pH and pK - AP Chemistry | Varsity Tutors

SECTION 1

Historical Context & Motivation

The chemistry of acids and bases has been central to human civilization for millennia—from the fermentation of vinegar to the extraction of dyes—yet a quantitative framework for measuring acidity did not emerge until the early twentieth century. Before that era, chemists relied on crude indicators such as litmus and taste tests to classify substances as acidic or basic, methods that were qualitative at best. The need for a precise, reproducible scale became urgent as industrial chemistry, biochemistry, and clinical medicine began demanding standardized measurements of hydrogen ion activity in solution.

1884 • Arrhenius Theory

Svante Arrhenius proposed that acids produce H⁺ ions and bases produce OH⁻ ions in water, providing the first molecular definition of acid–base behavior.

1909 • pH Scale Introduced

Søren Sørensen at the Carlsberg Laboratory in Copenhagen introduced the pH scale as the negative common logarithm of hydrogen ion concentration, enabling convenient numerical comparison of acidity.

1923 • Brønsted–Lowry Framework

Johannes Brønsted and Thomas Lowry independently redefined acids as proton donors and bases as proton acceptors, expanding acid–base theory beyond aqueous solutions and introducing the concept of conjugate acid–base pairs.

1920s–1930s • pKₐ Becomes Standard

As equilibrium constants for acid dissociation (Kₐ) were tabulated for hundreds of compounds, the logarithmic pKₐ scale became the standard shorthand, analogous to how pH simplified [H⁺].

1966 • IUPAC Formal Definition

IUPAC refined pH in terms of hydrogen ion activity rather than concentration, aligning the scale with thermodynamic rigor and ensuring accuracy in solutions of varying ionic strength.

The central question these developments addressed was deceptively simple: How can we express the enormous range of hydrogen ion concentrations (from roughly 10⁰ to 10⁻¹⁴ M) in a compact, intuitive way, and how can we similarly rank the intrinsic strength of an acid or base? The pH and pK scales answer both questions through the elegance of the common logarithm.

SECTION 2

Core Principles & Definitions

At the heart of the pH and pK framework lie a few interconnected ideas that convert unwieldy powers of ten into a manageable numerical scale. Understanding these principles is essential before tackling calculations, buffer design, or titration analysis.

pH — Measure of [H⁺]

pH = −log[H⁺]. A low pH means high hydronium concentration (acidic); a high pH means low hydronium concentration (basic). Each unit change represents a tenfold change in [H⁺].

pOH — Measure of [OH⁻]

pOH = −log[OH⁻]. At 25 °C, pH + pOH = 14.00 because

K_w

= [H⁺][OH⁻] = 1.0 × 10⁻¹⁴.

Kₐ and pKₐ — Acid Strength

Kₐ is the equilibrium constant for acid dissociation: HA ⇌ H⁺ + A⁻. pKₐ = −log Kₐ. A smaller pKₐ means a stronger acid.

Kb and pKb — Base Strength

Kb is the equilibrium constant for base hydrolysis: B + H₂O ⇌ BH⁺ + OH⁻. pKₐ + pKb = 14.00 for a conjugate acid–base pair at 25 °C.

Henderson–Hasselbalch

pH = pKₐ + log([A⁻]/[HA]). This equation links the pH of a buffer to the ratio of conjugate base to acid, making pKₐ the pivot point where [HA] = [A⁻].

SECTION 3

Visual Explanation — The pH Scale

The pH gradient bar spans 0 to 14, with representative substances and their approximate pH values. Below, the three key equations linking pH, pOH, and pK are summarized. Notice that [H⁺] values span 14 orders of magnitude while pH compresses this to a simple 0–14 number line.

The diagram above illustrates why the logarithmic transformation is so powerful. Stomach acid at pH ≈ 1 has a hydrogen ion concentration of roughly 0.1 M, while household bleach at pH ≈ 13 has [H⁺] ≈ 10⁻¹³ M—a difference of twelve orders of magnitude compressed into just 12 pH units. The color gradient reinforces the continuous nature of the scale; there is no sharp boundary between "acidic" and "basic," only a smooth transition through neutrality at pH 7. Keep this visual in mind whenever you convert between [H⁺] and pH—it grounds the abstraction in physical intuition.

SECTION 4

Mathematical Framework

The mathematical backbone of pH and pK calculations rests on equilibrium expressions, logarithmic identities, and the autoionization constant of water. Mastering the equations below will equip you for buffer problems, titration curves, and the quantitative free-response questions on the AP exam.

DEFINITION OF pH

pH = −log₁₀[H⁺] ⟺ [H⁺] = 10⁻ᵖᴴ

[H⁺] is the molar concentration of hydrogen (hydronium) ions. The base-10 logarithm converts multiplicative changes into additive ones: a tenfold increase in [H⁺] decreases pH by exactly 1.

