pH and pKa

AP Chemistry· AP Chemistry CED — Acids and Bases· 14 min read

1. Core Definitions: pH and pKa★★☆☆☆⏱ 3 min

pH is a logarithmic scale developed to simplify describing the extremely wide range of hydronium ion concentrations in aqueous solution, which span roughly 14 orders of magnitude from concentrated strong acids to concentrated strong bases. pKa is the analogous logarithmic scale for acid dissociation constants $K_{a}$ , which also span many orders of magnitude.

📘 Definition

pH

$p H$

The negative base-10 logarithm of hydronium ion concentration in an aqueous solution, used to quantify acidity.

Example:

$p H = - lo g_{10} [H_{3} O^{+}]$

📘 Definition

pKa

$p K_{a}$

The negative base-10 logarithm of the acid dissociation constant $K_{a}$ , used to quantify acid strength.

Example:

$p K_{a} = - lo g_{10} K_{a}$

The core intuition that trips up many new students is: lower pH = higher $[H_{3} O^{+}]$ = more acidic solution, and lower pKa = larger $K_{a}$ = stronger acid. This topic is heavily tested on the AP Chemistry exam, appearing in both multiple-choice and free-response sections.

2. Conversions and Pure Weak Acid pH★★★☆☆⏱ 4 min

✓ Calculator OK

All p-scale values follow the same fundamental rule: $pX = - lo g_{10} X$ , so the inverse conversion (from pX back to X) is always $X = 1 0^{- pX}$ . This rule works for pH, pKa, pOH, pKb, and any other p-scale value you will encounter on the exam.

The logarithmic scale simplifies working with very small or very large values: a 10-fold increase in $K_{a}$ (a 10x stronger acid) translates to a 1-unit decrease in pKa, which is far easier to compare than working with exponents in scientific notation. For context: strong acids have $K_{a} > 1$ , so their pKa values are negative, while weak acids have $K_{a} < 1$ , so their pKa values are positive.

For pure dilute weak acid solutions where the 5% rule holds (dissociation is less than 5% of the initial acid concentration), we can use a simplified pH formula that avoids solving a quadratic equation:

pH = \frac{1}{2}\left(pK_a - \log[HA]\right)

📐 Worked Example

A 0.10 M aqueous solution of propanoic acid ( $H C_{3} H_{5} O_{2}$ ) has a $K_{a} = 4.5 \times 1 0^{- 5}$ . Calculate (a) the pKa of propanoic acid, and (b) the pH of the solution, confirming your assumption is valid.

1
Use the definition of pKa to convert from $K_{a}$ :
$pK_a = -\log_{10}(4.5 \times 10^{-5}) = 4.35$
2
Confirm this is a pure weak acid with no added conjugate base, so the shortcut formula applies.
3
Substitute values: $[H A] = 0.10 M$ , so $lo g [H A] = - 1$ :
$pH = \frac{1}{2}(4.35 - (-1)) = 2.67 \approx 2.7$
4
Check the 5% rule to confirm the approximation is valid:
$[H_3O^+] = 10^{-2.7} = 2.0 \times 10^{-3} M; \frac{2.0 \times 10^{-3}}{0.10} \times 100\% = 2\% < 5\%$

3. pKa and Acid Strength★★☆☆☆⏱ 3 min

pKa is the standard way to compare the strength of weak acids, because the logarithmic scale eliminates the need to compare negative exponents for $K_{a}$ . By definition, since $p K a = - lo g K_{a}$ , a lower pKa always corresponds to a larger $K_{a}$ , which means the acid dissociates more completely in water, so it is a stronger acid.

This relationship is tested conceptually as often as it is tested numerically: AP questions frequently ask you to rank acids by strength given pKa values, or predict the direction of a proton transfer reaction based on pKa. The rule for proton transfer is simple: an acid will donate a proton to any base whose conjugate acid has a higher pKa than the original acid, because equilibrium always favors formation of the weaker (higher pKa) acid.

