Acid Dissociation Constant and pKa

HSC Chemistry Syllabus

• calculate and apply the dissociation constant (K_a) and pK_a (pK_a = –log10 (K_a)) to determine the difference between strong and weak acids (ACSCH098)

pK_a and pK_b

This video discusses the theory and calculations surrounding pK_a and pK_b, including the use of and to determine the difference between strong and weak acids.

 

Acid Dissociation Constant (K_a)

The strength of weak acids can be measured and quantified by the equilibrium constant of their dissociation reaction. For a generic weak acid, HA, dissolved in water

 

$$HA_{(aq)} + H_2O_{(l)} \rightleftharpoons A^-_{(aq)} + H_3O^+_{(aq)}$$

 

The equilibrium constant or acid dissociation constant is

 

$$K_{eq} = \frac{[A^-][H_3O^+]}{[HA][H_2O]}$$

 

Since the concentration of water is large and remains relatively constant compared to a weak acid, [H₂O] can be removed from the expression

 

$$K_a = \frac{[A^-][H_3O^+]}{[HA]}$$

 

The dissociation constant for ethanoic acid is:

$$CH_3COOH_{(aq)} + H_2O{(l)} \rightleftharpoons CH_3COO^-_{(aq)} + H_3O^+_{(aq)}$$

 

$$K_a = \frac {[CH_3COO^-][H_3O^+]}{[CH_3COOH]}$$

 

Acid strength is proportional to the value of K_a. A largerK_a suggests a greater ratio hydrogen (hydronium) ions and conjugate base to unionised acid. This corresponds to a greater degree of ionisation and hence acid strength. 

 

The table shows K_a values of some simple weak acids, listed in decreasing strength.

 

Weak acid	Ka
Hydrofluoric acid	5.6 x 10–4
Methanoic acid	1.6 x 10–4
Ethanoic acid	1.7 x 10–5
Hydrogen sulfide	8.9 x 10–8

 

What is pK_a?

The pK_a of an acid is given by 

 

$$pK_a = -log_{10}(K_a)$$

 

The K_a of an acid can be calculated if its pK_a is given:

 

$$K_a = 10^{-pK_a}$$

 

K_a values vary greatly over numerous magnitudes. As a result, pK_a is used to measure acid strength on a logarithmic scale. pK_a range widely, can be as small as – 4 and as large as 50.

 

A higher value of K_a corresponds to a lower pK_a value. This means stronger acids have lower pK_a values.

Weak acids	Ka	pKa
Ascorbic acid	1.0 x 10–4	3.1
Hydrofluoric acid	5.6 x 10–4	3.3
Methanoic acid	1.6 x 10–4	3.8
Ethanoic acid	1.7 x 10–5	4.8
Carbonic acid	4.2 x 10–7	6.4
Hydrogen sulfide	8.9 x 10–8	7.1

  

Strong acids usually have negative pK_a values

 

Strong acids	Ka	pKa
Hydrochloric acid	104	–4
Sulfuric acid	103	–3
Nitric acid	1	–1

 

A change in pK_a value of 1 corresponds to a ten-fold change in dissociation constant K_a. For example, methanoic acid (pK_a = 3.8) has a K_a value about 10 times higher than ethanoic acid (pK_a = 4.8).

Relationship between pK_a and pH

pK_a is a more suitable for comparing the strength of various acids than pH because pK_a is unaffected by the concentration of solution and dilution. In other words, adding water to an acid solution alters [H⁺] and thus pH, but does not change its acid dissociation constant. Therefore, pK_a and pH are independent on one another.

When pH = pK_a of an acid, [H⁺] = K_a since –log ([H⁺]) = –log (K_a).
Consequently, [acid] = [conjugate base].

This means pK_a is also the pH at which an acid becomes 50% dissociated or ionised.

 

$$Ka = \frac{[H^+][A^-]}{[HA]}$$

 

$$1 = \frac{[A^-]}{[HA]}$$

 

$$[HA] = [A^-]$$

Calculations involving pH and pK_a

For a generic weak acid HA:

 

$$HA_{(aq)} + H_2O_{(l)} \rightleftharpoons A^-_{(aq)} + H_3O^+_{(aq)}$$

 

When acid molecules dissociate, the concentration of hydrogen ions, [H₃O⁺] equals [A^–] at equilibrium. 

 

Common assumptions when performing calculations involving pH and pK_a 

1. The self-ionisation of water does not affect [H₃O⁺] of the solution and therefore, it does not contribute to the pH. This is because the equilibrium constant for this reaction is relatively small. 

