pH, Ka, pKa & Kw Calculations (CIE A Level Chemistry)

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

Worked Example: Calculating the pH of acids

Answer

pH = -log [H⁺]

= -log 1.32 x 10^-3

= 2.9

K_a & pK_a

• The K_a 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 K_a 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 K_a indicates the extent of dissociation
  • A high value of K_a means that:
    • The equilibrium position lies to the right
    • The acid is almost completely ionised
    • The acid is strongly acidic

  • A low value of K_a 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 K_a values of many weak acids are very low, pK_a values are used instead to compare the strengths of weak acids with each other

pK_a = -log₁₀ K_a

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

Worked Example: Calculating the K_a & pK_a of weak acids

Answer

• 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 K_a

• 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:

• Step 4: Substitute the values into the expression to find K_a

= 1.74 x 10^-5

• Step 5: Determine the units of K_a

= mol dm^-3

The value of K_a is therefore 1.74 x 10^-5 mol dm^-3

• Step 6: Find pK_a

pK_a = - log₁₀ K_a

= - log₁₀ (1.74 x 10^-5)

= 4.76

K_w

• The K_w 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 K_w is as follows:

H₂O (l) ⇌ H⁺ (aq) + OH^- (aq)

• As the extent of ionisation is very low, only small amounts of H⁺ and  OH^- ions are formed
• The concentration of H₂O can therefore be regarded as constant and removed from the K_w expression
• The equilibrium expression therefore becomes:

K_w = [H⁺] [OH^-]

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

K_w = [H⁺]²

Worked Example: Calculating the concentration of H⁺ of pure water

Answer

• 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 K_w

• Step 3: Simplify the expression

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

K_w = [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:

K_w = [H⁺]²

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

K_w = [H⁺]²

• Step 6: Substitute the values into the expression to find K_w

= 1.00 x 10^-7 mol dm^-3