Ionic Product of Water and pH

Learning Objectives (OCR Spec 5.1.3)

By the end of this lesson you should be able to:

• Define the ionic product of water Kw and state its value at 298 K
• Use Kw to relate [H+] and [OH-] in aqueous solutions
• Define pH mathematically and calculate pH from [H+]
• Calculate [H+] from pH (reverse logarithm)
• Apply the Kw expression to both acidic and alkaline solutions
• Explain how Kw varies with temperature (in outline)

Self-Ionisation of Water

Even pure water contains a tiny concentration of ions. Water is amphoteric, so one water molecule can donate a proton to another:

2H2O(l) <=> H3O+(aq) + OH-(aq)

Simplifying H3O+(aq) to H+(aq):

H2O(l) <=> H+(aq) + OH-(aq)

This equilibrium lies very far to the left - at 298 K only about 1 in every 500 million water molecules is ionised at any instant, but even that tiny fraction is enough to give pure water a measurable conductivity and a definite pH.

The Ionic Product of Water Kw

The equilibrium constant expression for the ionisation of water at constant temperature is:

Kc = [H+][OH-] / [H2O]

Because water is the solvent, [H2O] is essentially constant (~55.5 mol dm-3) - it barely changes as ions form. We absorb it into the constant and define a new constant, the ionic product of water:

Kw = [H+][OH-]

At 298 K (25 degC):

Kw = 1.00 x 10^-14 mol^2 dm^-6

The units follow directly from multiplying two concentrations: (mol dm-3)(mol dm-3) = mol^2 dm^-6.

pH of Pure Water at 298 K

In pure water the self-ionisation produces equal numbers of H+ and OH-. Let [H+] = [OH-] = x. Then:

x^2 = Kw = 1.00 x 10^-14 x = sqrt(1.00 x 10^-14) = 1.00 x 10^-7 mol dm-3

So in pure water [H+] = [OH-] = 1.00 x 10^-7 mol dm-3.

Definition of pH

pH is defined as:

pH = -log10[H+]

where [H+] is the hydrogen ion concentration in mol dm-3. The minus sign ensures that pH values for typical solutions are positive. The logarithm is base 10.

For pure water at 298 K:

pH = -log10(1.00 x 10^-7) = 7.00

A neutral solution is one in which [H+] = [OH-]. At 298 K this corresponds to pH = 7.00, but only at 298 K - at other temperatures the neutral pH is different because Kw changes with temperature.

Acidic, Neutral and Alkaline Solutions (at 298 K)

Classification	Condition	Example
Acidic	[H+] > 1.00 x 10^-7	pH < 7
Neutral	[H+] = [OH-] = 1.00 x 10^-7	pH = 7
Alkaline	[H+] < 1.00 x 10^-7	pH > 7

Because pH uses a logarithmic scale, each unit change in pH corresponds to a 10-fold change in [H+]. A solution of pH 3 has 100x the [H+] of a pH 5 solution.

Calculating pH from [H+]

Worked Example 1: A solution has [H+] = 2.50 x 10^-3 mol dm-3. Calculate its pH.

pH = -log10(2.50 x 10^-3) pH = -(log10 2.50 + log10 10^-3) pH = -(0.398 + (-3)) = -(-2.602) pH = 2.60 (2 dp)

Worked Example 2: A solution has [H+] = 4.80 x 10^-9 mol dm-3. Calculate its pH.

pH = -log10(4.80 x 10^-9) = 8.32 (2 dp)

The solution is alkaline because pH > 7.

Note on significant figures: OCR convention is to quote pH to 2 decimal places, because the number of decimal places in a log corresponds to the number of significant figures in the original value.

Calculating [H+] from pH (Reverse Logarithm)

Taking antilogs: [H+] = 10^-pH

Worked Example 3: A solution has pH = 4.72. Calculate [H+].

[H+] = 10^-4.72 = 1.91 x 10^-5 mol dm-3

Worked Example 4: A solution has pH = 10.40. Calculate [H+].

[H+] = 10^-10.40 = 3.98 x 10^-11 mol dm-3

Using Kw for Alkaline Solutions

For an alkaline solution, we often know [OH-] (from the base concentration) and want to find [H+] so we can calculate pH. We use:

[H+] = Kw / [OH-]

Worked Example 5: 0.0200 mol dm-3 NaOH. NaOH is a strong base, so it dissociates fully: [OH-] = 0.0200 mol dm-3.

[H+] = Kw / [OH-] = (1.00 x 10^-14) / (0.0200) = 5.00 x 10^-13 mol dm-3 pH = -log10(5.00 x 10^-13) = 12.30 (2 dp)

Worked Example 6: 1.25 x 10^-3 mol dm-3 KOH.

[H+] = (1.00 x 10^-14) / (1.25 x 10^-3) = 8.00 x 10^-12 mol dm-3 pH = -log10(8.00 x 10^-12) = 11.10 (2 dp)