📐 Worked Example

Given the following pKa values: formic acid = 3.75, hypochlorous acid = 7.46, hydrazoic acid = 4.75. (a) Rank the three acids from weakest to strongest. (b) Predict whether the reaction $H N_{3} (a q) + C l O^{-} (a q) ⇌ N_{3}^{-} (a q) + H C l O (a q)$ favors reactants or products at equilibrium.

1
Recall that lower pKa = stronger acid, so weakest to strongest means ordering from highest pKa to lowest pKa.
2
Order the pKa values: 7.46 (HClO) > 4.75 ( $H N_{3}$ ) > 3.75 (formic acid). The rank from weakest to strongest is: hypochlorous acid < hydrazoic acid < formic acid.
3
Identify the acid on each side of the reaction: reactant acid is $H N_{3}$ (pKa = 4.75), product acid is $H C l O$ (pKa = 7.46).
4
Equilibrium favors the side with the weaker acid (higher pKa). The product acid is weaker, so the reaction favors products at equilibrium.

4. The Henderson-Hasselbalch Equation★★★☆☆⏱ 4 min

✓ Calculator OK

The Henderson-Hasselbalch (HH) equation is the core tool for calculating the pH of buffer solutions, which contain a weak acid and its conjugate base in roughly equal concentrations. It is derived directly from the $K_{a}$ equilibrium expression:

🔬 Derivation

Goal:

Derive the Henderson-Hasselbalch equation for buffer pH

Starting from:

K_a = \frac{[H_3O^+][A^-]}{[HA]}

1
Take the negative base-10 logarithm of both sides:
$-\log K_a = -\log [H_3O^+] - \log\left(\frac{[A^-]}{[HA]}\right)$
2
Substitute $p K_{a} = - lo g K_{a}$ and $p H = - lo g [H_{3} O^{+}]$ , then rearrange terms:
$pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)$

Result:

This final form is the Henderson-Hasselbalch equation, used exclusively for buffer solutions.

The most important relationship from this equation is: when $[A^{-}] = [H A]$ , the ratio $\frac{[ A ^{-} ]}{[ H A ]} = 1$ , $lo g (1) = 0$ , so $p H = p K_{a}$ . This is why at the half-equivalence point of a weak acid-strong base titration, the pH of the solution equals the pKa of the weak acid, which is the standard experimental method for measuring pKa.

📐 Worked Example

A buffer is prepared by dissolving 0.12 moles of benzoic acid ( $p K_{a} = 4.20$ ) and 0.24 moles of sodium benzoate in enough water to make 2.00 L of solution. Calculate the pH of the buffer.

1
Confirm this is a buffer: it contains a weak acid (benzoic acid) and its conjugate base (benzoate from sodium benzoate), so HH applies.
2
Calculate concentrations (note that total volume cancels in the ratio, so moles can be used directly):
$[HA] = 0.060 M; [A^-] = 0.12 M$
3
Substitute into the HH equation:
$pH = 4.20 + \log\left(\frac{0.12}{0.060}\right) = 4.20 + 0.30 = 4.50$
4
Check intuition: there is more conjugate base than acid, so pH should be higher than pKa, which matches our result.

5. AP-Style Concept Check★★★★☆⏱ 3 min

✓ Quick check

Test your understanding with these worked practice problems:

Given the following pKa values: $H B r$ : $p K_{a} = - 9$ , $C H_{3} C O O H$ : $p K_{a} = 4.8$ , $C H_{3} C O O N a$ : conjugate acid $C H_{3} C O O H$ pKa 4.8, $N H_{3}$ : conjugate acid $N H_{4}^{+}$ pKa 9.25. All solutions are 0.10 M. Which correctly ranks the solutions from lowest pH to highest pH?
- A) $H B r < C H_{3} C O O H < N H_{3} < C H_{3} C O O N a$
- B) $C H_{3} C O O H < H B r < C H_{3} C O O N a < N H_{3}$
- C) $H B r < C H_{3} C O O H < C H_{3} C O O N a < N H_{3}$
- D) $N H_{3} < C H_{3} C O O N a < C H_{3} C O O H < H B r$
Reveal answer
C —
Lower pH = more acidic solution. $H B r$ (strong acid, lowest pKa) has lowest pH, followed by weak acid $C H_{3} C O O H$ . $C H_{3} C O O N a$ is weakly basic, and $N H_{3}$ is a stronger base with higher pH, so order C is correct.