WATER AUTOIONIZATION

K_w = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ (at 25 °C)

Taking −log of both sides: pK_w = pH + pOH = 14.00. This relationship holds in any aqueous solution at 25 °C, whether acidic, basic, or neutral.

ACID DISSOCIATION CONSTANT

Kₐ = [H⁺][A⁻] / [HA] → pKₐ = −log Kₐ

For the generic weak acid HA ⇌ H⁺ + A⁻, a larger Kₐ (smaller pKₐ) indicates a stronger acid. Strong acids like HCl have negative pKₐ values (Kₐ ≫ 1).

HENDERSON–HASSELBALCH EQUATION

pH = pKₐ + log([A⁻] / [HA])

Derived by taking −log of the Kₐ expression. When [A⁻] = [HA], log(1) = 0, so pH = pKₐ. This is the half-equivalence point in a titration and the point of maximum buffer capacity.

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Derivation Sketch

SECTION 5

pKₐ Values, Buffers, and Titration Curves

A substance's pKₐ is the single most informative number for predicting its acid–base behavior: it tells you the pH at which the acid is exactly half-dissociated, the effective pH range of any buffer prepared from it, and the shape of its titration curve. In this section we examine how pKₐ values classify acids, how buffers work, and how a titration curve encodes pKₐ graphically.

Acid	Formula	Kₐ	pKₐ	Classification
Hydrochloric	HCl	≈ 10⁶	≈ −6	Strong acid
Hydrofluoric	HF	6.6 × 10⁻⁴	3.18	Weak acid
Acetic	CH₃COOH	1.8 × 10⁻⁵	4.74	Weak acid
Carbonic (1st)	H₂CO₃	4.3 × 10⁻⁷	6.37	Weak acid
Ammonium ion	NH₄⁺	5.6 × 10⁻¹⁰	9.25	Very weak acid

Selected pKₐ values at 25 °C. A smaller pKₐ corresponds to a stronger acid.

Titration curve for acetic acid with NaOH. At the half-equivalence point (25 mL added), pH equals pKₐ (4.74) and the buffer capacity is maximized. At the equivalence point (50 mL), all CH₃COOH has been converted to CH₃COO⁻, and the solution is basic (pH ≈ 8.72) because the conjugate base hydrolyzes.

The buffer region on a titration curve is the relatively flat zone surrounding the half-equivalence point, typically spanning pH = pKₐ ± 1. Within this window, the concentrations of HA and A⁻ are both substantial, so the solution can neutralize added acid or base with only modest pH changes. Selecting an appropriate buffer for a target pH therefore reduces to choosing an acid whose pKₐ is close to that target—ideally within one unit.

SECTION 6

Worked Example — Buffer pH Calculation

A researcher prepares a buffer by dissolving 0.30 mol of acetic acid (CH₃COOH, pKₐ = 4.74) and 0.15 mol of sodium acetate (CH₃COONa) in enough water to make 1.00 L of solution. What is the pH of this buffer?

Worked example

1
Step 1 — Identify the conjugate pair and given quantities

The weak acid is CH₃COOH (HA) and its conjugate base is CH₃COO⁻ (A⁻) provided by the sodium acetate salt. We have [HA] = 0.30 mol / 1.00 L = 0.30 M and [A⁻] = 0.15 mol / 1.00 L = 0.15 M. The pKₐ is 4.74.
2
Step 2 — Apply the Henderson–Hasselbalch equation

pH = pKₐ + log([A⁻]/[HA]) = 4.74 + log(0.15/0.30). The ratio [A⁻]/[HA] = 0.50.
3
Step 3 — Evaluate the logarithm

log(0.50) = log(1/2) = −log 2 ≈ −0.301. Therefore pH = 4.74 + (−0.301) = 4.74 − 0.30.
4
Step 4 — State the final answer and check reasonableness

pH ≈ 4.44. Because there is more acid than conjugate base ([HA] > [A⁻]), the pH is below pKₐ, which is physically reasonable—the excess acid tips the equilibrium toward higher [H⁺].

pH = 4.44

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Quick Reasonableness Check

SECTION 7

Strengths, Limitations, and Common Pitfalls

Aspect	Strength / Applicability	Limitation / Pitfall
pH scale range	Compresses 14 orders of magnitude into 0–14; intuitive and universal.	pH can be negative (very concentrated strong acids) or >14 (very concentrated strong bases)—the 0–14 range is not absolute.
Henderson–Hasselbalch	Quick, closed-form buffer pH; no ICE table needed when percent dissociation is small.	Assumes x ≪ [HA] and [A⁻] (≤5% dissociation). Breaks down in very dilute solutions or near equivalence points.
pKₐ as constant	Tabulated values allow rapid prediction of acid behavior.	Kₐ (and therefore pKₐ) varies with temperature and ionic strength. Published values are typically for 25 °C and low ionic strength.
pH + pOH = 14	Simple interconversion at 25 °C.	At other temperatures, Kw changes (e.g., Kw ≈ 5.5 × 10⁻¹⁴ at 37 °C), so pH + pOH ≠ 14 exactly.
Activity vs. concentration	Concentration-based pH works well for dilute ideal solutions.	In concentrated or high-ionic-strength solutions, activity coefficients deviate significantly from unity, and rigorous pH requires activity.

