4.5.4 pH Calculations of Acids

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^-3sulfuric 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:

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

K_a x [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^-3ethanoic 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 = $\frac{[H^{+}] [{CH}_{3} {COO}^{-}]}{[{CH}_{3} COOH]}$

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 = $\frac{{[H^{+}]}^{2}}{[{CH}_{3} COOH]}$

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^-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.

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
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.

Edexcel International A Level Chemistry

Revision Notes

Acids - pH Calculations

Strong acids

Worked example

Weak acids

Worked example

Exam Tip

You've read 0 of your 0 free revision notes

Get unlimited access

Join the 100,000+ Students that ❤️ Save My Exams

Author: Sonny