Buffer Calculations (Edexcel International A Level Chemistry)

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Buffer Calculations

  • The pH of a buffer solution can be calculated using:
    • The Ka 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

Calculating pH of Buffer Solutions equation 1

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

Calculating pH of Buffer Solutions equation 2

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

  • This is known as the Hendersen-Hasselbalch equation

Worked example

Calculate the pH of a buffer solution containing 0.305 mol dm-3 of ethanoic acid and 0.520 mol dm-3 sodium ethanoate.

The Ka of ethanoic acid  = 1.74 × 10-5 mol dm-3 at 298 K

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 Ka

Calculating pH of Buffer Solutions equation 4

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

Calculating pH of Buffer Solutions equation 5

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)] = Kafraction numerator left square bracket acid right square bracket over denominator left square bracket base right square bracket end fraction

  • We will use ethanoic acid as our weak acid of choice, with a Ka 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 10-5fraction numerator left square bracket acid right square bracket over denominator left square bracket base right square bracket end fraction

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

fraction numerator left square bracket acid right square bracket over denominator left square bracket base right square bracket end fraction= 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

fraction numerator 0.2875 over denominator 0.500 end fraction= 0.575

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Sonny

Author: Sonny

Expertise: Chemistry

Sonny graduated from Imperial College London with a first-class degree in Biomedical Engineering. Turning from engineering to education, he has now been a science tutor working in the UK for several years. Sonny enjoys sharing his passion for science and producing engaging educational materials that help students reach their goals.