4.5.9 Buffer Calculations

• The pH of a buffer solution can be calculated using:
  • The K_a of the weak acid
  • The equilibrium concentration of the weak acid and its conjugate base (salt)

• To determine the pH, the concentration of hydrogen ions is needed which can be found using the equilibrium expression

• To simplify the calculations, logarithms are used such that the expression becomes:

• Since -log₁₀ [H⁺] = pH, the expression can also be rewritten as:

• This is known as the Hendersen-Hasselbalch equation

Answer

Ethanoic acid is a weak acid that ionises as follows:

CH₃COOH (aq) ⇌ H⁺ (aq) + CH₃COO^- (aq)

Step 1: Write down the equilibrium expression to find K_a

Step 2: Rearrange the equation to find [H⁺]

Step 3: Substitute the values into the expression

   = 1.02 x 10^-5 mol dm^-3

Step 4: Calculate the pH

   pH = - log [H+]

   = -log 1.02 x 10^-5

   = 4.99

How to make a buffer solution with a required pH

• To make a buffer solution with a pH of less than 7, you need to use a mixture of a weak acid and its conjugate base
  • Conversely, you can make a buffer solution with a pH greater than 7 by using a mixture of a weak base and its conjugate acid
• Imagine we want to make a buffer solution with a pH of 5.00 at a temperature of 298K
• This would require a hydrogen ion concentration of:

[H+(aq)] = 1.00 x 10^-5 mol dm^-3 

• The hydrogen ion concentration of a buffer solution of a weak acid and its conjugate base is calculated using the formula:

[H⁺ (aq)] = K_a x 

• We will use ethanoic acid as our weak acid of choice, with a K_a value of 1.74 x 10^-5 mol dm^-3 
• Substituting our known values into the equation we get:

1.00 x 10^-5 = 1.74 x 10^-5 x 

• This gives a value for the ratio of the concentrations of acid and base needed in our buffer solution:

= 0.575

• Mixing an equal volume of ethanoic acid with a concentration of 0.575 mol dm^-3 and a sodium ethanoate solution of 1.00 mol dm^-3 would allow us to make this buffer solution
• This would give a solution with an acid concentration of 0.2875 mol dm^-3 and a salt concentration of 0.500 mol dm^-3

= 0.575