The pH Scale, Strong Acids and Stong Bases

This lesson covers: 

1. What the pH scale measures
2. How to calculate pH from [H⁺] and vice versa
3. The ionic product of water (K_w)
4. Using K_w to find the pH of strong bases

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.

Worked example 1 - Calculating the pH of a HCl solution

Calculate the pH of a 0.005 mol dm^-3 HCl solution.

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.

Step 1: Equation

pH = -log₁₀([H⁺])

Step 2: Rearrange equation

[H⁺] = 10^-pH

Step 3: Substitution and correct evaluation

[H+]=10−1.60=2.51×10−2 mol dm−3

[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 3 - Calculating the pH of a H₂SO₄ solution

Calculate the pH of a 0.003 mol dm^-3 H₂SO₄ solution.

Step 1: Determine [H⁺]

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

This means [H+]=2×[H2​SO4​]=2×0.003=0.006 mol dm−3

Step 2: Equation

pH = -log₁₀([H⁺])

Step 3: Substitution and correct evaluation

pH = -log₁₀(0.006) = 2.22

The ionic product of water (K_w)

Water undergoes self-ionisation to a small extent:

H₂O_(l) ⇌ H⁺_(aq) + OH^-_(aq)

The equilibrium constant for this process is the ionic product of water, K_w:

K_w = [H⁺][OH^-]

At 25°C, Kw​=1.0×10−14 mol2 dm−6

In pure water, [H⁺] = [OH^-] due to the 1:1 dissociation ratio, so:

K_w = [H⁺]²

At 25°C, the concentrations of H⁺ and OH^- ions in pure water are equal and very low, at 1.0 × 10^-7 mol dm^-3, resulting in a neutral pH of 7.

Knowing K_w for pure water at a given temperature allows you to calculate [H⁺] and, subsequently, the pH.

Use K_w to find the pH of a strong base

Strong bases, like sodium hydroxide (NaOH) and potassium hydroxide (KOH), fully ionise in water:

NaOH_(aq) ➔ Na⁺_(aq) + OH^-_(aq)

KOH_(aq) ➔ K⁺_(aq) + OH^-_(aq)

They contribute one mole of OH^- per mole of base, so [OH^-] equals the base concentration.

To find the pH, use K_w to calculate [H⁺]:

K_w = [H⁺][OH^-]

Worked example 4 - Calculating the pH of a NaOH solution

Calculate the pH of a 0.025 mol dm^-3 NaOH solution at 25°C, given that K_w at this temperature is 1.0×10−14 mol2 dm−6.

Step 1: Determine [OH^-]

NaOH completely dissociates in water to release one OH^- ion per molecule, meaning [OH^-] = [NaOH] = 0.025 mol dm^-3

Step 2: Rearrange K_w equation

[H+]=[OH−]Kw​​

Step 3: Substitution and correct evaluation

[H+]=0.025(1.0×10−14)​=4.0×10−13 mol dm−3

Step 4: Calculate pH

pH =−log10​([H+])=−log10​(4.0×10−13)=12.40