16.1: Acid-Ionization Equilibria

Recap:

In pure water, [H₃0⁺] = [OH^-] = 1 x 10^-7 M and pH 7 is neutral.

Acidic Solution:
[H₃0⁺] is greater than [OH^-]
[H₃0⁺] is greater than 1 x 10^-7 M
[OH^-] is less than 1 x 10^-7 M
pH is less than 7

Basic Solution:
[OH^-] is greater than [H₃0⁺]
[H₃0⁺] is less than 1 x 10^-7 M
[OH^-] is greater than 1 x 10^-7 M
pH is greater than 7

Weak Acids/Bases:

Example:

A strong acid, HCl dissociates in water 100%, so a concentration of 0.0100 M HCl would yield +0.0100 M H₃O⁺, with a pH of -log(0.0100) = 2.
A weak acid dissociates less than 100%. So 0.0100 M of a weak acid would yield less than 0.0100 M H₃O⁺. For a weak acid, the concentrations of ions in solution are determined from the acid-ionization constant.

Acid-ionization constant: the equilibrium constant for the ionization of a weak acid. Denoted K_a.

Shorthand for a weak acid is HA (where H is the hydrogen and A is the rest) for a monoprotic acid,
H₂A for a diprotic acid, and H₃A for a triprotic acid. So:

HA (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + A^- (aq)




where [H₃0⁺] and [H⁺] are interchangeable.

Example:

HC₆H₄NO₂ (aq) + H₂0 (l) ⇌ H₃O⁺ (aq) + C₆H₄NO₂^- (aq)
is the same as
HA (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + A^- (aq)

[HA] = 0.012 M
pH = 3.39
What is the K_a and what is the degree of ionization?

Degree of Ionization: the fraction of molecules that react with water to give ions. Given as a percentage it is called the percent ionization.




Since we know the pH is 3.39, we can find [H₃O⁺].

[H₃O⁺] = 10^-pH
[H₃O⁺] = 4.07 x 10^-4

So, x = 4.07 x 10^-4

K_a = (4.07 x 10^-4)(4.07 x 10^-4) / (0.0120)
K_a = 1.38 x 10^-5

To obtain the degree of ionization, notice that 4.07 x 10^-4 mol out of 0.0120 mol of acid ionizes, so

degree of ionization = (4.07 x 10^-4) / (0.0120) = 0.0339
percent ionization = 0.0339 x 100 = 3.39 %

Approximation Method:

Let C_a represent the initial concentration of the weak acid.

There will be less than a 5% error with the approximation method if C_a / K_a is greater than 100.

Example:

What is the pH for 0.500 M of acetic acid? K_a = 1.7 x 10^-5




K_a = 1.7 x 10^-5 = (x²) / (0.500 - x)

To solve this, we need to use the quadratic equation. Or, we can test to see if we are able to apply the approximation method and use that instead, since it is much quicker and easier.

Is C_a / K_a greater than 100?

(0.500) / (1.7 x 10^-5) = 29411.8

Yes.

Here is the Approximation: in the term (0.500 - x), the x will be so small that subtracting it from 0.500 will be negligible, and the term is roughly the same as simply 0.500. This approximation greatly simplifies the equation:

1.7 x 10^-5 = (x²) / (0.500)
x = 0.00292
pH = -log(0.00292) = 2.54

Remember, this substitution only works if C_a / K_a is greater than 100. If it's not, then the quadratic equation must be used.