The ionic product of water (Kw)

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, ${\text{K}}_{\text{w}}=1.0\times {10}^{-14}{\text{ mol}}^2{\text{ dm}}^{-6}$

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.

The relationship between Kw and pKw

The relationship between K_w and pK_w is given by the equation:

pK_w = $-$logK_w

At 298 K, when K_w = 1.00 x 10^-14 mol² dm^-6, pK_w = 14.00

Use Kw 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 1 - 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 $1.0\times {10}^{-14}{\text{ mol}}^2{\text{ dm}}^{-6}$. Give your answer to 2 decimal places.

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

$[{\text{H}}^{+}]=\frac{{\text{K}}_{\text{w}}}{[{\text{OH}}^{-}]}$

Step 3: Substitution and correct evaluation

$[{\text{H}}^{+}]=\frac{(1.0\times {10}^{-14})}{0.025}=4.0\times {10}^{-13}{\text{ mol dm}}^{-3}$

Step 4: Calculate pH

$\text{pH }=-{\text{log}}_{10}([{\text{H}}^{+}])=-{\text{log}}_{10}(4.0\times {10}^{-13})=12.40$