When dealing with range of concentrations (such as these of H^{+} (aq) ions) that spans many powers of ten, it is convenient to represent them on a more compressed logarithmic scale. By convention, we use the pH scale for denoting the **concentration of hydrogen ions**.

pH notation was devised by the Danish biochemist Soren Sorensen in 1909. The term pH means "power of hydrogen". The pH is the logarithm of the reciprocal of the hydrogen ion concentration.

pH = log 1/[H^{+}]

Alternately, the pH is the negative logarithm of the hydrogen ion concentration.

pH = – log [H^{+}]

Because of the negative sign in the expression, if [H^{+}] increases, pH would decrease and if it decreases, pH would increase.

In pure water at 25° (298 K)

[H+] = 1.0 × 10^{–7} mol L^{–1}

log [H^{+}] = log (10^{–7}) = – 7

pH = – log [H^{+}] = –(–7)

pH = 7

Since in pure water at 25°C (298 K)

[OH^{–}] = 1.0 × 10^{–7} mol L^{–1}

Also, pOH = 7

Since, K_{w} = 1.0 × 10^{–14}

pK_{w} = 14

The relationship between pK_{w}, pH and pOH is

pK_{w} = pH + pOH

At 25°C (298 K)

14 = pH + pOH

**Example 1: Calculate the pH of 0.001 molar solution of HCl.**

HCl is a strong acid and is completely dissociated in its solutions according to the process:

HCl (aq) → H^{+} (aq) + Cl^{–} (aq)

From this process, one mole of HCl would give one mole of H^{+} ions. Therefore, the concentration of H^{+} ions would be equal to that of HCl i.e. 0.001 molar or 1.0 × 10^{–3} mol L^{–1}.

Thus, [H^{+}] = 1 × 10^{–3} mol L^{–1}

pH = – log [H^{+}] = – (log 10^{–3})

= – (– 3 × log 10) = – (3 × 1) = 3

Thus, pH = 3

**Example 2: What would be the pH of an aqueous solution of sulphuric acid which is 5 × 10 ^{–5} mol L^{–1} in concentration.**

Sulphuric acid dissociates in water as:

H_{2}SO_{4} (aq) → 2H^{+} (aq) + SO_{4}^{2–} (aq)

Each mole of sulphuric acid gives two mole of H^{+} ions in the solution. One litre of 5 × 10^{–5} mol L^{–1} solution contains 5 × 10^{–5} moles of H_{2}SO_{4} which would give 2 × 5 × 10^{–5} = 10 × 10^{–5} or 1.0 × 10^{–4} moles of H^{+} ion in one litre solution.

Therefore,

[H^{+}] = 1.0 × 10^{–4} mol L^{–1}

pH = – log[H^{+}] = – log 10^{–4} = – (– 4 × log 10)

= – (– 4 × 1) = 4

**Example 3: Calculate the pH of 1 × 10 ^{–4} molar solution of NaOH.**

NaOH is a strong base and dissociate in its solution as:

NaOH (aq) → Na^{+} (aq) + OH^{–} (aq)

One mole of NaOH would give one mole of OH^{–} ions. Therefore,

[OH^{–}] = 1 × 10^{–4} mol L^{–1}

pOH = – log [OH^{–}] = – log × 10^{–4} = – (– 4)

= 4

Since pH + pOH = 14

pH = 14 – pOH = 14 – 4 = 10

**Example 4: Calculate the pH of a solution in which the concentration of hydrogen ions is 1.0 × 10 ^{–8} mol L^{–1}.**

Here, although the solution is extremely dilute, the concentration given is not of acid or base but that of H^{+} ions. Hence, the pH can be calculated from the relation:

pH = – log [H^{+}]

Given [H^{+}] = 1.0 × 10^{–8} mol L^{–1}

Therefore, pH = – log 10^{–8} = – (– 8 × log 10)

= – (– 8 × 1) = 8