What Is pOH and How Is It Related to pH? (HL)

pOH measures hydroxide ion concentration on a logarithmic scale. At 298 K, it is related to pH by $\text{pH}+\text{pOH}=14.00$, so either value can be used to calculate the other.

The Relationship Between pOH and pH

The definitions are:

$$\text{pOH}=-\log_{10}[\text{OH}^-]$$ $$\text{pH}=-\log_{10}[\text{H}^+]$$

Because these scales are logarithmic, a decrease of one pOH unit represents a tenfold increase in hydroxide ion concentration. A solution with pOH 2 therefore has ten times the hydroxide ion concentration of a solution with pOH 3.

Water undergoes slight self-ionization. Its ion product constant, $K_{\text{w}}$, is defined as:

$$K_{\text{w}}=[\text{H}^+][\text{OH}^-]$$

Taking the negative logarithm of both sides gives:

$$\text{p}K_{\text{w}}=\text{pH}+\text{pOH}$$

At 298 K, $K_{\text{w}}=1.00\times10^{-14}$, so $\text{p}K_{\text{w}}=14.00$. Consequently, at this temperature.

This relationship connects the two concentration scales rather than defining acidity and basicity independently. When hydroxide ion concentration increases, pOH decreases; through the ion product, hydrogen ion concentration decreases and pH increases. For example, a pOH of 4.00 corresponds to a pH of 10.00 at 298 K. The solution is therefore basic because hydroxide ions are more concentrated than hydrogen ions. Always distinguish concentration from strength: pOH alone does not show whether a base is strong or weak.

For example, if $[\text{OH}^-]=2.5\times10^{-4}\text{ mol dm}^{-3}$:

$$\text{pOH}=-\log_{10}(2.5\times10^{-4})=3.60$$ $$\text{pH}=14.00-3.60=10.40$$

The solution is basic because its pH is greater than 7 at 298 K. The calculated pOH has two decimal places because the concentration has two significant figures.

A common misconception is that $\text{pH}+\text{pOH}=14$ at every temperature. The general relationship is $\text{pH}+\text{pOH}=\text{p}K_{\text{w}}$. It equals 14.00 only at 298 K because changes with temperature. Therefore, pH 7 is neutral only when corresponds to that temperature.

Exam Technique

This is Additional Higher Level content in Reactivity 3.1.9. For a “calculate” question, show the logarithmic substitution and each conversion step. Use $\text{mol dm}^{-3}$ for concentration, retain appropriate significant figures, and confirm that the temperature is 298 K before using 14.00.