pH and pKa — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Definitions of pH and pKa, the inverse relationship between acid strength and pKa, pH calculations for weak acids, conversion between Ka and pKa, the Henderson-Hasselbalch equation, and buffer pH prediction for weak acid-conjugate base systems.

You should already know: Bronsted-Lowry definitions of acids and bases, equilibrium constant expressions for weak acid dissociation, the autoionization of water and $K_w$.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is 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. Per the AP Chemistry CED, this topic makes up ~15-20% of Unit 8 (Acids and Bases) exam weight, and appears in both multiple-choice (MCQ) and free-response (FRQ) sections of the exam. It is the foundation for buffer problems, acid-base titrations, and even solubility questions in FRQs, and is a frequent source of concept and calculation MCQs. Formally, pH is defined as $pH=-{\log }_{10}[H_3{O}^{+}]$, and pKa is defined as $p{K}_a=-{\log }_{10}{K}_a$, where $K_a$ is the acid dissociation equilibrium constant for a weak acid. 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.

2. Conversions Between $K_a$, $p{K}_a$, $pH$, and $[H_3{O}^{+}]$

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, which matches most standard pKa data tables. For pure dilute weak acid solutions where the 5% rule holds (dissociation is less than 5% of the initial acid concentration), we can derive a simplified pH formula from the Ka equilibrium expression: $$pH=\frac{1}{2}\left(p{K}_a-\log [HA]\right)$$ This formula is a shortcut for calculating pH of a pure weak acid without solving a quadratic equation, which is acceptable for most AP problems when the 5% assumption is valid.

Worked Example

A 0.10 M aqueous solution of propanoic acid ($H{C}_3{H}_5{O}_2$) has a $K_a=4.5\times 10^{-5}$. 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 $K_a$: $p{K}_a=-{\log }_{10}(K_a)=-{\log }_{10}(4.5\times 10^{-5})=-(0.65-5)=4.35$.
2. Confirm this is a pure weak acid with no added conjugate base, so the shortcut formula applies: $pH=\frac{1}{2}\left(p{K}_a-\log [HA]\right)$.
3. Substitute values: $[HA]=0.10M$, so $\log [HA]=\log (0.10)=-1$. Plugging in: $pH=\frac{1}{2}(4.35-(-1))=\frac{1}{2}(5.35)=2.67\approx 2.7$.
4. Check the 5% rule: $[H_3{O}^{+}]=10^{-2.7}=2.0\times 10^{-3}M$. Percent dissociation = $\frac{2.0\times 10^{-3}}{0.10}\times 100\%=2\%<5\%$, so the assumption is valid.

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 $[H_3{O}^{+}]$. For example, $K_a=4.5\times 10^{-5}$ (two sig figs) gives pKa = 4.35 (two decimal places).

3. pKa and Acid Strength: Interpreting Relative 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. Common pKa values range from ~-10 for the strongest mineral acids to ~20 for very weak organic acids. 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 $H{N}_3(aq)+Cl{O}^{-}(aq)\rightleftharpoons N_3^{-}(aq)+HClO(aq)$ favors reactants or products at equilibrium.

1. Recall that lower pKa = stronger acid, so weakest to strongest means order of highest pKa to lowest pKa.
2. Order the pKa values from highest to lowest: 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. For the reaction, identify the acid on each side: reactant acid is $H{N}_3$ (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, just remember: the weaker the acid (higher pKa), the stronger its conjugate base. This will always hold for Bronsted-Lowry acid-base pairs.

4. 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: Starting with $K_a=\frac{[H_3{O}^{+}][A^{-}]}{[HA]}$, take the negative log of both sides: $$-\log K_a=-\log [H_3{O}^{+}]-\log \left(\frac{[A^{-}]}{[HA]}\right)$$ Rearranging and substituting $p{K}_a=-\log K_a$ and $pH=-\log [H_3{O}^{+}]$ gives the final form: $$pH=p{K}_a+\log \left(\frac{[A^{-}]}{[HA]}\right)$$ The most important relationship from this equation is: when $[A^{-}]=[HA]$, the ratio $\frac{[A^{-}]}{[HA]}=1$, $\log (1)=0$, so $pH=p{K}_a$. 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. The HH equation only applies to buffer systems where the initial concentrations of acid and conjugate base are much larger than the $H^{+}$ or $O{H}^{-}$ from water autoionization or dissociation, which is true for all AP buffer problems.

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 the Henderson-Hasselbalch equation applies.
2. Calculate concentrations: $[benzoicacid]=\frac{0.12mol}{2.00L}=0.060M$, $[benzoate]=\frac{0.24mol}{2.00L}=0.12M$. Note that the total volume is the same for both species, so the volume cancels out in the ratio, and we could use moles directly for the calculation.
3. Substitute into the HH equation: $pH=4.20+\log \left(\frac{0.12}{0.060}\right)=4.20+\log (2)=4.20+0.30=4.50$.
4. Check intuition: there is more conjugate base than acid, so pH should be higher than pKa, which matches 4.50 > 4.20, so the answer makes sense.

