Question

A weak acid (HA) has a pKa of 4.679. If a solution of this acid has a pH of 4.876, what percentage of the acid is not ionized? (Assume all H in solution came from the ionization of HA.)

Answer 1

Concepts and reason

Problem is based on the concept of chemical equilibrium. Equilibrium is attained by a reaction if rate of change of concentration of reactant is equal to rate of change of concentration of product.

For a solution, pH is the measure of acidity/basicity of a compound in that solution.

Percentage ionization is the extent of dissociated form of an acid/base in a solution.

Fundamentals

For a solution, pH is equal to the negative of log of concentration of ions in a solution.

For acid HA,

Acid dissociation constant is calculated as follows:

Here, is concentration of hydrogen ion, is concentration of anion and is concentration of undissociated acid in solution.

Percentage ionization is calculated as follows:

Here, is concentration of hydrogen ion and is concentration of undissociated acid in solution.

For an acid,

Substitute, thus,

$-log(K)=4.679 $

Or,

Calculate the concentration of ions from pH as follows:

Substitute, 4.876 for pH thus,

$4.876=-log[H] $

Or,

The dissociation reaction of acid is as follows;

Concentration of ions is. ICE table for this reaction is as follows:

Now,

Substitute, for , for , for and for

Rearrange,

Or,

Calculate the percentage ionization as follows:

Substitute,for and for thus,

Calculate unionized percent of acidas follows:

Substitute, for thus,

Ans:

The unionized percent of acid is 39 %