pH Calculations of Acids (Edexcel IAL Chemistry)

Acids - pH Calculations

Spec point — 14.7

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^-⇌ SO₄^2-+ 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 K_a value of the acid From the K_aexpression we can see that there are three variables:

• 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

K_ax [HA] = [H⁺]²

[H⁺]² = K_a x [HA]

• Taking the square roots of each side

[H⁺] = √(K_a x [HA])

• Then take the negative logs

pH = -log[H⁺] = -log√(K_a x [HA])

Worked Example pH calculations of weak acids

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

K_a = 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 K_a* *

K_a = $\frac{[ \mathrm{H} ^{ + } ] [ CH_{3} COO^{ - } ]}{[ CH_{3} COOH ]}$

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:

K_a = $\frac{[ \mathrm{H} ^{ + } ] ^{2} }{[ CH_{3} COOH ]}$

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

[H⁺] = $\sqrt{K_{a} \times [ CH_{3} COOH ] }$

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

[H⁺] = $\sqrt{ ( 1.74 \times 10^{ - } 5 ) \times [ 0.100 ] }$ = 1.32 x 10^-3mol dm^-3

**Step 5: **Find the pH

pH = -log[H⁺]

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

Exam Hint 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.