6. Common Pitfalls

Wrong move:

Dropping the negative sign in the p-scale definition, writing $p K_{a} = lo g K_{a}$ or $p H = lo g [H_{3} O^{+}]$

Why:

Students rush calculations and forget the negative sign that is core to all p-scale definitions

Correct move:

Always write the full definition $pX = - lo g X$ on your scratch paper before starting any calculation

Wrong move:

Reporting one decimal place for pKa when $K_{a}$ has two significant figures (e.g. writing $p K_{a} = 4.6$ for $K_{a} = 2.3 \times 1 0^{- 5}$ )

Why:

Students confuse sig fig rules for logarithmic and linear values, applying standard whole-number sig fig rules instead of the p-scale rule

Correct move:

For any p-scale value, the number of decimal places equals the number of significant figures in the original value

Wrong move:

Flipping the ratio in the Henderson-Hasselbalch equation, writing $lo g (\frac{[ H A ]}{[ A ^{-} ]})$

Why:

Students memorize the equation incorrectly or mix up which species is the conjugate base

Correct move:

Quickly rederive the ratio from the $K_{a}$ expression to confirm: $K_{a} = \frac{[ H ^{+} ] [ A ^{-} ]}{[ H A ]} \to p H = p K_{a} + lo g (\frac{[ A ^{-} ]}{[ H A ]})$

Wrong move:

Claiming a higher pKa means a stronger acid

Why:

The negative log flips the order of $K_{a}$ , so students forget the inverse relationship

Correct move:

Every time you rank acid strength, remember the mnemonic: 'Lower pKa = stronger acid'

Wrong move:

Using the Henderson-Hasselbalch equation to calculate the pH of a pure weak acid with no added conjugate base

Why:

Students memorize HH and overuse it, forgetting it requires comparable concentrations of both acid and conjugate base

Correct move:

Only use HH for buffers; use the $p H = \frac{1}{2} (p K_{a} - lo g [H A])$ approximation for pure weak acids

7. Quick Reference Cheatsheet

Category	Formula / Rule	Notes
pH definition	$p H = - lo g_{10} [H_{3} O^{+}]$	Inverse: $[H_{3} O^{+}] = 1 0^{- p H}$ ; applies to all solutions
pKa definition	$p K_{a} = - lo g_{10} K_{a}$	Inverse: $K_{a} = 1 0^{- p K_{a}}$ ; for any acid
Acid strength rule	Lower $p K_{a}$ = stronger acid	Negative pKa = strong acid; positive pKa = weak acid
pH of pure weak acid (5% rule)	$p H = \frac{1}{2} (p K_{a} - lo g [H A])$	Only for pure weak acid; valid if % dissociation <5%
Henderson-Hasselbalch	$p H = p K_{a} + lo g (\frac{[ A ^{-} ]}{[ H A ]})$	Only for buffers; volume cancels, use moles directly
Half-equivalence point	$p H = p K_{a}$	At half-titration, $[H A] = [A^{-}]$ so pH = pKa
Proton transfer rule	Equilibrium favors higher pKa acid	Proton transfer always forms the weaker acid
p-scale sig figs	Decimal places = sig figs in original value	$K_{a} = 2.3 \times 1 0^{- 5}$ (2 sig figs) → pKa = 4.64 (2 decimals)

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Rank acid strength by pKa values
2022 · FRQ
Calculate buffer pH with HH equation
2021 · MCQ
pH at half-equivalence point

What's Next

Mastery of pH and pKa is the foundational prerequisite for all remaining topics in Unit 8 Acids and Bases, and it is also critical for Unit 9 Applications of Thermodynamics, specifically solubility equilibria. Next, you will apply the relationship between pH and pKa to solve buffer capacity problems and acid-base titration curve problems; without correctly calculating pH from pKa and interpreting the pH = pKa half-equivalence rule, you will not be able to analyze titration data or select appropriate buffer systems for a given pH. pH and pKa also underpin acid-base reactivity in all contextual problems that appear on the AP exam, and they are central to calculating pH of salt solutions after titration.