Exam tip: 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.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Dropping the negative sign in the p-scale definition, writing $p{K}_a=\log K_a$ or $pH=\log [H_3{O}^{+}]$. Why: Students rush calculations and forget the negative sign that is core to all p-scale definitions. Correct move: Always write the full definition $pX=-\log X$ 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 $K_a=2.3\times 10^{-5}$). 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 $\log \left(\frac{[HA]}{[A^{-}]}\right)$. 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: $K_a=\frac{[H^{+}][A^{-}]}{[HA]}\to [H^{+}]=K_a\frac{[HA]}{[A^{-}]}\to pH=p{K}_a+\log \left(\frac{[A^{-}]}{[HA]}\right)$.
• 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, say the mnemonic "Lower pKa = stronger acid" out loud (or in your head) before ordering.
• 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 $pH=\frac{1}{2}(p{K}_a-\log [HA])$ approximation for pure weak acids.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

Given the following pKa values: $HBr$: $p{K}_a=-9$, $C{H}_{3}COOH$: $p{K}_a=4.8$, $C{H}_{3}COONa$: conjugate acid $C{H}_{3}COOH$ pKa 4.8, $N{H}_3$: conjugate acid $N{H}_4^{+}$ pKa 9.25. All solutions are 0.10 M. Which of the following correctly ranks the solutions from lowest pH to highest pH? A) $HBr<C{H}_{3}COOH<N{H}_3<C{H}_{3}COONa$ B) $C{H}_{3}COOH<HBr<C{H}_{3}COONa<N{H}_3$ C) $HBr<C{H}_{3}COOH<C{H}_{3}COONa<N{H}_3$ D) $N{H}_3<C{H}_{3}COONa<C{H}_{3}COOH<HBr$

Worked Solution: Lower pH corresponds to more acidic solutions. $HBr$ is a strong acid with the lowest pKa, so it has the lowest pH, eliminating options B and D. $C{H}_{3}COOH$ is a weak acid, so it has the next lowest pH after $HBr$. $C{H}_{3}COONa$ is the salt of a weak acid and strong base, so it forms a weakly basic solution (pH >7), while $N{H}_3$ is a stronger weak base than $C{H}_{3}CO{O}^{-}$, so it has a higher pH than $C{H}_{3}COONa$. The correct order from lowest to highest pH is $HBr<C{H}_{3}COOH<C{H}_{3}COONa<N{H}_3$. Correct answer: C

Question 2 (Free Response)

A student prepares two aqueous solutions at 25°C: Solution 1: 0.020 M hydrocyanic acid ($HCN$), $K_a=6.2\times 10^{-10}$ Solution 2: A buffer containing 0.20 mol $HCN$ and 0.40 mol $NaCN$ in 1.00 L total solution (a) Calculate the pKa of hydrocyanic acid. (b) Calculate the pH of Solution 1, and use the 5% rule to justify your approximation. (c) Calculate the pH of Solution 2 using the Henderson-Hasselbalch equation, and explain why the 5% approximation is not needed here.

Worked Solution: (a) By definition: $p{K}_a=-{\log }_{10}(K_a)=-{\log }_{10}(6.2\times 10^{-10})=-(0.79-10)=9.21$. Two decimal places match the two significant figures in $K_a$, so $p{K}_a=9.21$. (b) For a pure weak acid: $pH=\frac{1}{2}(p{K}_a-\log [HA])=\frac{1}{2}(9.21-\log (0.020))=\frac{1}{2}(9.21-(-1.70))=5.46$. Check 5% rule: $[H_3{O}^{+}]=10^{-5.46}=3.5\times 10^{-6}M$. Percent dissociation = $\frac{3.5\times 10^{-6}}{0.020}\times 100\%=0.0175\%<5\%$, so the approximation is valid. (c) Use the Henderson-Hasselbalch equation: $pH=p{K}_a+\log \left(\frac{[C{N}^{-}]}{[HCN]}\right)=9.21+\log \left(\frac{0.40}{0.20}\right)=9.21+0.30=9.51$. The 5% approximation is not needed because the common ion effect from added $NaCN$ suppresses dissociation of $HCN$ almost completely: the change in $[HCN]$ and $[C{N}^{-}]$ from dissociation is negligible compared to the initial concentrations, so the HH approximation holds automatically.

Question 3 (Application / Real-World Style)

The cytoplasm of most mammalian cells is buffered to pH 7.2 by the phosphate buffer system, with a pKa of 7.21 for dihydrogen phosphate ($H_{2}P{O}_4^{-}$). The total concentration of phosphate species ($H_{2}P{O}_4^{-}+HP{O}_4^{2-}$) in cytoplasm is 0.010 M. Calculate the concentration of the basic component ($HP{O}_4^{2-}$) in normal cytoplasm, and interpret the result.

Worked Solution: Let $x=[HP{O}_4^{2-}]$, so $[H_{2}P{O}_4^{-}]=0.010-x$. Substitute into HH: $7.2=7.21+\log \left(\frac{x}{0.010-x}\right)$. Rearrange: $-0.01=\log \left(\frac{x}{0.010-x}\right)$, so $10^{-0.01}=0.977=\frac{x}{0.010-x}$. Solve for x: $0.977(0.010-x)=x\to 0.00977=1.977x\to x=0.0049M$. The concentration of $HP{O}_4^{2-}$ is ~0.0049 M, and the concentration of $H_{2}P{O}_4^{-}$ is ~0.0051 M. The nearly equal concentrations of the acid and base components means the cytoplasm buffer has maximum capacity to neutralize both excess acid and excess base added to the system, which is ideal for maintaining constant pH for cellular reactions.

7. Quick Reference Cheatsheet

8. 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. The follow-on topics you will study next build directly on the skills you mastered in this chapter: Buffer pH and Capacity Acid-Base Titrations pH of Salt Solutions Solubility Equilibria