The pH scale measures [H+]

The pH scale is a logarithmic scale that measures the concentration of hydrogen ions (H⁺) in a solution.

It ranges from 0 to 14:

• pH < 7 indicates an acidic solution - This means the concentration of H⁺ ions is greater than the concentration of OH^- ions.
• pH = 7 is a neutral solution - This is because the concentration of H⁺ ions equals the concentration of OH^- ions.
• pH > 7 indicates an alkaline solution - Here, the concentration of OH^- ions is greater than the concentration of H⁺ ions.

To calculate the pH of a solution from its hydrogen ion concentration, use the equation:

pH = $-$log₁₀[H⁺]

To find the hydrogen ion concentration from the pH, use the inverse equation:

[H⁺] = 10^-pH

These equations show that for each unit decrease in pH, the H⁺ ion concentration increases by a factor of 10. Conversely, [H⁺] decreases by a factor of 10 for each unit increase in pH.

[H+] equals [acid] for strong monoprotic acids

Strong acids like hydrochloric acid (HCl) and nitric acid (HNO₃) completely ionise in solution:

HCl_(aq) ➔ H⁺_(aq) + Cl^-_(aq)

HNO_3(aq) ➔ H⁺_(aq) + NO₃^-_(aq)

As they are monoprotic, meaning each acid molecule donates one H⁺ ion, the concentration of H⁺ equals the concentration of the acid.

[H+] equals 2[acid] for strong diprotic acids

For strong diprotic acids like sulfuric acid (H₂SO₄), each acid molecule donates two H⁺ ions:

H₂SO_4(aq) + H₂O_(l) ➔ 2H⁺_(aq) + SO₄^2-_(aq)

So, [H⁺] is double the acid concentration.

Worked example 1 - Calculating the pH of a HCl solution

Calculate the pH of a 0.005 mol dm^-3 HCl solution. Give your answer to 2 decimal places.

Step 1: Determine [H⁺]

HCl is a strong monoprotic acid that completely dissociates in water, meaning [H⁺] = [HCl] = 0.005 mol dm^-3

Step 2: Equation

pH = $-$log₁₀([H⁺])

Step 3: Substitution and correct evaluation

pH = $-$log₁₀(0.005) = 2.30

Worked example 2 - Calculating [H+] of a HCl solution

Calculate the hydrogen ion concentration of a HCl solution with a pH of 1.60. Give your answer to 3 significant figures.

Step 1: Equation

pH = $-$log₁₀([H⁺])

Step 2: Rearrange equation

[H⁺] = 10^-pH

Step 3: Substitution and correct evaluation

$[{\text{H}}^{+}]={10}^{-1.60}=2.51\times {10}^{-2}{\text{ mol dm}}^{-3}$

Worked example 3 - Calculating the pH of a H2SO4 solution

Calculate the pH of a 0.003 mol dm^-3 H₂SO₄ solution. Give your answer to 2 decimal places.

Step 1: Determine [H⁺]

H₂SO₄ is a strong diprotic acid that completely dissociates in water to release two H⁺ ions per molecule.

This means $[{\text{H}}^{+}]=2\times [{\text{H}}_2{\text{SO}}_4]=2\times 0.003=0.006{\text{ mol dm}}^{-3}$

Step 2: Equation

pH = $-$log₁₀([H⁺])

Step 3: Substitution and correct evaluation

pH = $-$log₁₀(0.006) = 2.22

Ka is the acid dissociation constant

Weak acids, such as ethanoic acid (CH₃COOH), only partially dissociate in aqueous solution. This means the concentration of H⁺ ions is less than the initial concentration of the acid.

To measure how much a weak acid dissociates, we use the acid dissociation constant, K_a.

For a generic weak acid HA, the dissociation equilibrium is represented as:

HA_(aq) ⇌ H⁺_(aq) + A^-_(aq)

The formula for K_a based on this equilibrium is:

${\text{K}}_{\text{a }}=\frac{[{\text{H}}^{+}][{\text{A}}^{-}]}{[\text{HA}]}$

Where:

• [HA] is the concentration of the acid that has not dissociated.
• [H⁺] is the concentration of hydrogen ions.
• [A^-] is the concentration of the conjugate base.

K_a has units of mol dm^-3.

The larger the K_a value, the stronger the weak acid.

Assumptions for weak acids

When calculating the pH of a weak acid (HA) based on its concentration and K_a value, we typically make two important assumptions:

1. [HA]_equilibrium ≈ [HA]_initial - This is because the ionisation of a weak acid is so small that the concentration of undissociated HA molecules present at equilibrium is approximately the same as the initial concentration of the acid.
2. [H⁺]_equilibrium ≈ [A^-]_equilibrium - This is because the ionisation of water is negligible, so the concentration of H⁺ ions produced by the ionisation of water molecules present in the solution is ignored.