 

2. When the dissociation constant, K_a, is extremely small, it can be assumed that the initial concentration of acid, HA, equals to its concentration at equilibrium. In other words, the change in [HA] during dissociation is negligible.

 

3. If the acid is diprotic or triprotic, it can be usually assumed that the second and third ionisation occur to a smaller extent than the first ionisation. In other words, if K_a of first ionisation is already small, K_a of subsequent ionisation reactions can be disregarded. For example, hydrogen sulfide is diprotic and its two ionisation reactions are as follows:

 

$$H_2S(aq) + H_2O(l) \leftrightharpoons HS^-(aq) + H_3O^+(aq) \hspace{1cm} K_a1 = 8.9 \times 10^{-8}$$

$$HS^-(aq) + H_2O(l) \leftrightharpoons S^{2-}(aq) + H_3O^+(aq) \hspace{1cm} K_a2 = 1.3 \times 10^{-14}$$

 

This assumption does not apply to all weak acids and certainly not strong acids. Some weak acids ionise to a considerable extent, to which its initial concentration differs substantially with that at equilibrium

Calculating pH from pK_a

The pK_a of hypochlorous acid, HOCl, is 7.46.

A solution of hypochlorous acid, HOCl, is made by adding 0.0400 moles of HOCl to 500.0 mL of distilled water.

What is the pH of the solution?

Step 1: Calculate K_a from  pK_a

 

$$K_a = 10^{-pK_a}$$

 

$$K_a = 10^{-7.46} = 3.5 \times 10^{-8}$$

 

Step 2: Write a balanced equation and an expression for K_a

 

$$ HOCl(aq) \leftrightharpoons H^+(aq) + OCl^-(aq)$$

 

$$K_a = \frac{[H^+][OCl^-]}{[HOCl]}$$

Step 3: Set-up an ICE table

 

	[HOCl]	[H+]	[OCl–]
Initial	$$\frac{0.0400}{0.5000} = 0.00800 \hspace{0.2cm} mol \hspace{0.1cm} L^{-1}$$	0	0
Change	–x	+x	+x
Equilibrium	0.0800 –x	x	x

 

Step 3: Substitute equilibrium concentration expressions (in terms of x) into K_a expression:

 $$3.5 \times 10^{-8} = \frac{(x)(x)}{(0.0800-x)}$$

 

Step 4: By making the assumption: x is negligible as K_a is small, simplify the equation to find x:

$$3.5 \times 10^{-8} = \frac{x^2}{0.0800}$$

 

$$x = [H^+] = \sqrt{0.0800 \times 3.5 \times 10^{-8}}$$

 

Step 5: Calculate pH from [H⁺]:

 $$pH = -log_{10}(5.3 \times 10^{-5})$$

 

pH = 4.28 (2 s. f.)

 

Note: when presenting pH as the final answer, the number of significant figures equals to the number of decimal places.

Effect of Dilution on pH, Degree of Ionisation and K_a

• Adding water to an acid solution reduces the concentration of all chemical species in the solution. As [H⁺] decreases, pH increases.

 

• However, Le Chatelier’s principle also applies when dilution occurs as water (reactant) is added to the system. This will shift the equilibrium to the right (product) side to form more H⁺/H₃O⁺ ions in order to increase the concentration of ions in the system. As a result, the degree of dissociation increases upon dilution of acid.

 

• Besides K_a, the degree of dissociation or ionisation of an acid is another quantitative measure of its strength. Degree ionisation is the percentage of weak acid (HA) molecules that ionises in solution.

 

 The degree of dissociation is given by the formula: 

$$\frac{[H^+]}{initial[HA]} \times 100$$

 

• Although acid dissociates more upon dilution, the effect of dilution, compared to equilibrium shift, has a much greater effect on [H⁺]. This means pH always increases upon dilution.

 

Table: the effect of adding water or acid to a solution of acid on its pH and degree of dissociation

Action	Effect on pH	Effect on degree of dissociation	Effect on K­a
Adding water	Increases	Increases	Unchanged
Adding acid	Decreases	Decreases	Unchanged

 

• Although the addition of water increases the degree of ionisation of an acid, the K_a remains unchanged since the ratio of generic acid [HA] to [H⁺] and conjugate base [A^-] is the same. 
• While it may not seem this way, the change in concentration between the different species is proportional to one another

$$[HA] \propto [H^+] \propto [A^-]$$ 

• The only factor change which affects `K_a` is a change in temperature

 

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