L6 - pH of Weak Acids

objectives

● calculate pH, pOH, hydrogen ion concentration ([H+ ]) and hydroxide ion concentration ([OH– ]) for a range of solutions

The Acid Dissociation Constant - K_a

The acid dissociation constant, Ka, comes from the equilibrium constant for the breakdown of an acid in aqueous solution:

HA + H₂O ⇋ A^- + H₃O⁺

Where H3O+ is the hydrogen ion is solution, it may also be written H+(aq). The equilibrium law for this dissociation is:

K_a = [H₃O⁺] [A^-]

[HA] [H₂O]

As the concentration of water is effectively constant, a new constant, the acid dissociation constant, Ka, is defined as:

K_a = [H⁺] [A^-]

[HA]

The acid dissociation constant, Ka gives a measure of the extent of the dissociation.

If Ka is a large value then the acid is strong and dissociates into ions easily.

These constants are only useful for weak acids.

Please be aware that the Water being termed 'effectively constant' is one assumption of Ka

The above table has some examples of K_a for a range of different weak acids. From the list it may be seen that methanoic acid is a stronger acid than ethanoic acid, i.e. it dissociates further releasing more hydrogen ions in solution.

Calculating pH of a weak acid - method

Step 1 - Define Ka

K_a = [H⁺] [A^-]

[HA]

Step 2 - [H⁺] = [A^-] - As a weak acid only ever dissociates in a 1:1 ratio into it's ions we can ASSUME that [H⁺] = [A^-]

Therefore we can rewrite the expression as:

K_a = [H⁺]²

[HA]

Step 3 - Rearrange to find H⁺

Therefore:

√(K_a x [HA]) = [H⁺]

Step 4 - Use the pH calculation to find pH

Worked Example

Example: Hydrogen ion concentration from Ka.

Calculate the [H⁺(aq)] of 0.2 M ethanoic acid (K_a = 1.78 x 10^-5)

Ethanoic acid is a weak acid. It only partially dissociates according to the equation:

CH₃COOH ⇋ CH₃COO^- + H⁺

Therefore the acid dissociation constant is:

K_a = [H⁺] [CH₃COO^-]

[CH₃COOH]

We can assume that as the acid only slightly dissociates then the concentration of the acid at equilibrium is the same (to a close approximation) as the concentration of the original acid (in this case = 0.2 M)

Therefore:

And as the hydrogen ion concentration equals the ethanoate ion concentration, then:

0.2 x (1.78x10^-5) = [H⁺]²

[H⁺] = √ (3.56 x 10^-6)

[H⁺] =1.89 x 10^-3

pK_a

If we use the same mathematical function for Ka as we do for pH we find we get the following -log₁₀(K_a).

In this instance you can see that the SMALLER pKa values correspond to the stronger acids. Methanoic acid has a SMALLER pKa value, therefore it is a stronger acid than ethanoic acid.