# pH and pKa

> AP Chemistry · Unit 8: Acids and Bases
> Source: https://www.owlsprep.com/study/ap-chemistry-u8-ph-and-pka/

This guide covers core definitions of pH and pKa, key inverse relationships, conversions between $K_a/pK_a/pH/[H_3O^+]$, pH calculation for weak acids, and application of the Henderson-Hasselbalch equation for buffer systems.

**Prerequisites:** Bronsted-Lowry definitions of acids and bases; Equilibrium constant expressions for weak acid dissociation; Autoionization of water and $K_w$

## Learning objectives

- Define pH and pKa using logarithmic scale definitions
- Convert between $K_a$, $pK_a$, $[H_3O^+]$, and $pH$
- Relate pKa to acid strength and predict proton transfer equilibrium direction
- Calculate pH of pure weak acids and buffer solutions
- Apply the Henderson-Hasselbalch equation correctly

## Core Definitions: pH and pKa

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.

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

*Notation:* $pH$

*Example:* $pH = -\log_{10} [H_3O^+]$

**pKa** — The negative base-10 logarithm of the acid dissociation constant $K_a$, used to quantify acid strength.

*Notation:* $pK_a$

*Example:* $pK_a = -\log_{10} K_a$

The core intuition that trips up many new students is: lower pH = higher $[H_3O^+]$ = 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.

## Conversions and Pure Weak Acid pH

All p-scale values follow the same fundamental rule: $pX = -\log_{10} X$, so the inverse conversion (from pX back to X) is always $X = 10^{-pX}$. 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 ($HC_3H_5O_2$) has a $K_a = 4.5 \times 10^{-5}$. 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: $[HA] = 0.10 M$, so $\log[HA] = -1$:

   $$pH = \frac{1}{2}(4.35 - (-1)) = 2.67 \approx 2.7$$
4. Check the 5% rule to confirm the approximation is valid:

   $$[H_3O^+] = 10^{-2.7} = 2.0 \times 10^{-3} M; \frac{2.0 \times 10^{-3}}{0.10} \times 100\% = 2\% < 5\%$$

> **Exam Tip**
>
> Always follow the sig fig rule for p-scale values: the number of decimal places in pKa/pH equals the number of significant figures in the original $K_a$ or $[H_3O^+]$. For example, $K_a = 4.5 \times 10^{-5}$ (two sig figs) gives pKa = 4.35 (two decimal places).

*Calculator:* allowed

## pKa and Acid Strength

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 $pKa = -\log K_a$, 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 $HN_3(aq) + ClO^-(aq) \rightleftharpoons N_3^-(aq) + HClO(aq)$ 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 ($HN_3$) > 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 $HN_3$ (pKa = 4.75), product acid is $HClO$ (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.

> **Exam Tip**
>
> If you get confused about conjugate strengths, always remember: the weaker the acid (higher pKa), the stronger its conjugate base. This relationship holds for all Bronsted-Lowry acid-base pairs.

## The Henderson-Hasselbalch Equation

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:** 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 $pK_a = -\log K_a$ and $pH = -\log [H_3O^+]$, then rearrange terms:

   $$pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)$$

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

The most important relationship from this equation is: when $[A^-] = [HA]$, the ratio $\frac{[A^-]}{[HA]} = 1$, $\log(1) = 0$, so $pH = pK_a$. 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 ($pK_a = 4.20$) 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.

> **Common Mistake**
>
> Always double-check that you put conjugate base in the numerator and acid in the denominator of the log term. Flipping the ratio is the most common mistake on HH problems.

*Calculator:* allowed

## AP-Style Concept Check

**Check your understanding**

Test your understanding with these worked practice problems:

1. Given the following pKa values: $HBr$: $pK_a = -9$, $CH_3COOH$: $pK_a = 4.8$, $CH_3COONa$: conjugate acid $CH_3COOH$ pKa 4.8, $NH_3$: conjugate acid $NH_4^+$ pKa 9.25. All solutions are 0.10 M. Which correctly ranks the solutions from lowest pH to highest pH?

