A LevelChemistryCIERevision Notes25. Equilibria (AL)25.1 Acids & BasespH & [H+] Calculations

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pH & [H+] Calculations (CIE A Level Chemistry)

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[H+] & pH Calculations

  • If the concentration of H+ of an acid or alkali is known, the pH can be calculated using the equation:

pH = -log [H+]

  • Similarly, the concentration of H+ of a solution can be calculated if the pH is known by rearranging the above equation to:

[H+] = 10-pH

Strong acids

  • Strong acids are completely ionised in solution

HA (aq) → H+ (aq) + A- (aq)

  • Therefore, the concentration of hydrogen ions ([H+]) is equal to the concentration of acid ([HA])
  • The number of hydrogen ions ([H+]) formed from the ionisation of water is very small relative to the [H+] due to the ionisation of the strong acid and can therefore be neglected
  • The total [H+] is therefore the same as the [HA]

Worked example

pH calculations of a strong acid

For a solution of hydrochloric acid, calculate the following:

  1. The pH when the hydrogen ion concentration is 1.6 x 10-4 mol dm-3.
  2. The hydrogen ion concentration when the pH is 3.1.

Answer

  • Hydrochloric acid is a strong monobasic acid

HCl (aq) → H+ (aq) + Cl- (aq)

Answer 1

  • The pH of the solution is:
    • pH = -log [H+]
    • pH = -log 1.6 x 10-4
    • pH = 3.80

Answer 2

  • The hydrogen concentration can be calculated by rearranging the equation for pH:
    • pH = -log [H+]
    • [H+] = 10-pH
    • [H+] = 10-3.1
    • [H+] = 7.9 x 10-4 mol dm-3

Strong alkalis

  • Strong alkalis are completely ionised in solution

BOH (aq) → B+ (aq) + OH- (aq)

  • Therefore, the concentration of hydroxide ions ([OH-]) is equal to the concentration of base ([BOH])
    • Even strong alkalis have small amounts of H+ in solution which is due to the ionisation of water
  • The concentration of OH- in solution can be used to calculate the pH using the ionic product of water

Kw = [H+] [OH-]

[H+] = fraction numerator bold K subscript bold w over denominator stretchy left square bracket OH to the power of minus stretchy right square bracket end fraction

  • Since Kw is 1.00 x 10-14 mol2 dm-6

[H+] = fraction numerator bold 1 bold. bold 00 bold cross times bold 10 to the power of bold minus bold 14 end exponent over denominator stretchy left square bracket OH to the power of minus stretchy right square bracket end fraction

  • Once the [H+] has been determined, the pH of the strong alkali can be found using pH = -log[H+]
  • Similarly, the ionic product of water can be used to find the concentration of OH- ions in solution if [H+] is known

[OH–] = fraction numerator bold 1 bold. bold 00 bold cross times bold 10 to the power of bold minus bold 14 end exponent over denominator stretchy left square bracket H to the power of plus stretchy right square bracket end fraction

Worked example

pH calculations of a strong alkali

For a solution of sodium hydroxide, calculate the following:

  1. The pH when the hydrogen ion concentration is 3.5 x 10-11 mol dm-3.
  2. The hydroxide ion concentration when the pH is 12.3.

Answer

  • Sodium hydroxide is a strong base which ionises as follows:

NaOH (aq) → Na+ (aq) + OH- (aq) 

Answer 1

  • The pH of the solution is:
    • pH = -log [H+]
    • pH = -log 3.5 x 10-11
    • pH = 10.5

Answer 2

  • Step 1: Calculate the hydrogen concentration by rearranging the equation for pH:
    • pH = -log [H+]
    • [H+] = 10-pH
    • [H+] = 10-12.3
    • [H+] = 5.01 x 10-13 mol dm-3
  • Step 2: Rearrange the ionic product of water  to find the concentration of hydroxide ions:
    • Kw = [H+] [OH-]
    • [OH–] = begin mathsize 14px style fraction numerator K subscript straight w over denominator open square brackets straight H to the power of plus close square brackets end fraction end style
  • Step 3: Substitute the values into the expression to find the concentration of hydroxide ions:
    • Since Kw is 1.00 x 10-14 mol2 dm-6
      • [OH–] = fraction numerator 1.00 cross times 10 to the power of negative 14 end exponent over denominator 5.01 cross times 10 to the power of negative 13 end exponent end fraction
      • [OH–] = 0.0199 mol dm-3

Weak acids

  • The pH of weak acids can be calculated when the following is known:
    • The concentration of the acid
    • The Ka value of the acid

Worked example

pH calculations of weak acids

Calculate the pH of 0.100 mol dm-3 ethanoic acid at 298K with a Ka value of 1.74 x 10-5 mol dm-3.

Answer

  • Ethanoic acid is a weak acid which 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 straight H plus close square brackets space 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
  • Step 2: Simplify the expression:
    • The ratio of H+ to CH3COO- ions is 1:1
    • Therefore the concentration of H+ and CH3COO- ions are the same
    • So, the expression can be simplified to:
    • Ka = fraction numerator open square brackets straight H plus close square brackets squared over denominator open square brackets CH subscript 3 COOH close square brackets end fraction
  • Step 3: Rearrange the expression to find [H+]:
    • [H+] = square root of K subscript straight a cross times open square brackets CH subscript 3 COOH close square brackets end root
  • Step 4: Substitute the values into the expression to find [H+]:
    • [H+] = square root of open parentheses 1.74 cross times 10 to the power of negative 5 end exponent close parentheses cross times 0.100 end root
    • [H+] = 1.32 x 10-3 mol dm-3
  • Step 5: Find the pH:
    • pH = -log10 [H+]
    • pH = -log10 1.32 x 10-3
    • pH = 2.88

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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.