SECTION 8

Connection to Advanced Theory

The pH and pK concepts you have learned form the gateway to several more sophisticated topics encountered in upper-division chemistry and biochemistry. Below is a comparison of the AP-level treatment with the fuller thermodynamic picture.

Concept	AP Chemistry Treatment	Advanced / Thermodynamic Treatment
pH definition	pH = −log[H⁺] using molar concentration	pH = −log aH⁺, where aH⁺ = γ·[H⁺] and γ is the activity coefficient from Debye–Hückel theory
Polyprotic acids	Treat each Kₐ sequentially (Kₐ₁ ≫ Kₐ₂); often ignore second dissociation for [H⁺]	Simultaneous equilibria solved via alpha (α) fraction diagrams or computational methods
Temperature dependence	Assume Kw = 10⁻¹⁴ and constant Kₐ at 25 °C	Van 't Hoff equation: ln(K₂/K₁) = −ΔH°/R × (1/T₂ − 1/T₁); Kₐ shifts with T
Buffer capacity	Qualitative: effective within pKₐ ± 1	Quantitative: β = 2.303 × C × Kₐ[H⁺]/(Kₐ + [H⁺])², where C is total buffer concentration

For the AP exam, focus on the left column; however, being aware of the right column will deepen your understanding and prepare you for college-level analytical chemistry and biochemistry. Polyprotic systems, in particular, appear on the AP exam and require careful attention to the relative magnitudes of Kₐ₁ and Kₐ₂ when determining dominant species at a given pH.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A student prepares two solutions: Solution X has pH = 3 and Solution Y has pH = 5. How many times greater is the hydrogen ion concentration of Solution X compared to Solution Y?

A. 2 times
B. 20 times
C. 100 times
D. 1000 times

Show answer (C)

The correct answer is C. [H⁺] for Solution X = 10⁻³ M and for Solution Y = 10⁻⁵ M. The ratio is 10⁻³ / 10⁻⁵ = 10² = 100. Each pH unit represents a tenfold difference, and the solutions differ by 2 pH units, so the factor is 10² = 100. A is wrong because it confuses the pH difference (2) with the concentration ratio. B is wrong because it multiplies the pH difference by 10 rather than raising 10 to the power of the difference. D is wrong because 1000 = 10³ would require a 3-unit pH difference, not 2.

Question 2: BASIC CALCULATION

What is the pH of a 0.0025 M HNO₃ solution at 25 °C? (HNO₃ is a strong acid.)

A. 2.60
B. 3.40
C. 11.40
D. 2.00

Show answer (A)

The correct answer is A. HNO₃ is a strong acid and dissociates completely, so [H⁺] = 0.0025 M = 2.5 × 10⁻³ M. pH = −log(2.5 × 10⁻³) = −(log 2.5 + log 10⁻³) = −(0.40 − 3) = 2.60. B is wrong because it would correspond to [H⁺] ≈ 4.0 × 10⁻⁴ M, not matching the given concentration. C is wrong because 11.40 is the pOH, not the pH; the solution is acidic, not basic. D is wrong because pH = 2.00 corresponds to [H⁺] = 0.010 M, which is four times greater than 0.0025 M.

Question 3: INTERMEDIATE

A buffer is prepared from 0.20 M NH₃ (Kb = 1.8 × 10⁻⁵) and 0.10 M NH₄Cl. What is the pH of this buffer at 25 °C?

A. 8.95
B. 9.55
C. 4.44
D. 10.25

Show answer (B)

The correct answer is B. First, find pKb = −log(1.8 × 10⁻⁵) = 4.74. Then pKₐ of NH₄⁺ = 14.00 − 4.74 = 9.26. Apply Henderson–Hasselbalch with the conjugate acid NH₄⁺ as HA and NH₃ as A⁻: pH = pKₐ + log([NH₃]/[NH₄⁺]) = 9.26 + log(0.20/0.10) = 9.26 + 0.30 = 9.56 ≈ 9.55. A is wrong because it omits the log correction and uses only the pKₐ minus a rounding error. C is wrong because 4.44 corresponds to using pKb directly in the equation as though the system were an acidic buffer. D is wrong because 10.25 would imply log([A⁻]/[HA]) ≈ 1.0, meaning a 10:1 ratio, which is not the case here.

