Topic 8.2 Notes – pH and pOH of Strong Acids and Bases

1. What pH and pOH Mean

At the core, pH measures acidity and pOH measures basicity.

pH

$${pH = -\log[H_{3}O^{+}]}$$

• Log is base 10.
• Larger $[H_{3}O^{+}]$ → smaller pH.
• A 1-unit change in pH means a 10× change in $[H_{3}O^{+}]$.

Example:
If $[H_{3}O^{+}] = 1.0 \times 10^{-3}\,\text{M}$,
$$\text{pH} = -\log(1.0 \times 10^{-3}) = 3.00$$

That decimal place matters. On AP problems, the number of decimal places in pH = significant figures in concentration.

pOH

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

• Larger $[OH^{-}]$ → smaller pOH.
• Same log rules apply.

Autoionization of Water

Water naturally forms small amounts of ions:

$$\ce{2H2O(l) <=> H3O^{+}(aq) + OH^{-}(aq)}$$

$${K_w = [H_{3}O^{+}][OH^{-}] = 1.0 \times 10^{-14} \text{ at 25°C}}$$

Taking the negative log of both sides gives:

$${pH + pOH = 14}$$

That 14 only applies at 25°C, which is what AP Chemistry assumes unless told otherwise.

Quick classification at 25°C:

• pH < 7 → acidic
• pH = 7 → neutral
• pH > 7 → basic

Now let’s connect this to strong acids and bases.

2. Strong Acids and Strong Bases

The defining feature is complete ionization. There is no equilibrium calculation. No ICE table.

Strong Acids

Memorize these:

• $\ce{HCl}$
• $\ce{HBr}$
• $\ce{HI}$
• $\ce{HNO3}$
• $\ce{HClO4}$
• $\ce{H2SO4}$ (first proton only)

Example reaction:

$$\ce{HNO3(aq) + H2O(l) -> H3O^{+}(aq) + NO3^{-}(aq)}$$

Because ionization is complete:

$${[H_{3}O^{+}] = initial acid concentration}$$

If you have 0.020 M $\ce{HBr}$:

$$[H_{3}O^{+}] = 0.020\,\text{M}$$ $$\text{pH} = -\log(0.020) = 1.70$$

Notice we did not set up an equilibrium expression.

Strong Bases

Group I hydroxides:

• $\ce{NaOH}$, $\ce{KOH}$, $\ce{LiOH}$

$$\ce{NaOH(aq) -> Na^{+}(aq) + OH^{-}(aq)}$$

$${[OH^{-}] = initial base concentration}$$

Group II hydroxides:

• $\ce{Ca(OH)2}$, $\ce{Sr(OH)2}$, $\ce{Ba(OH)2}$

$$\ce{Ba(OH)2(aq) -> Ba^{2+}(aq) + 2OH^{-}(aq)}$$

Each formula unit gives two hydroxides:

$${[OH^{-}] = 2 \times initial concentration}$$

This is one of the most common quiz mistakes.

If you have 0.015 M $\ce{Ba(OH)2}$:

$$[OH^{-}] = 2(0.015) = 0.030\,\text{M}$$

$$\text{pOH} = -\log(0.030) = 1.52$$

$$\text{pH} = 14 - 1.52 = 12.48$$

3. How to Calculate pH and pOH

Here’s the clean process your brain should follow.

Strong Acid

1. Write dissociation equation.
2. Determine $[H_{3}O^{+}]$ from stoichiometry.
3. Calculate $\text{pH} = -\log[H_{3}O^{+}]$.
4. Use $\text{pOH} = 14 - \text{pH}$ if needed.

If dilution is involved, calculate new molarity first using $M_{1}V_{1} = M_{2}V_{2}$. Then take the log.

Strong Base

1. Write dissociation equation.
2. Determine $[OH^{-}]$ (double it for Group II).
3. Calculate $\text{pOH} = -\log[OH^{-}]$.
4. Convert using $\text{pH} = 14 - \text{pOH}$.

If a problem gives you pH and asks for $[OH^{-}]$, go backward:

• Find pOH.
• Use inverse log: $[OH^{-}] = 10^{-\text{pOH}}$.

That reversal step shows up a lot in multiple choice.

4. Common AP Traps

• Treating strong acids as equilibria problems.
• Forgetting to double hydroxide for Group II.
• Mixing up pH and pOH in the log step.
• Taking $-\log$ twice.
• Ignoring significant figures in pH.
• Forgetting that $\ce{H2SO4}$ is only fully strong for the first proton in this course.

If it says “strong,” assume complete dissociation immediately.