# Acid Dissociation Constant and pKa

This is part of the HSC Chemistry course under the topic Quantitative Analysis.

### HSC Chemistry Syllabus

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

### pKa and pKb

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

### Acid Dissociation Constant (Ka)

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 Ka. A largerKa 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 Ka 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 pKa?

The pKof an acid is given by

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

The Kof an acid can be calculated if its pKis given:

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

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

A higher value of Ka corresponds to a lower pKa value. This means stronger acids have lower pKa 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 pKa values

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

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

### Relationship between pKa and pH

pKa is a more suitable for comparing the strength of various acids than pH because pKa 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, pKa and pH are independent on one another.

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

This means pKa 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 pKa

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

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.

1. When the dissociation constant, Ka, 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.

1. 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 Ka of first ionisation is already small, Ka 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 pKa

The pKa 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 Ka from  pKa

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

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

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

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

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

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 Ka expression:

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

Step 4: By making the assumption: x is negligible as Ka 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 Ka

• 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 Ka, 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 Ka 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