Defining and Calculating pOH Revision Notes for HSC SSCE Chemistry

Defining and Calculating pOH

Just as pH quantifies the concentration of hydrogen ions in a solution, pOH provides a measure of hydroxide ion concentration. Understanding pOH is essential for working with basic solutions and allows you to calculate properties of alkaline substances using the same logarithmic approach used for pH.

What is pOH?

pOH is defined as the negative logarithm (to base 10) of the hydroxide ion concentration in an aqueous solution. This mathematical relationship allows you to express very small hydroxide concentrations in a more convenient numerical scale.

The two key formulas you need to know are:

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

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

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These equations work in exactly the same way as the pH formulas, but measure hydroxide ions instead of hydronium ions. When you know the hydroxide ion concentration, you can calculate pOH using the first formula. When you know the pOH, you can work backwards to find the hydroxide ion concentration using the second formula.

Calculating pOH from hydroxide concentration

When calculating pOH, you follow the same logical steps as calculating pH. The process involves selecting the appropriate formula and substituting the given values carefully. Let's look at some worked examples to understand the process.

Example calculations

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Worked Example 1: Calculating pOH from hydroxide concentration

Problem: Calculate pOH when [OH⁻] = 1.5 × 10⁻⁴ mol L⁻¹

Solution:

Use: pOH = -log₁₀[OH⁻]
Substitute: pOH = -log₁₀(1.5 × 10⁻⁴)
Result: pOH = 3.82

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Worked Example 2: Calculating hydroxide concentration from pOH

Problem: Calculate [OH⁻] when pOH = 5.8

Solution:

Use: [OH⁻] = 10⁻ᵖᴼᴴ
Substitute: [OH⁻] = 10⁻⁵·⁸
Result: [OH⁻] = 1.6 × 10⁻⁶ mol L⁻¹

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Worked Example 3: Strong base concentration from pOH

Problem: Calculate the concentration of a NaOH solution with the same pOH as Problem 2

Solution:

This problem requires understanding how strong bases dissociate. Sodium hydroxide (NaOH) is a strong base that dissociates completely in water according to this equation:

$\text{NaOH(aq)} \rightarrow \text{Na}^+\text{(aq)} + \text{OH}^-\text{(aq)}$

From the balanced equation, you can see that one mole of NaOH produces one mole of OH⁻ ions. Therefore, [NaOH] = [OH⁻] = 1.6 × 10⁻⁶ mol L⁻¹.

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This stoichiometric relationship is crucial when working with strong bases. Always write the dissociation equation to determine the ratio between the base concentration and hydroxide ion concentration.

Relating pH and pOH

The relationship between pH and pOH comes from the water dissociation constant (Kw). At 298 K (25°C), water establishes an equilibrium described by:

$K_w = [\text{OH}^-][\text{H}_3\text{O}^+] = 1.0 \times 10^{-14}$

From this relationship, you can derive an extremely useful equation:

$\text{pH} + \text{pOH} = 14$

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This equation applies to all aqueous solutions at 25°C and provides an alternative method for calculations. When you know the pH of a basic solution, you can calculate pOH by subtraction. Similarly, if you know the pOH of an acidic solution, you can find its pH.

Using the pH + pOH relationship

Understanding how to interconvert between pH and pOH is essential for solving problems involving basic solutions. Let's work through examples that demonstrate this powerful relationship.

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Worked Example 1: Finding pH of a strong base solution

Problem: Find the pH of a 0.02 mol L⁻¹ solution of sodium hydroxide

Solution:

Step 1: Recognize that NaOH is a strong base that dissociates completely: $\text{NaOH(aq)} \rightarrow \text{Na}^+\text{(aq)} + \text{OH}^-\text{(aq)}$

Step 2: Determine [OH⁻]. Since NaOH dissociates 100%, [OH⁻] equals the original [NaOH] = 0.02 mol L⁻¹.

Step 3: Calculate pOH: $\text{pOH} = -\log_{10}(0.02) = 1.7$

Step 4: Use the relationship pH + pOH = 14: $\text{pH} = 14 - 1.7 = 12.3$

Final Answer: pH = 12.3

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Worked Example 2: Finding concentration from pH

Problem: If the pH of a Ba(OH)₂ solution is 8.3, what is the concentration?

