The \(\text{pOH}\) scale measures the alkalinity, or basicity, of an aqueous solution. This measurement is directly related to the concentration of hydroxide ions (\(\text{OH}^-\)) present in the water. The \(\text{pOH}\) value is derived mathematically using the negative logarithm of this ion concentration. It is useful when analyzing solutions containing bases, as the hydroxide ion is the defining chemical species for alkalinity.
Calculating pOH Directly from Hydroxide Concentration
The most direct method for determining the \(\text{pOH}\) of a solution is by measuring the concentration of hydroxide ions. This calculation uses the formula \(\text{pOH} = -\log[\text{OH}^-]\). The brackets \([\text{OH}^-]\) represent the concentration of the ion in moles per liter, or molarity (\(\text{M}\)).
For instance, consider a solution where the concentration of \(\text{OH}^-\) is \(1.0 \times 10^{-5} \text{ M}\). Applying the negative logarithm function to this value yields a result of \(5.0\). A solution with this concentration has a \(\text{pOH}\) of \(5.0\).
For another example, if a strong base solution yields an \([\text{OH}^-]\) of \(4.82 \times 10^{-5} \text{ M}\), the calculation is \(\text{pOH} = -\log(4.82 \times 10^{-5})\). Entering this into a scientific calculator gives a \(\text{pOH}\) value of \(4.32\). Lower \(\text{pOH}\) numbers correspond to higher basicity.
Calculating pOH from pH
If the \(\text{pH}\) of an aqueous solution is already known, the \(\text{pOH}\) can be calculated using a simple algebraic relationship. This method relies on the fact that the sum of \(\text{pH}\) and \(\text{pOH}\) is a constant value in water-based systems. At the standard temperature of \(25^{\circ}\text{C}\), this constant sum is \(14.00\).
The relationship can be expressed as the formula \(\text{pH} + \text{pOH} = 14.00\). To isolate the \(\text{pOH}\) value, the formula can be rearranged to \(\text{pOH} = 14.00 – \text{pH}\). This equation allows for a quick conversion between the two scales without needing to calculate ion concentrations.
For instance, consider a solution with a measured \(\text{pH}\) of \(8.5\). Applying the conversion formula, the \(\text{pOH}\) is calculated as \(14.00 – 8.5\), resulting in a \(\text{pOH}\) of \(5.5\). Similarly, a more acidic solution with a \(\text{pH}\) of \(4.0\) would have a \(\text{pOH}\) of \(10.0\).
This calculation illustrates the inverse nature of the two scales: as the \(\text{pH}\) value increases, the corresponding \(\text{pOH}\) value decreases. The constant sum of \(14.00\) ensures that a change in the concentration of one ion is always balanced by an opposite change in the other. This simple subtraction is often the quickest path to finding \(\text{pOH}\) when a \(\text{pH}\) meter reading is available.
Understanding the Constant: The Ion Product of Water
The constant relationship between \(\text{pH}\) and \(\text{pOH}\) originates from the autoionization of water molecules. In pure water, a small number of molecules spontaneously react with each other, resulting in the formation of hydronium ions (\(\text{H}^+\)) and hydroxide ions (\(\text{OH}^-\)). This chemical reaction is an equilibrium process.
The equilibrium constant for this specific reaction is called the Ion Product of Water, symbolized as \(K_w\). The \(K_w\) value represents the product of the molar concentrations of the two ions: \(K_w = [\text{H}^+][\text{OH}^-]\). At \(25^{\circ}\text{C}\), the concentration of both ions in pure water is \(1.0 \times 10^{-7} \text{ M}\), which gives \(K_w\) a standard value of \(1.0 \times 10^{-14}\).
The \(\text{pH} + \text{pOH} = 14\) rule is derived directly from this \(K_w\) value through a logarithmic transformation. Taking the negative logarithm of the \(K_w\) expression yields the equation \(-\log(K_w) = -\log([\text{H}^+]) + (-\log([\text{OH}^-]))\). This simplifies to \(\text{p}K_w = \text{pH} + \text{pOH}\). Since \(-\log(1.0 \times 10^{-14})\) equals \(14\), the sum of \(\text{pH}\) and \(\text{pOH}\) is \(14\) at \(25^{\circ}\text{C}\).
\(K_w\) is an equilibrium constant, and its value changes with temperature. Increasing the temperature causes more water molecules to ionize, increasing the concentration of both \(\text{H}^+\) and \(\text{OH}^-\), thus increasing the value of \(K_w\). For example, at \(60^{\circ}\text{C}\), the sum \(\text{pH} + \text{pOH}\) is approximately \(13.0\). While the \(\text{pH} + \text{pOH} = \text{p}K_w\) relationship is always true, the use of the number \(14\) is only accurate at \(25^{\circ}\text{C}\).