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A LevelChemistryCIERevision Notes25. Equilibria (AL)25.1 Acids & BasesBuffer Calculations

Buffer Calculations (CIE A Level Chemistry)

Revision Note

Philippa Platt

Last updated

Buffer Solution pH 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:

Ka = fraction numerator stretchy left square bracket salt stretchy right square bracket bold space stretchy left square bracket H to the power of plus stretchy right square bracket over denominator stretchy left square bracket acid stretchy right square bracket end fraction

  • This can be rearranged to determine the concentration of hydrogen ions:

[H+] = bold italic K subscript bold a bold cross times fraction numerator stretchy left square bracket acid stretchy right square bracket over denominator stretchy left square bracket salt stretchy right square bracket end fraction

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

–log10 [H+] = bold minus bold log subscript bold 10 bold italic K subscript bold a bold cross times bold minus bold log subscript bold 10 fraction numerator stretchy left square bracket acid stretchy right square bracket over denominator stretchy left square bracket salt stretchy right square bracket end fraction

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

bold pH bold equals bold p bold italic K subscript bold italic a bold a bold plus bold log subscript bold 10 fraction numerator stretchy left square bracket salt stretchy right square bracket over denominator stretchy left square bracket acid stretchy right square bracket end fraction

Worked example

Calculating the pH of a buffer solution

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

The Ka of ethanoic acid = 1.43 x 10-5 mol dm-3.

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:
    • Ka = fraction numerator open square brackets CH subscript 3 COO to the power of minus close square brackets space open square brackets straight H to the power of plus close square brackets over denominator open square brackets CH subscript 3 COOH close square brackets end fraction
  • Step 2: Rearrange the equation to find [H+]:
    • [H+] = Ka x fraction numerator open square brackets CH subscript 3 COOH close square brackets over denominator open square brackets CH subscript 3 COO to the power of minus close square brackets end fraction
  • Step 3: Substitute the values into the expression:
    • [H+] = 1.43 x 10-5 x begin mathsize 14px style fraction numerator open square brackets 0.305 close square brackets over denominator open square brackets 0.520 close square brackets end fraction end style
    • [H+] = 8.39 x 10-6 mol dm-3
  • Step 4: Calculate the pH:
    • pH = –log10 [H+]
    • pH = –log 8.39 x 10-6
    • pH = 5.08

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    Philippa Platt

    Author: Philippa Platt

    Expertise: Chemistry

    Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener.