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Buffer Calculations (HL) (DP IB Chemistry): Revision Note

Philippa Platt

Written by: Philippa Platt

Reviewed by: Richard Boole

Updated on

Buffer calculations

  • The pH of a buffer depends on:

    • The pKa (or pKb) of the weak acid (or base)

    • The ratio of the concentration of acid to conjugate base, or base to conjugate acid

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

    • K subscript straight a space equals space fraction numerator open square brackets salt close square brackets open square brackets straight H to the power of plus close square brackets over denominator open square brackets acid close square brackets end fraction space which space can space be space rearranged space to space open square brackets straight H to the power of plus close square brackets space equals space K subscript straight a space fraction numerator open square brackets acid close square brackets over denominator open square brackets salt close square brackets end fraction

  • For acidic buffers (e.g. CH3COOH + CH3COO-)

pH space equals space straight p K subscript straight a space plus space log subscript 10 fraction numerator open square brackets salt close square brackets over denominator open square brackets acid close square brackets end fraction

  • For basic buffers (e.g. NH3 + NH4+)

 pOH space equals space straight p K subscript straight b plus log subscript 10 fraction numerator open square brackets salt close square brackets over denominator open square brackets base close square brackets end fraction

pH = 14 - pOH

  • This is known as the Hendersen-Hasselbalch equation

Calculating the pH of a buffer

  • The pKa or pKb values of its component acid or base

  • The ratio of initial concentrations of acid and salt used to prepare the buffer

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 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

K subscript straight a space equals space fraction numerator open square brackets CH subscript 3 COO to the power of minus close square brackets 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+]

open square brackets straight H to the power of plus close square brackets space equals italic space K subscript straight a space cross times space 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

open square brackets straight H to the power of plus close square brackets space equals space 1.74 space cross times 10 to the power of negative 5 end exponent cross times fraction numerator 0.305 over denominator 0.520 end fraction = 1.02 x 10-5 mol dm-3

  • Step 4: Calculate the pH

pH = - log [H+]   

pH = -log 1.02 x 10-5 = 4.99

Worked Example

A buffer solution has a pH of 4.76. It contains 0.100 mol of ethanoic acid (CH3COOH) and 0.100 mol of sodium ethanoate (CH3COONa) in 1.00 dm³ of solution.

Calculate the new pH after 0.010 mol of HCl is added to the buffer.

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

Answer:

Step 1: Write down the Henderson–Hasselbalch equation

pH space equals space straight p K subscript straight a space plus space log subscript 10 fraction numerator open square brackets salt close square brackets over denominator open square brackets acid close square brackets end fraction

Step 2: Account for acid added

  • H+ from HCl reacts with CH3COO⁻:

CH3COO⁻ + H+→CH3COOH

n(CH3COOH) = 0.100 + 0.010 = 0.110 mol

n(CH3COO⁻) = 0.100 − 0.010 = 0.090 mol

Step 3: Substitute into the equation

pH space equals space 4.76 space plus space log subscript 10 fraction numerator open square brackets 0.090 close square brackets over denominator open square brackets 0.110 close square brackets end fraction

pH = 4.76 + log10(0.818)

pH = 4.76 - 0.088

pH = 4.67

Factors that can influence buffers

Dilution

  • Diluting a buffer reduces the concentration of both the weak acid/base and its conjugate:

  • The ratio of fraction numerator open square brackets salt close square brackets over denominator open square brackets acid close square brackets end fractionstays almost the same

  • Ka and Kb are equilibrium constants so are not changed by dilution

  • Therefore, the pH changes only very slightly so the buffer still works

  • However, buffer capacity decreases, it can now resist less added acid or base before the pH changes significantly

    • This is because the concentrations are lower so there are fewer particles of buffer components available to neutralise added H+ or OH-

Temperature

  • A constant temperature must be maintained when using buffers as temperature will influence the pH of the solution

  • Temperature affects the values of Ka and Kb 

  • In medical procedures, temperature fluctuations should be avoided due to the effect on the buffers in the blood

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    Author
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    Philippa Platt

    Author: Philippa Platt

    Expertise: Chemistry Content Creator

    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

    Richard Boole

    Reviewer: Richard Boole

    Expertise: Chemistry Content Creator

    Richard has taught Chemistry for over 15 years as well as working as a science tutor, examiner, content creator and author. He wasn’t the greatest at exams and only discovered how to revise in his final year at university. That knowledge made him want to help students learn how to revise, challenge them to think about what they actually know and hopefully succeed; so here he is, happily, at SME.