactions.

The proton transfer reactions of aqueous acids and bases are generally rapid and take place stoichiometrically whenever possible. In all cases, the reaction proceeds from strength to weakness, in that the resultant acid or base is always weaker than the species from which it originated. This has to be the case for, if the resultant acid or base were stronger, it would promptly donate the proton to (or, in the case of the base, acquire the proton from) the species which originally did (or, in the case of the base, did not) possess it. Thus, strong acids react with strong bases to give a neutral solution because the hydrogen ions and hydroxide ions neutralize each other to give water. A strong acid reacts with a weak base to yield the weak acid which is the conjugate acid of the weak base. Likewise, a strong base reacts with a weak acid to yield the weak base which is the conjugate base of the weak acid. A weak acid and a weak base react with each other, generally partially, yielding a mixture of two weak acid - weak base conjugate pairs. The last of these cases are rarely of interest as reactions although they do find use as buffer systems.

The pH of a solution made up by placing together species which react with each other depends upon the species remaining in solution after the proton transfers have taken place. These final solutions or equilibrium solutions will be aqueous solutions of strong acids, weak acids, strong bases, weak bases, or buffer mixtures. In the special case of a strong acid and a strong base added in exactly equal stoichiometric amounts, the solution will be that of a neutral salt because no acid or base other than water will remain. The only other special case is a solution of an amphiprotic substance, where two ionization constants must be considered because the substance engages in acid-base reactions with itself.


Example. Let us add 10 mL of 0.10 molar HCl to 10 mL of 0.2 molar ethanolamine and calculate the solution pH. Before reaction there are 10 mL x 0.10 molar = 1.0 mmol HCl and 10 mL x 0.2 molar = 2.0 mmol HOCH2CH2NH2. After reaction there are 1.0 - 1.0 = 0.0 mmol HCl; 2.0 - 1.0 = 1.0 mmol HOCH2CH2NH2; and 0.0 + 1.0 = 1.0 mmol HOCH2CH2NH3+.

A solution containing both a weak acid and its conjugate base forms a buffer mixture, whose pH can be calculated from the ionization constant of the acid, which is the ethanolammonium ion:

Ka = [H3O+][HOCH2CHNH2]/[HOCH2CH2NH3+] Since here the number of moles of acid and base forms are equal, their concentrations are equal also and so Ka = [H3O+] = 3.17 x 10-10, pH = -(-10 + 0.50) = 9.50.


The student will find it useful to repeat the above example, using sodium hydroxide in place of hydrochloric acid and acetic acid in place of monoethanolamine.


Example. Let us add 10 mL of 0.2 molar HCl to 10 mL of 0.1 molar NaOH and calculate the solution pH. Before reaction there are 10 mL x 0.2 molar = 2 mmol HCl and 10 mL x 0.1 molar = 1 mmol NaOH. After reaction, there are 2 - 1 = 1 mmol HCl; 1 - 1 = 0 mmol NaOH; and 0 + 1 = 1 mmol NaCl. A solution of a neutral salt and a strong acid has its pH determined by the strong acid:

1 mmol/20 mL = 0.05 molar HCl

[H3O+] = 5 x 10-2, pH = -(-2 + 0.70) = 1.30


Example. Let us add 10 mL of 0.1 molar HCl to 10 mL of 0.1 molar NaOH and calculate the solution pH. Before reaction there are 10 mL x 0.1 molar = 1 mmol HCl and 10 mL x 0.1 molar = 1 mmol NaOH. After reaction, there are 1 - 1 = 0 mmol HCl; 1 - 1 = 0 mmol NaOH; and 0 + 1 = 1 mmol NaCl. A solution containing only a neutral salt has a pH of exactly 7.00.


Example. Let us add 10 mL of 0.1 molar HCl to 10 mL of 0.2 molar NaOH and calculate the solution pH. Before reaction there are 10 mL x 0.1 molar = 1 mmol HCl and 10 mL x 0.2 molar = 2 mmol NaOH. After reaction, there are 1 - 1 = 0 mmol HCl; 2 - 1 = 1 mmol NaOH; and 0 + 1 = 1 mmol NaCl. A solution containing a neutral salt and a strong base has its pH determined by the strong base: 1 mmol/20 mL = 0.05 molar NaOH; [OH-] = 5 x 10-2, pOH = 1.30, pH = 12.70


Example. Let us add 10 mL of 0.2 molar HCl to 10 mL of 0.1 molar ethanolamine, HOCH2CH2NH2, and calculate the solution pH. Before reaction there are 10 mL x 0.2 molar = 2 mmol HCl and 10 mL x 0.1 molar = 1 mmol HOCH2CH2NH2. After reaction, there are 2 - 1 = 1 mmol HCl; 1 - 1 = 0 mmol HOCH2CH2NH2; and 0 + 1 = 1 mmol HOCH2CH2NH3+. A solution containing a strong acid and a weak acid has its pH determined by the strong acid essentially as if the strong acid were present alone: 1 mmol/20 mL = 0.05 molar HCl, pH = 1.30.


Example. Let us add 10 mL of 0.1 molar HCl to 10 mL of 0.1 molar ethanolamine and calculate the solution pH. Before reaction there are 10 mL x 0.1 molar = 1.0 mmol HCl and 10 mL x 0.1 molar = 1.0 mmol ethanolamine. After reaction there are 0 + 1 = 1 mmol ethanolammonium ion in 20 mL of solution, giving a concentration of 0.05 mol/liter. This is a solution of a weak acid, so its pH can be calculated from the dissociation of the acid:

Ka = [H3O+][HOCH2CH2NH2]/[HOCH2CH2NH3+]

3.17 x 10-10 = [H3O+]2/5 x 10-2; [H3O+] = 3.98 x 10-6, pH = 5.40


Copyright 1997 James R. Fromm