# 5.6.1 Acid Dissociation Constant

### Acid Dissociation Constant

#### pH

• The pH indicates the acidity or basicity of an acid or alkali
• The pH scale goes from 0 to 14
• Acids have pH between 0-7
• Pure water is neutral and has a pH of 7
• Bases and alkalis have pH between 7-14
• The pH can be calculated using: pH = -log10 [H+]

where [H+] = concentration of H+ ions (mol dm-3)

• The pH can also be used to calculate the concentration of H+ ions in solution by rearranging the equation to:

[H+] = 10-pH

#### Calculating the pH of acids

Calculate the pH of ethanoic acid at 298 K when the hydrogen concentration is 1.32 x 10-3 mol dm-3

pH = -log [H+]

= -log 1.32 x 10-3

= 2.9

#### Ka & pKa

• The Ka is the acidic dissociation constant
• It is the equilibrium constant for the dissociation of a weak acid at 298 K
• For the partial ionisation of a weak acid HA the equilibrium expression to find Ka is as follows:

HA (aq) ⇌ H+ (aq) + A (aq) • When writing the equilibrium expression for weak acids, the following assumptions are made:
• The concentration of hydrogen ions due to the ionisation of water is negligible
• The dissociation of the weak acid is so small that the concentration of HA is approximately the same as the concentration of A
• The value of Ka indicates the extent of dissociation
• A high value of Ka means that:
• The equilibrium position lies to the right
• The acid is almost completely ionised
• The acid is strongly acidic
• A low value of Ka means that:
• The equilibrium position lies to the left
• The acid is only slightly ionised (there are mainly HA and only a few H+ and A ions)
• The acid is weakly acidic
• Since Ka values of many weak acids are very low, pKa values are used instead to compare the strengths of weak acids with each other

pKa = -log10 Ka

• The less positive the pKa value the more acidic the acid is

#### Calculating the Ka & pKa of weak acids

Calculate the Ka & pKa values of 0.100 mol dm-3 ethanoic acid at 298 K which forms 1.32 x 10-3 mol dm-3 of H+ ions in solution.

• Step 1: Write down the equation for the partial dissociation of ethanoic acid

CH3COOH (aq) ⇌ H+ (aq) + CH3COO (aq)

• Step 2: Write down the equilibrium expression to find Ka • Step 3: Simplify the expression

The ratio of H+ to CH3COO is 1:1

The concentration of H+ and CH3COO is, therefore, the same

The equilibrium expression can be simplified to: #### Kw

• The Kw is the ionic product of water
• It is the equilibrium constant for the dissociation of water at 298 K
• Its value is 1.00 x 10-14 mol2 dm-6
• For the ionisation of water the equilibrium expression to find Kw is as follows:

H2O (l) ⇌ H+ (aq) + OH (aq) • As the extent of ionisation is very low, only small amounts of H+ and  OHions are formed
• The concentration of H2O can therefore be regarded as constant and removed from the Kw expression
• The equilibrium expression therefore becomes:

Kw = [H+] [OH]

• As the [H+] = [OH+] in pure water, the equilibrium expression can be further simplified to:

Kw = [H+]2

#### Calculating the concentration of H+ of pure water

Calculate the concentration of H+ in pure water using the ionic product of water.

• Step 1: Write down the equation for the partial dissociation of water

In pure water, the following equilibrium exists:

H2O (l) ⇌ H+ (aq) + OH (aq)

• Step 2: Write down the equilibrium expression to find Kw • Step 3: Simplify the expression

Since the concentration of H2O is constant, this expression can be simplified to:

Kw = [H+] [OH]

• Step 4: Further simplify the expression

The ratio of H+ to OH is 1:1

The concentration of H+ and OH is, therefore, the same and the equilibrium expression can be further simplified to:

Kw = [H+]2

• Step 5: Rearrange the equation to find [H+]

Kw = [H+]2

• Step 6: Substitute the values into the expression to find Kw = 1.00 x 10-7 mol dm-3

#### Exam Tip

Remember:

The greater the Ka value, the more strongly acidic the acid is.

The greater the pKa value, the less strongly acidic the acid is.

Also, you should be able to rearrange the following expressions:

pH = -log10 [H+] TO [H+] = 10-pH pKa = – log10 Ka TO Ka = 10-pKa

### pH of Weak Acids

pH of Weak Acids
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