These assumptions simplify the K_a formula to:

${\text{K}}_{\text{a }}=\frac{{[{\text{H}}^{+}]}^2}{[\text{HA}]}$

These assumptions are valid only for weak acids because stronger acids dissociate more, significantly affecting the initial and equilibrium concentrations of HA.

Calculating pH of weak acids using K_a

The K_a value for a weak acid is constant at a specific temperature and does not depend on the concentration. This property allows us to calculate the pH of a weak acid solution if we know the K_a value and the initial concentration of the acid.

Worked example 4 - Calculating the pH of a weak acid solution

Calculate the pH of a 0.0100 mol dm^-3 solution of ethanoic acid, given that its K_a is $1.75\times {10}^{-5}$ mol dm^-3 at 298 K. Give your answer to 2 decimal places.

Step 1: K_a equation

${\text{K}}_{\text{a }}=\frac{{[{\text{H}}^{+}]}^2}{[{\text{CH}}_3\text{COOH}]}$

Step 2: Rearrange K_a equation

$[{\text{H}}^{+}]=\sqrt{{\text{ K}}_{\text{a }}\times [{\text{CH}}_3\text{COOH}]}$

Step 3: Substitution and correct evaluation

$[{\text{H}}^{+}]=\sqrt{(1.75\times {10}^{-5})\times 0.0100}=4.18\times {10}^{-4}{\text{ mol dm}}^{-3}$

Step 4: Calculate pH

$\text{pH }=-{\text{log}}_{10}[{\text{H}}^{+}]=-{\text{log}}_{10}(4.18\times {10}^{-4})=3.38$

Thus, the pH of the 0.01 mol dm^-3 ethanoic acid solution is 3.38.

Determining acid concentration or K_a from pH

We can use the same principles to find either the starting concentration of a weak acid or its K_a value if the pH is given.

Worked example 5 - Calculating the concentration of propanoic acid from pH

Given a propanoic acid solution's pH is 2.89 and its K_a is $1.34\times {10}^{-5}$ mol dm^-3 at 298 K, calculate the acid's concentration. Give your answer to 3 significant figures.

Step 1: Calculate [H⁺]

[H⁺] = 10^-pH = 10^-2.89 = 1.29 x 10^-3 mol dm^-3

Step 2: K_a equation

${\text{K}}_{\text{a }}=\frac{{[{\text{H}}^{+}]}^2}{[{\text{CH}}_3{\text{CH}}_2\text{COOH}]}$

Step 3: Rearrange K_a equation

$[{\text{CH}}_3{\text{CH}}_2\text{COOH}]=\frac{{[{\text{H}}^{+}]}^2}{{\text{K}}_{\text{a}}}$

Step 4: Substitution and correct evaluation

$[{\text{CH}}_3{\text{CH}}_2\text{COOH}]=\frac{{(1.29\times {10}^{-3})}^2}{(1.34\times {10}^{-5})}=0.124{\text{ mol dm}}^{-3}$

Hence, the concentration of the propanoic acid solution is 0.124 mol dm^-3.

The relationship between pKa and Ka

pK_a offers another way to express the acid dissociation constant, defined as:

pK_a = $-$log₁₀(K_a)

Conversely, we can find K_a from pK_a through:

K_a = 10^-pKa

The smaller the pK_a value, the stronger the weak acid.

Worked example 6 - Calculating the pH of a benzoic acid solution

Calculate the pH of a 0.0500 mol dm^-3 solution of benzoic acid, given its pK_a is 4.20. Give your answer to 2 decimal places.

Step 1: Calculate K_a

K_a = 10^-pKa = 10^-4.20 = 6.31 x 10^-5 mol dm^-3

Step 2: K_a equation

${\text{K}}_{\text{a }}=\frac{{[{\text{H}}^{+}]}^2}{[{\text{C}}_6{\text{H}}_5\text{COOH}]}$

Step 3: Rearrange K_a equation

$[{\text{H}}^{+}]=\sqrt{{\text{ K}}_{\text{a }}\times [{\text{C}}_6{\text{H}}_5\text{COOH}]}$

Step 4: Substitution and correct evaluation

$[{\text{H}}^{+}]=\sqrt{(6.31\times {10}^{-5})\times 0.0500}=1.78\times {10}^{-3}{\text{ mol dm}}^{-3}$

Step 5: Calculate pH

$\text{pH }=-{\text{log}}_{10}[{\text{H}}^{+}]=-{\text{log}}_{10}(1.78\times {10}^{-3})=2.75$

Therefore, the pH of the 0.0500 mol dm^-3 benzoic acid solution is 2.75.