   - A) $HBr < CH_3COOH < NH_3 < CH_3COONa$
   - B) $CH_3COOH < HBr < CH_3COONa < NH_3$
   - C) $HBr < CH_3COOH < CH_3COONa < NH_3$
   - D) $NH_3 < CH_3COONa < CH_3COOH < HBr$

   *Why:* Lower pH = more acidic solution. $HBr$ (strong acid, lowest pKa) has lowest pH, followed by weak acid $CH_3COOH$. $CH_3COONa$ is weakly basic, and $NH_3$ is a stronger base with higher pH, so order C is correct.

## Common pitfalls

- **Wrong:** Dropping the negative sign in the p-scale definition, writing $pK_a = \log K_a$ or $pH = \log [H_3O^+]$
  - Why it fails: Students rush calculations and forget the negative sign that is core to all p-scale definitions
  - Correct: Always write the full definition $pX = -\log X$ on your scratch paper before starting any calculation
- **Wrong:** Reporting one decimal place for pKa when $K_a$ has two significant figures (e.g. writing $pK_a = 4.6$ for $K_a = 2.3 \times 10^{-5}$)
  - Why it fails: Students confuse sig fig rules for logarithmic and linear values, applying standard whole-number sig fig rules instead of the p-scale rule
  - Correct: For any p-scale value, the number of decimal places equals the number of significant figures in the original value
- **Wrong:** Flipping the ratio in the Henderson-Hasselbalch equation, writing $\log\left(\frac{[HA]}{[A^-]}\right)$
  - Why it fails: Students memorize the equation incorrectly or mix up which species is the conjugate base
  - Correct: Quickly rederive the ratio from the $K_a$ expression to confirm: $K_a = \frac{[H^+][A^-]}{[HA]} \rightarrow pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)$
- **Wrong:** Claiming a higher pKa means a stronger acid
  - Why it fails: The negative log flips the order of $K_a$, so students forget the inverse relationship
  - Correct: Every time you rank acid strength, remember the mnemonic: 'Lower pKa = stronger acid'
- **Wrong:** Using the Henderson-Hasselbalch equation to calculate the pH of a pure weak acid with no added conjugate base
  - Why it fails: Students memorize HH and overuse it, forgetting it requires comparable concentrations of both acid and conjugate base
  - Correct: Only use HH for buffers; use the $pH = \frac{1}{2}(pK_a - \log[HA])$ approximation for pure weak acids

## Cheatsheet

| Category | Formula / Rule | Notes |
| --- | --- | --- |
| pH definition | $pH = -\log_{10}[H_3O^+]$ | Inverse: $[H_3O^+] = 10^{-pH}$; applies to all solutions |
| pKa definition | $pK_a = -\log_{10} K_a$ | Inverse: $K_a = 10^{-pK_a}$; for any acid |
| Acid strength rule | Lower $pK_a$ = stronger acid | Negative pKa = strong acid; positive pKa = weak acid |
| pH of pure weak acid (5% rule) | $pH = \frac{1}{2}\left(pK_a - \log[HA]\right)$ | Only for pure weak acid; valid if % dissociation <5% |
| Henderson-Hasselbalch | $pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)$ | Only for buffers; volume cancels, use moles directly |
| Half-equivalence point | $pH = pK_a$ | At half-titration, $[HA] = [A^-]$ 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 | $K_a = 2.3 \times 10^{-5}$ (2 sig figs) → pKa = 4.64 (2 decimals) |

## 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.

- [Buffer Capacity](https://www.owlsprep.com/study/ap-chemistry-u8-buffer-capacity/)
- [Applications of Thermodynamics](https://www.owlsprep.com/study/ap-chemistry-u9-overview/)
- [Entropy and Gibbs Free Energy](https://www.owlsprep.com/study/ap-chemistry-u9-entropy-and-gibbs-free-energy/)

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