# pKa

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

# 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

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

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:

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

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

• Acid strength is proportional to the value of Ka. Higher the value indicates higher concentration of hydrogen ions and more acids ionised. This corresponds to a stronger acid

• This is because the Ka value reveals to us information regarding the ratio of ions to unionised acid.

•  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

## Calculations involving pH and Acid Dissociation Constant Ka

• For a generic weak acid HA:

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

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

Common assumptions

1. It can be assumed, during calculation, that self-ionisation of water does not affect [H3O+] of the solution and therefore, it does not contribute to the pH.

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

## Example 1 - Calculating pH from Ka

The Ka of hypochlorous acid, HOCl, is 3.5 × 10–8.

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: Write a balanced equation and an expression for Ka

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

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

Step 2: Set-up an ICE table

 [HOCl] [H+] [OCl–] Initial $$\frac{0.0400}{0.50000} = 0.00800 mol \hspace{0.1cm} L^{-1}$$ 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 small 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. after decimal place)

# Introduction to pKa

• Values of dissociation constant vary greatly over numerous magnitudes. As a result, pKa is used to measure the variety on a logarithmic scale. pKa range widely, can be as small as – 4 and as large as 50.

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

• A higher value of Ka corresponds to a lower pKa value. This means stronger acids have lower pKa values.
 Weak acid 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 acid 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).
• pKa is a more suitable indicator for acids’ strength 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.

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

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

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