The pOH Scale and Its Relationship with pH
- The pOH scale, like the pH scale, is logarithmic.
- However, instead of measuring the concentration of hydrogen ions $ [H^+] $, it measures the concentration of hydroxide ions $ [OH^-] $.
- The pOH of a solution is defined by the formula: $$\text{pOH} = -\log_{10}[\text{OH}^-]$$
- Here, $ [OH^-] $ represents the molar concentration of hydroxide ions.
Note
This formula allows us to express very small concentrations of $ [OH^-] $ as manageable numbers.
Example
A solution with $ [OH^-] = 1 \times 10^{-3} \, \text{mol dm}^{-3} $ has a pOH of:
$$\text{pOH} = -\log_{10}(1 \times 10^{-3}) = 3$$
Tip
- The pOH scale is inversely proportional to $ [OH^-] $.
- As $ [OH^-] $ increases, the pOH decreases.
The pH-pOH Relationship
- The pOH scale is closely linked to the pH scale through the ionic product of water ($ K_w $), which describes the equilibrium constant for water's self-ionization: $$\text{K}_w = [H^+][OH^-]$$
- At 298 K (25°C), $ K_w = 1.00 \times 10^{-14} \, \text{mol}^2 \, \text{dm}^{-6} $. Taking the negative logarithm of both sides gives: $$\text{pH} + \text{pOH} = 14 \quad \text{(at 298 K)}$$
Hint
This formula allows you to convert between pH and pOH:
- If you know the pH: $$\text{pOH} = 14 - \text{pH}$$
- If you know the pOH: $$\text{pH} = 14 - \text{pOH}$$
Example
- A solution has a pH of 11. What is its pOH?
- Using the relationship $ \text{pH} + \text{pOH} = 14 $: $$\text{pOH} = 14 - 11 = 3$$
- This low pOH indicates the solution is strongly basic.
Calculating $[OH^-]$ and $[H^+]$ from pOH and pH
- The pOH scale is particularly useful for basic solutions.
- Once you know the pOH, you can calculate $ [OH^-] $ using the formula: $$\text{[OH}^-] = 10^{-\text{pOH}}$$
- Similarly, you can calculate $ [H^+] $ using $ K_w = [H^+][OH^-] $, or equivalently: $$\text{[H}^+] = \frac{K_w}{[\text{OH}^-]}$$
Example question
A solution has a pOH of 4.5. Calculate $ [OH^-] $ and $ [H^+] $.
- Find $[OH^-]$
- Find $[H^+]$
Solution
- $\text{[OH}^-] = 10^{-\text{pOH}} = 10^{-4.5} \approx 3.16 \times 10^{-5} \, \text{mol dm}^{-3}$
- Using $ K_w = [H^+][OH^-] $: $$\text{[H}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1.00 \times 10^{-14}}{3.16 \times 10^{-5}} \approx 3.16 \times 10^{-10} \, \text{mol dm}^{-3}$$
Common Mistake
- Be careful with units!
- Always express concentrations in $ \text{mol dm}^{-3} $.
Using pOH and pH to Analyze Solutions
By combining pH and pOH, you can classify solutions:
- Acidic: $ \text{pH}< 7 $, $ \text{pOH} >7 $
- Neutral: $ \text{pH} = 7 $, $ \text{pOH} = 7 $
- Basic: $ \text{pH} > 7 $, $ \text{pOH}< 7 $
Example
- A solution has $ [OH^-] = 2.00 \times 10^{-3} \, \text{mol dm}^{-3} $.
- To calculate the pOH: $$\text{pOH} = -\log_{10}(2.00 \times 10^{-3}) \approx 2.70$$
- To calculate the pH: $$\text{pH} = 14 - \text{pOH} = 14 - 2.70 = 11.30$$
- This solution is basic because $ \text{pH} > 7 $.
