Buffer Capacity (College Board AP Chemistry)

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Buffer Capacity

  • Buffer capacity tells you how much acid or base can be added to the buffer without the pH changing
  • The greater the buffer capacity, the more acid or base that can be added before the pH starts to change
  • The buffer capacity depends on the concentrations of both the weak acid and its conjugate base (or weak base and its conjugate acid

  • A buffer is formed from ethanoic acid and sodium ethanoate
    • If acid, H+, is added the following reaction will occur

CH3COO (aq) + H+ (aq) rightwards harpoon over leftwards harpoon CH3COOH (aq) 

the equilibrium will shift to the right-hand side as H+ ions are 'mopped' up

    • If a base, OH, is added the following reaction will occur

CH3COOH (aq) + OH– (aq) rightwards harpoon over leftwards harpoon CH3COO (aq) + H2O (l)

the equilibrium will shift to the right-hand side as OH ions are 'mopped up'

  • Using the different concentrations of ethanoic acid and sodium ethanoate we can demonstrate the difference in buffer capacity (the pKa of ethanoic acid is 4.74)
  Buffer 1 - higher concentration Buffer 2 - lower concentration
 Initial pH of buffer solution

 [CH3COOH] = 0.80 M

 [CH3COO] = 0.40 M

 pH = pKa + log10 begin mathsize 14px style fraction numerator open square brackets CH subscript 3 COO to the power of minus close square brackets over denominator open square brackets CH subscript 3 COOH close square brackets end fraction end style

 pH = pKa + log10 fraction numerator 0.40 over denominator 0.80 end fraction= 5.04

 [CH3COOH] = 0.080 M

 [CH3COO] = 0.040 M

pH = pKa + log10 fraction numerator stretchy left square bracket CH subscript 3 COO to the power of minus stretchy right square bracket over denominator stretchy left square bracket CH subscript 3 COOH stretchy right square bracket end fraction= 5.04

 pH = pKa + log10 fraction numerator 0.04 over denominator 0.08 end fraction= 5.04

 pH on addition of 0.05 M NaOH

 [CH3COOH] = 0.80 M - 0.05 M = 0.75 M

 [CH3COO] = 0.40 M + 0.05 M = 0.45 M

 pH = pKa + log10 fraction numerator open square brackets CH subscript 3 COO to the power of minus close square brackets over denominator open square brackets CH subscript 3 COOH close square brackets end fraction

 pH = 4.74 + log10 fraction numerator 0.45 over denominator 0.75 end fraction= 5.56

  [CH3COOH] = 0.080 M - 0.05 M = 0.075 M

 [CH3COO] = 0.040 M + 0.05 M = 0.045 M

 pH = pKa + log10 fraction numerator stretchy left square bracket CH subscript 3 COO to the power of minus stretchy right square bracket over denominator stretchy left square bracket CH subscript 3 COOH stretchy right square bracket end fraction

 pH = 4.74 + log10 fraction numerator 0.045 over denominator 0.075 end fraction= 6.18

 

  • Buffer 1 has a higher capacity than buffer 2

Worked example

Two different buffer solutions are formed. Buffer 1 contains 50 mL of 0.100 M NaOH and 100 mL of 0.100 M HNO2 (aq) and buffer 2 contains 0.100 mol NaNO2 (s) and 100 mL of 1.00 M HNO2.

Which buffer is more resistant to changes in pH? Justify your answer.

Answer:

  • Buffer 2

Justification:

  • Buffer 2 contains a higher concentration of HNO2 AND NO2 to react with added H+ and OH ions.

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Philippa

Author: Philippa

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.