International A LevelChemistryEdexcelRevision Notes4: Rates, Equilibria & Further Organic Chemistry4.5 Acid-base Equilibria4.5.4 pH Calculations of Acids

pH Calculations of Acids (Edexcel International A Level Chemistry)

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Acids - pH Calculations

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 formed from the ionisation of water is very small relative to the [H+] due to ionisation of the strong acid and can therefore be neglected
  • The total [H+] is therefore the same as the [HA]

Worked example

What is the pH of 0.01 mol dm-3 hydrochloric acid?

Answer

   [HCl] = [H+] = 0.01 mol dm-3

        pH = - log[H+]

        pH = - log[0.01] = 2.00

The pH of dibasic acids

  • Dibasic or diprotic acids have two replaceable protons and will react in a 1:2 ratio with bases
  • Sulfuric acid is an example

   H2SO4 (aq)  + 2NaOH (aq) → Na2SO4 (aq) + 2H2O (l)

  • You might think that being a strong acid it is fully ionised so the concentration of the hydrogen is double the concentration of the acid
  • This would mean that 0.1 mol dm-3 would be 0.2 mol dm-3 in [H+] and have a pH of 0.69
  • However, measurements of the pH of  0.1 mol dm-3 sulfuric acid show that it is actually about pH 0.98, which indicates it is not fully ionised
  • The ionisation of sulfuric acid occurs in two steps

H2SO4 → HSO4- + H+

HSO4- ⇌ SO42- + H+

  • Although the first step is thought to be fully ionised, the second step is suppressed by the abundance of hydrogen ions from the first step creating an equilibrium
  • The result is that the hydrogen ion concentration is less than double the acid concentration

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
  • From the Ka expression we can see that there are three variables:

The acid dissociation constant, downloadable AS & A Level Chemistry revision notes

  • However, the equilibrium concentration of [H+] and [A-] will be the same since one molecule of HA dissociates into one of each ion
  • This means you can simplify and re-arrange the expression to

Ka x [HA] = [H+]2

[H+]2 = Ka x [HA] 

  • Taking the square roots of each side

[H+] = √(Ka x [HA])

  • Then take the negative logs

pH = -log[H+] = -log√(Ka x [HA])

Worked example

pH calculations of weak acids

Calculate the pH of 0.100 mol dm-3 ethanoic acid at 298 K.

Ka = 1.74 × 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 left square bracket straight H to the power of plus right square bracket space left square bracket CH subscript 3 COO to the power of minus right square bracket over denominator left square bracket CH subscript 3 COOH right square bracket end fraction

   Step 2: Simplify the expression

    • The ratio of H+ to CH3COO- ions is 1:1
    • The concentration of H+ and CH3COO- ions are therefore the same
    • The expression can be simplified to:

Ka = fraction numerator left square bracket straight H to the power of plus right square bracket squared over denominator left square bracket CH subscript 3 COOH right square bracket end fraction

   Step 3: Rearrange the expression to find [H+]

[H+] = square root of K subscript a cross times left square bracket CH subscript 3 COOH right square bracket end root

   Step 4: Substitute the values into the expression to find [H+]

[H+] = square root of left parenthesis 1.74 cross times 10 to the power of negative 5 end exponent right parenthesis cross times left square bracket 0.100 right square bracket end root = 1.32 x 10-3 mol dm-3

   Step 5: Find the pH

pH = -log[H+]

pH = -log(1.32 x 10-3) = 2.88

Exam Tip

Assumptions 

Sometimes exam questions will ask what assumptions are made in calculating the pH of a weak acid. There are two being made in these calculations.

  1. We assume the concentration of the anion, [A-], is the same as [H+] when dissociation occurs, which is why we can simplify [H+][A-] to [H+]2
  2. We assume the concentration of [HA]eqm is the same as [HA]initial, since the degree of dissociation is very small. In other words we say the dissociation of [HA] is negligible or so small it can be ignored.

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