Solution:

Step 1: Calculate pOH using pH + pOH = 14: $\text{pOH} = 14 - 8.3 = 5.7$

Step 2: Calculate [OH⁻]: $[\text{OH}^-] = 10^{-5.7} = 2.0 \times 10^{-6} \text{ mol L}^{-1}$

Step 3: Write the dissociation equation for barium hydroxide: $\text{Ba(OH)}_2\text{(aq)} \rightarrow \text{Ba}^{2+}\text{(aq)} + 2\text{OH}^-\text{(aq)}$

Step 4: Notice that one mole of Ba(OH)₂ produces two moles of OH⁻. Therefore: $[\text{Ba(OH)}_2] = \frac{1}{2}[\text{OH}^-] = \frac{1}{2} \times 2.0 \times 10^{-6} = 1.0 \times 10^{-6} \text{ mol L}^{-1}$

Final Answer: [Ba(OH)₂] = 1.0 × 10⁻⁶ mol L⁻¹

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The stoichiometry here is crucial. Because Ba(OH)₂ contains two hydroxide ions per formula unit, you must divide the hydroxide concentration by two to find the barium hydroxide concentration.

Diluting solutions and effects on pH and pOH

When you dilute a solution or allow it to evaporate, you change the concentration of ions present, which affects both pH and pOH. Understanding these effects is important for practical laboratory work.

Because pH measures [H₃O⁺] concentration, adding or removing solvent changes the pH. A useful rule to remember is:

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Dilution Rule

A change in concentration by a factor of 10 results in a pH (or pOH) change of one unit. Similarly, a change by a factor of 100 results in a change of two units.

Dilution example

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Worked Example: Dilution effect on pH

Problem: If a solution of HF with pH = 3 is diluted by a factor of 100, what is the final pH?

Solution:

Logic:

A factor of 10 change in concentration equals 1 unit change in pH
A factor of 100 change equals 2 units change in pH
Dilution decreases [H₃O⁺], so pH increases
Final pH = 3 + 2 = 5

Evaporation example

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Worked Example: Evaporation effect on pH and pOH

Problem: A 500 mL solution of NaOH with pOH = 5 evaporates until the volume is 40 mL. Calculate:

a) The new pOH
b) The new pH

Solution:

Part a) First, find the initial hydroxide concentration: $[\text{OH}^-] = 10^{-\text{pOH}} = 10^{-5} \text{ mol L}^{-1}$

Use the dilution formula c₁V₁ = c₂V₂: $10^{-5} \times 500 = c_2 \times 40$ $c_2 = \frac{10^{-5} \times 500}{40} = 1.25 \times 10^{-4} \text{ mol L}^{-1}$

Calculate the new pOH: $\text{pOH} = -\log(1.25 \times 10^{-4}) = 3.90$

Answer: pOH = 3.90

Part b) Use pH + pOH = 14: $\text{pH} = 14 - 3.90 = 10.1$

Answer: pH = 10.1

Notice how evaporation increases the concentration, which decreases pOH (and increases pH for this basic solution).

Exam tips

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Key Tips for Success:

Always write dissociation equations for strong bases to determine the stoichiometric relationship between the base and OH⁻ ions
Remember that pH + pOH = 14 only applies at 25°C (298 K)
When working with polyprotic bases like Ba(OH)₂, pay careful attention to the number of OH⁻ ions produced per formula unit
For dilution problems, a factor of 10 change = 1 pH unit change; factor of 100 = 2 pH units change
Keep track of whether concentration is increasing (evaporation) or decreasing (dilution) to predict whether pH/pOH increases or decreases

Remember!

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Key Points to Remember:

pOH is defined as: pOH = -log₁₀[OH⁻], and [OH⁻] = 10⁻ᵖᴼᴴ
The pH-pOH relationship: pH + pOH = 14 at 25°C, derived from Kw = 1.0 × 10⁻¹⁴
Strong base dissociation: Always write the equation to find the relationship between base concentration and [OH⁻]. For NaOH: [NaOH] = [OH⁻]. For Ba(OH)₂: [Ba(OH)₂] = ½[OH⁻]
Dilution effects: A 10-fold dilution changes pH or pOH by 1 unit; a 100-fold dilution changes it by 2 units
Calculation pathway: You can find pH of a basic solution by calculating pOH first, then using pH + pOH = 14

Defining and Calculating pOH (HSC SSCE Chemistry): Revision Notes