Question 4: APPLIED

A biochemist needs a buffer at pH 7.40 (physiological pH) and has access to dihydrogen phosphate (H₂PO₄⁻, pKₐ₂ = 7.20) and its conjugate base hydrogen phosphate (HPO₄²⁻). (a) Write the equilibrium expression for the relevant dissociation of H₂PO₄⁻. (b) Using the Henderson–Hasselbalch equation, calculate the required mole ratio [HPO₄²⁻]/[H₂PO₄⁻]. (c) If the biochemist prepares 500 mL of buffer with total phosphate concentration 0.15 M, determine the molarity of each species. (d) Explain why this phosphate buffer is a better choice than an acetate buffer (pKₐ = 4.74) for maintaining pH 7.40.

Show answer

(a) H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻; Kₐ₂ = [H⁺][HPO₄²⁻]/[H₂PO₄⁻]. (b) pH = pKₐ + log([HPO₄²⁻]/[H₂PO₄⁻]) → 7.40 = 7.20 + log(ratio) → log(ratio) = 0.20 → ratio =

10^{0.20}

≈ 1.58. (c) Let x = [H₂PO₄⁻]. Then [HPO₄²⁻] = 1.58x and x + 1.58x = 0.15 M → 2.58x = 0.15 → x = 0.058 M, so [H₂PO₄⁻] = 0.058 M and [HPO₄²⁻] = 0.092 M. (d) An effective buffer operates within ±1 pH unit of pKₐ. Phosphate's pKₐ₂ (7.20) is only 0.20 units from the target 7.40, ensuring robust buffering. Acetate's pKₐ (4.74) is 2.66 units away, meaning virtually no buffer capacity exists at pH 7.40; the solution would behave essentially as a dilute NaOH solution with acetate as a spectator.

Question 5: CRITICAL THINKING

A student titrates 25.0 mL of an unknown monoprotic weak acid (HA) with 0.100 M NaOH. The following data are recorded: Volume NaOH (mL): 0, 5.0, 10.0, 15.0, 20.0, 25.0, 30.0 pH: 2.87, 4.14, 4.57, 4.92, 8.72, 12.05, 12.36 (a) Determine the volume of NaOH at the equivalence point and calculate the initial concentration of HA. (b) Identify the half-equivalence point and determine the pKₐ of the acid. (c) Using the pH at 0 mL, calculate Kₐ independently via an ICE table and compare with your pKₐ from part (b). (d) Explain why the equivalence-point pH is above 7, and predict how the equivalence-point pH would change if the unknown acid had a smaller pKₐ.

Show answer

(a) The equivalence point is at 20.0 mL because the pH jumps dramatically between 15.0 mL (pH 4.92, still buffering) and 20.0 mL (pH 8.72, basic — indicating the conjugate base A⁻ is now the dominant species). Moles NaOH = 0.100 M × 0.0200 L = 0.00200 mol = moles HA. [HA] = 0.00200 mol / 0.0250 L = 0.0800 M. (b) The half-equivalence point is at 10.0 mL (half of 20.0 mL). At this point, [HA] = [A⁻], so pH = pKₐ = 4.57. Thus Kₐ = 10⁻⁴·⁵⁷ = 2.69 × 10⁻⁵. (c) At 0 mL, pH = 2.87, so [H⁺] = 10⁻²·⁸⁷ = 1.35 × 10⁻³ M. ICE table for HA ⇌ H⁺ + A⁻ with initial [HA] = 0.0800 M: Kₐ = x²/(C − x) = (1.35 × 10⁻³)² / (0.0800 − 0.00135) = 1.82 × 10⁻⁶ / 0.0787 = 2.3 × 10⁻⁵, giving pKₐ = −log(2.3 × 10⁻⁵) = 4.64. This is in reasonable agreement with the half-equivalence value of 4.57; the small discrepancy arises from rounding and activity effects. (d) At equivalence, only A⁻ is in solution. This conjugate base hydrolyzes: A⁻ + H₂O ⇌ HA + OH⁻, producing OH⁻ and making the solution basic (pH 8.72). Verification: At equivalence the total volume is 45.0 mL, so [A⁻] = 0.00200 / 0.0450 = 0.0444 M. Kb = Kw/Ka = 1.0 × 10⁻¹⁴ / 2.69 × 10⁻⁵ = 3.72 × 10⁻¹⁰. [OH⁻] = √(0.0444 × 3.72 × 10⁻¹⁰) = √(1.65 × 10⁻¹¹) = 4.06 × 10⁻⁶ M, pOH = 5.39, pH = 8.61, which is consistent with the observed pH of 8.72 (small differences from rounding/activity). If the acid had a smaller pKₐ (stronger acid), its conjugate base would be weaker, hydrolyze less, and the equivalence-point pH would be lower (closer to 7). Conversely, a larger pKₐ (weaker acid) yields a stronger conjugate base and a higher equivalence-point pH.