The pOH scale represents solution basicity

The pOH scale is a way to express the concentration of hydroxide ions (OH^-) in aqueous solutions. It allows us to quantify the basicity of solutions over a wide range of hydroxide concentrations.

To calculate the pOH of a solution from its hydroxide ion concentration, use the equation:

$pOH = -log[OH^{-}]$

To find the hydroxide ion concentration from the pOH, use the inverse equation:

$[OH^{-}] = 10^{-pOH}$

The pOH scale, like the pH scale, is logarithmic. This means each decrease of one pOH unit represents a tenfold increase in [OH^-]. For example, a solution with a pOH of 2 has ten times more hydroxide ions than a solution with a pOH of 3. Similarly, a solution with a pOH of 4 has one-tenth the concentration of hydroxide ions compared to a solution with a pOH of 3.

The relationship between pOH and basicity:

• Solutions with a high [OH^-] (strongly basic) have a low pOH value.
• Solutions with a low [OH^-] (weakly basic or acidic) have a high pOH value.

How pOH relates to pH

In any aqueous solution at 25°C, the product of [H⁺] and [OH^-] always equals a constant value known as the ionic product for water (K_w):

[H⁺][OH^-] = K_w = 1.00 × 10^-14 mol² dm^-6

We can convert this equation to the pOH scale:

pOH + pH = pK_w = 14

This relationship allows us to easily calculate pOH from pH or vice versa:

$pOH = 14 - pH$ or $pH = 14 - pOH$

This means that if you know the pH of a solution, you can easily determine its pOH, and if you know the pOH, you can calculate the pH.

Calculating pOH for aqueous solutions

To calculate the pOH of an aqueous solution, follow these steps:

1. Write the dissociation equation for the compound.
2. Determine the concentration of OH^- ions from the dissociation equation and the given concentration of the compound.
3. Use the expression $pOH = -log[OH^{-}]$ to calculate the pOH value.

Alternatively, you can use the relationship pH + pOH = 14 if the pH is known or easier to calculate.

Worked example 1 - Calculating pOH

Calculate the pOH of a 0.030 mol dm^-3 solution of sodium hydroxide, NaOH. Give your answer to 1 decimal place.

Step 1: Dissociation equation for NaOH

NaOH_(aq) ➔ Na⁺_(aq) + OH^-_(aq)

Step 2: Determine [OH^-]

NaOH completely dissociates in water to release one OH^- ion per molecule, meaning [OH^-] = [NaOH] = 0.030 mol dm^-3

Step 3: Equation

$pOH = -log[OH^{-}]$

Step 4: Substitution and correct evaluation

pOH = $-$log(0.030) = 1.5

Worked example 2 - Calculating pOH

Calculate the pOH of a 0.0400 mol dm^-3 solution of nitric acid, HNO₃ at 25°C. Give your answer to 1 decimal place.

K_w = 1.00 x 10^-14 mol² dm^-6 at 25°C.

Step 1: Dissociation equation for HNO₃

HNO_3(aq) ➔ H⁺_(aq) + NO₃^-_(aq)

Step 2: Determine [H⁺]

HNO₃ completely dissociates in water to release one H⁺ ion per molecule, meaning [H⁺] = [HNO₃] = 0.0400 mol dm^-3

Step 3: Calculate [OH^-]

$[OH^{-}]=\frac{K_{w}}{[H^{+}]}=\frac{(1.00\times10^{-14})}{0.0400}=2.50\times10^{-13}\text{ mol dm}^{-3}$

Step 4: Equation

$pOH = -log[OH^{-}]$

Step 5: Substitution and correct evaluation

pOH = $-$log(2.50 x 10^-13) = 12.6

Alternatively, using pH and pOH relationship:

Step 3: Calculate pH

$pH = -log[H^{+}] = -log(0.0400) = 1.40$

Step 4: Equation

$pOH = 14 - pH$

Step 5: Substitution and correct evaluation

$pOH = 14 - 1.40 = 12.6$

Both methods give the same pOH value of 12.6.