Open with a real photo/discussion of battery acid (~5 M H₂SO₄) and drain cleaner (~2 M NaOH). Ask: 'What pH would you predict for each?' Take votes for the acid: pH 0, pH −1, or 'impossible, pH can't be negative.' Reveal that concentrated battery acid has a pH near −0.7 and 2 M NaOH has pH ~14.3. Tell them by the end of class they will calculate these values themselves and explain why the 0–14 rule they learned in first-year chem is a lie. Targets SP 1 and SP 6 — students commit to a claim they will later defend or revise.

1. 5m

   Autoionization of water and Kw

   Content

   Even perfectly pure water conducts electricity a tiny bit because water molecules transfer protons to each other: 2 H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq). This autoionization of water is an equilibrium, and its equilibrium expression is Kw = [H₃O⁺][OH⁻]. At 25 °C, Kw = 1.0 × 10⁻¹⁴. In pure water the two concentrations must be equal, so [H₃O⁺] = [OH⁻] = 1.0 × 10⁻⁷ M — that is what 'neutral' means. Kw is temperature-dependent: because autoionization is endothermic, heating water increases Kw. At 50 °C, Kw ≈ 5.5 × 10⁻¹⁴, so neutral [H₃O⁺] ≈ 2.3 × 10⁻⁷ M and neutral pH ≈ 6.63 — still neutral, just not 7.

   Delivery

   Emphasize that Kw is an equilibrium constant like any other; it just happens to have a special name. Pre-empt the misconception that 'pH 7 = neutral always' — neutral means [H₃O⁺] = [OH⁻], and only at 25 °C does that equal pH 7. Quick check: 'If I heat water to 50 °C and the pH drops to 6.63, is it acidic?' Expected answer: no, because [H₃O⁺] still equals [OH⁻]. Targets SP 4.

2. 5m

   Defining pH and pOH; the pH + pOH = 14 relationship

   Content

   pH is defined as pH = −log₁₀[H₃O⁺] and pOH = −log₁₀[OH⁻]. Taking −log of both sides of Kw = [H₃O⁺][OH⁻] gives pKw = pH + pOH. At 25 °C, pKw = −log(1.0 × 10⁻¹⁴) = 14.00, so pH + pOH = 14.00 — only at 25 °C. At 50 °C, pKw ≈ 13.26, so pH + pOH = 13.26 there. The scale is logarithmic: each unit of pH represents a factor of 10 in [H₃O⁺], so pH 2 is 100× more acidic than pH 4. Because pH is a log, pH values can absolutely be negative (12 M HCl → pH ≈ −1.08) or above 14 (10 M NaOH → pH ≈ 15). The 0–14 range is a convenient window for dilute solutions, not a law.

   Delivery

   Anchor to the number-line pH scale on the slide — point out where 5 M H₂SO₄ and 2 M NaOH from the hook sit (off the ends). Pre-empt two misconceptions: (1) 'pH can't be negative' — it can, log of a number bigger than 1 is positive so −log is negative; (2) 'pH + pOH = 14 always' — only at 25 °C. Ask a cold-call: 'If pH = 2.30, what is pOH at 25 °C?' Expected: 11.70. Targets SP 5.

3. 6m

   Strong acid/base calculations — including diprotic H₂SO₄

   Content

   For a strong monoprotic acid, complete dissociation means [H₃O⁺] equals the formal concentration directly. Example 1: 0.025 M HCl → [H₃O⁺] = 0.025 M → pH = −log(0.025) = 1.60. Correspondingly [OH⁻] = Kw / [H₃O⁺] = 1.0 × 10⁻¹⁴ / 0.025 = 4.0 × 10⁻¹³ M, pOH = 12.40, and 1.60 + 12.40 = 14.00 ✓. Example 2: strong base stoichiometry. 0.0050 M Ba(OH)₂ dissociates to give TWO OH⁻ per formula unit, so [OH⁻] = 2(0.0050) = 0.010 M, pOH = 2.00, pH = 12.00. Missing the factor of 2 gives pH 11.70 — wrong. Example 3 (H₂SO₄, the classic AP trap): 0.10 M H₂SO₄ is often taught as giving [H₃O⁺] = 0.20 M. Only the FIRST proton is truly strong. The second is HSO₄⁻ with Ka₂ ≈ 1.2 × 10⁻². At 0.10 M this is close enough to strong that AP accepts [H₃O⁺] ≈ 0.20 M and pH ≈ 0.70, but at dilute concentrations HSO₄⁻ does not fully dissociate. Point is: read the acid — do not blindly copy concentration into pH.

   Delivery

   Work all three examples on the board with students calling out steps. On Ba(OH)₂, deliberately do it wrong first (forget the 2), get pH 11.70, ask 'what did I skip?' — pre-empts the stoichiometry misconception directly. On H₂SO₄, name that this is the specific trap in AP FRQs. Targets SP 5 Mathematical Routines and SP 6 Argumentation.

4. 4m

   Significant figures in logs — the mantissa rule

   Content

   A pH value is a log, and logs have unusual sig-fig rules. Only the digits AFTER the decimal point (the mantissa) count as significant; digits before the decimal (the characteristic) fix the order of magnitude and don't count. So if [H₃O⁺] = 1.0 × 10⁻³ M has 2 sig figs, the correct pH is 3.00 — two decimal places — not 3.0 and not 3. Another example: [H₃O⁺] = 4.8 × 10⁻⁵ M (2 sig figs) → pH = 4.32 (2 decimal places). Going the other way, pH = 8.762 (3 decimal places) → [H₃O⁺] = 10⁻⁸·⁷⁶² = 1.73 × 10⁻⁹ M (3 sig figs). AP graders will dock a point for pH = 4 when they expect 4.32.

   Delivery

   Give students 30 seconds to convert [OH⁻] = 3.2 × 10⁻⁴ M into pOH with correct sig figs. Expected: pOH = 3.49 (not 3.5, not 3.495). Pre-empt: 'the 3 before the decimal is just telling us the exponent — it's free, it doesn't count.' Targets SP 5.

1. 30m

   Dilution Series Lab: measured vs. calculated pH of HCl and diprotic H₂SO₄Lab

   Groups of 3. Each group builds a five-point dilution series of HCl AND runs a single 0.050 M H₂SO₄ trial to expose the diprotic issue. They calculate expected pH, measure actual pH with a calibrated probe, then plot pH vs. −log[HCl] and argue whether the H₂SO₄ point sits where a monoprotic assumption predicts. