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Solving Acid-Base Dissociation Problems (HL) (HL IB Chemistry)

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Solving Acid-Base Dissociation Problems

Ka, pKa, Kb and pKb

  • In reactions of weak acids and bases, we cannot make the same assumptions as for the ionisation of strong acids and bases
  • For a weak acid and its conjugate base, we can use the equation:

 Kw = Ka x Kb 

  • By finding the -log of these, we can use:

pKw = pKa + pKb 

  • Remember, to convert these terms you need to use:

pKa = -logKa                 Ka= 10–pKa

pKb = -logKb                 Kb= 10–pKb

  • The assumptions we must make when calculating values for Ka, pKa, Kb and pKb are:
    • The initial concentration of acid ≈ the equilibrium concentration of acid
    • [A-] = [H+]
    • There is negligible ionisation of the water, so [H+] is not affected
    • The temperature is 298 K

Worked example

Calculate the acid dissociation constant, Ka, at 298 K for a 0.20 mol dm-3 solution of propanoic acid with a pH of 4.88.

 

Answer:

  • Step 1: Calculate [H+] using
    • [H+] = 10-pH
      • [H+] = 10-4.88
      • [H+] = 1.3183 x 10-5
  • Step 2: Substitute values into Ka expression 
    • Ka = begin mathsize 14px style fraction numerator open square brackets straight H to the power of plus close square brackets squared over denominator open square brackets CH subscript 3 CH subscript 2 COOH close square brackets end fraction end style
      • Ka = fraction numerator open parentheses 1.3182 space cross times 10 to the power of negative 5 end exponent close parentheses squared over denominator 0.2 end fraction
      • Ka = 8.70 × 10-10

Worked example

A 0.035 mol dm-3 sample of methylamine (CH3NH2) has pKb value of 3.35 at 298 K. Calculate the pH of methylamine.

 

Answer:

  • Step 1: Calculate the value for Kb using
    • Kb = 10–pKb
      • Kb= 10-3.35
      • Kb = 4.4668 x 10-4
  • Step 2: Substitute values into Kb expression to calculate [OH-]
    • Kb = begin mathsize 14px style fraction numerator open square brackets OH to the power of minus close square brackets squared over denominator open square brackets CH subscript 2 NH subscript 2 close square brackets end fraction end style
      • 4.4668 x 10-4 = fraction numerator open square brackets OH to the power of minus close square brackets over denominator 0.035 end fraction
      • [OH–] = square root of left parenthesis 4.4668 space cross times space 10 to the power of negative 4 end exponent space cross times space 0.035 right parenthesis end root
      • [OH–] = 3.9540 x 10-3
  • Step 3: Calculate the pH
    • [H+] = fraction numerator K subscript straight w over denominator open square brackets OH to the power of minus close square brackets end fraction
      • [H+] = (1 x 10-14) ÷ 3.9539 x 10-3
      • [H+] = 2.5291 x 10-12
    • pH = -log [H+]
      • pH = -log 2.5291 x 10-12
    • pH = 11.60 to 2 decimal places 

   OR

  • Step 3: Calculate pOH and therefore pH 
    • pOH = -log [OH–]
      • pOH = -log 3.9540 x 10-3
      • pOH = 2.4029
    • pH = 14 - pOH
      • pH = 14 - 2.4030
      • pH = 11.60 to 2 decimal places 

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    Philippa

    Author: Philippa

    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.