pH Buffer solution

What is a pH buffer solution?

The mission of the pH buffer solutions is to maintain the pH of the solution containing them invariable (constant) in the against of dilutions and addition of acids or bases.

These solutions are made up of mixtures of:

-Weak acid + salt of weak acid-strong base: CH3COOH + CH3COONa

-Weak base + salt of a strong acid-weak base: NH4OH + ClNH4

Example 1

Let us see how a buffer acts to maintain the pH of a solution:

Buffer solution CH3COOH/CH3COONa

Acetic acid dissociation

CH3COOH -> CH3COO + H+

Ka = [CH3COO]·[H+]/[CH3COOH]

(shifted to the left)

Dissociation of sodium acetate

CH3COONa -> CH3COO + Na+

(shifted to the left)

the dissociation of the acid can be considered to be strongly shifted to the left as a consequence of the presence of CH3COO from CH3COONa.

On the other hand, the hydrolysis process of the CH3COO

CH3COO + H2O -> CH3COOH + OH

Está muy desplazada a la izquierda debido a que coexiste en el medio CH3COOH

If a strong acid is added, the protons from the dissociation of the acid disappear from the medium because they react with the acetate ions to form poorly dissociated acetic acid, the CH3COO ions constituting the so-called alkaline reserve of the solution.

CH3COO + H+ -> CH3COOH

On the other hand, if a strong base is added, the OH ions disappear and are neutralized by the acetic acid, the CH3COOH molecules constituting the acidic reserve of the solution.

CH3COOH + OH -> CH3COO + H2O

Let's see what is the pH of this buffer solution:

Ka = [CH3COO]·[H+]/[CH3COOH]

[H+] = Ka·[CH3COOH]/[CH3COO]

pH = -log Ka –log[CH3COOH]/[ CH3COO]

= pKa + log([CH3COO]/[ CH3COOH])

in general

pH = pKto + log([sal]/[ácido])

Example 2

Buffer solution NH4OH/ClNH4

-dissociation of the base:

NH4OH -> NH4+ + OH

Kb = [NH4+]·[OH]/[ NH4OH]

(shifted to the left because of the presence of NH4+ from total dissociation of ClNH4

ClNH4 -> NH4+ + Cl

(shifted to the left)

El proceso de hidrólisis del NH4+

NH4+ + H2O -> NH4OH + H+

It is highly shifted to the left because it coexists in the medium NH4OH

If a strong acid is added, the process takes place:

NH4OH + H+ -> NH4+ + H2O

NH4+ ions constitute the so-called alkaline reserve of the solution.

If a strong base is added:

NH4+ + OH -> NH4OH

the NH4+ molecules constitute the acid reserve of the solution.

Let's see what is the pH of this buffer solution:

Kb = [NH4+]·[OH]/[NH4OH]

[OH] = Kb·[NH4OH]/[NH4+]

pOH = -log Kb –log[NH4OH]/[NH4+]

= pKb + log([NH4+]/[NH4OH])

14 –pH = pKb + log([NH4+]/[NH4OH])

pH = 14 – pKb + log([NH4OH]/[NH4+])

in general

pH = 14 – pKb + log([base]/[salt])