Ionization of Weak Acids & Weak Bases (College Board AP® Chemistry)

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Ionization of a Weak Acid

Weak acids will only partially ionize in aqueous solutions
- Examples include ethanoic acid and carbonic acid
An equilibrium is established between the un-ionized acid and its conjugate base (HA represents the acid molecule):

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

HA (aq) ⇌ A^- (aq) + H⁺ (aq)

The position of the equilibrium lies to the left whereby most of the acid molecules remain unionized

Diagram to show the ionization of a weak acid

Diagram showing how weak acids partially dissociate

The diagram shows the partial dissociation of a weak acid in aqueous solution

The equations above can be used to write the following equilibrium expression:

Ka = [A-][H+][HA]

K_a is called the acid dissocation constant
In the context of ethanoic acid, CH₃COOH, the equation for the ionization of the acid would be:

CH₃COOH (aq) + H₂O (l) ⇌ CH₃COO^- (aq) + H₃O⁺ (aq)

CH₃COOH (aq) ⇌ CH₃COO^- (aq) + H⁺ (aq)

These could be used to deduce an equilibrium expression of:

$K_{a} = \frac{[{CH}_{3} {COO}^{-}] [H^{+}]}{[{CH}_{3} COOH]}$

When writing the equilibrium expression for weak acids, we assume that the concentration of H⁺(aq) due to the ionisation of water is negligible
K_a values are very small so we often use pK_a values when we are calculating pH values
To convert between K_aand pK_a we do the following:

pK_a= –logK_a

K_a= 10^–p^K^_a

The larger the K_avalue the stronger the acid
The larger the pK_avalue the weaker the acid
The pH of a weak acid solution can be determined from the initial acid concentration and the pK_a

Worked Example

Phenol, a weak acid, partially ionizes in water according to the equation below.

C₆H₅OH (aq) + H₂O (l) ⇌ C₆H₅O^– (aq) + H₃O⁺ (aq)

What is the pH of a 0.85 M C₆H₅OH (aq) solution at 25 °C?

Answer:

Start by drawing an ICE table and inputting the initial concentrations

	C₆H₅OH	+	H₂O (l)	⇌	C₆H₅O^- (aq)	+	H₃O⁺
Initial Conc.	0.85		-		0.00		0.00
Change			-
Equilibrium			-

In the change row we use x's to represent the relationship between the reactants and products
Any x's under reactants will be negative and under the products will be positive

	C₆H₅OH	+	H₂O (l)	⇌	C₆H₅O^- (aq)	+	H₃O⁺
Initial Conc.	0.85		-		0.00		0.00
Change	- x		-		+ x		+ x
Equilibrium			-

The sum of the initial concentration and change rows are used to fill in the equilibrium row

	C₆H₅OH	+	H₂O (l)	⇌	C₆H₅O^– (aq)	+	H₃O⁺
Initial Conc.	0.85		-		0.00		0.00
Change	- x		-		+ x		+ x
Equilibrium	0.85 - x		-		+ x		+ x

The equilibrium expression for the equation is:

$K_{a} = \frac{[C_{6} H_{5} O^{-}] [H_{3} O^{+}]}{[C_{6} H_{5} OH]}$

The terms from the equilibrium row can be inputted into the equilibrium expression with the value for K_a you were given in the question:
$1.12 \times 10^{- 10} = \frac{x^{2}}{0.85 - x}$
$x^{2} = 9.52 \times 10^{- 11}$
$x = \sqrt{9.52 \times 10^{- 11}}$
x = 9.56 x 10^-6M
The pH can be calculated using -log[H⁺]
- –log[9.56 x 10^-6] = 5.01

Ionization of a Weak Base

Weak bases will also partially ionize in aqueous solutions
- Examples include ammonia and methylamine
There will be a lower concentration of OH^–ions so the pH of a weak base is lower than that of a strong base
Again, an equilibrium is established between the un-ionized base and its conjugate acid with its position lying to the left so that a large proportion of base molecules remain un-ionizied

Diagram to show the ionization of a weak base

Equilibria Dissociation of a Weak Base, downloadable AS & A Level Chemistry revision notes

The diagram shows the partial dissociation of a weak base in aqueous solution

The equilibrium established is:

B (aq) + H₂O (l) ⇌ BH⁺ (aq) + OH^- (aq)

The base dissociation constant, K_b is deduced from this as:

$K_{b} = \frac{[{BH}^{+}] [{OH}^{-}]}{[B]}$

In the context of 1-phenylmethanamine, C₆H₅CH₂NH₂ (aq) this dissociates as follows:

C₆H₅CH₂NH₂ (aq) + H₂O (l) ⇌ C₆H₅CH₂NH₃⁺ (aq) + OH^– (aq)

This leads to the K_b expression:

$K_{b} = \frac{[C_{6} H_{5} {CH}_{2} {NH}_{3}^{+}] [{OH}^{-}]}{[C_{6} H_{5} {CH}_{2} {NH}_{2}]}$

K_b values are very small so we often use pK_b values when we are calculating pH values
To convert between K_band pK_bwe do the following:

pK_b= –logKb

K_b= 10^–p^Kb

The pH of a weak base solution can be determined from the initial base concentration and the pK_b

Examiner Tip

To calculate the pH for a weak base, ICE tables can be used as they were to calculate the pH for a weak acid.

Percentage Ionization Calculations

The percentage ionization of a weak acid / base represents the extent to which the acid / base dissociates into ions in a solution
To calculate the percentage ionization of a weak acid use:

$% ionization = \frac{[H_{3} O^{+}]}{HA} x 100$

To calculate the percentage ionization of a weak base (A-) use:

$% ionization = \frac{[{OH}^{-}]}{A^{-}} x 100$

Worked Example

Calculate the percentage ionization for a 0.1 M solution of acetic acid with a pH of 2.5.

Answer:

Convert pH to [H₃O⁺]

2.5 = -log[H₃O⁺]
[H⁺] = 10^−pH
[H⁺] = 10^-2.5
[H⁺] = 0.00316 M

Calculate the percentage ionization

$% ionization = \frac{[H_{3} O^{+}]}{HA} x 100$
$% ionization = \frac{0.00316}{0.10} x 100$
% ionization = 3.16 %

This indicates that 3.16% of acetic acid molecules ionize into H₃O⁺ ions in the solution

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Ionization of Weak Acids & Weak Bases (College Board AP® Chemistry)

Study Guide

Ionization of a Weak Acid

Diagram to show the ionization of a weak acid

Worked Example

Ionization of a Weak Base

Diagram to show the ionization of a weak base

Examiner Tip

Percentage Ionization Calculations

Worked Example

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Author: Alexandra Brennan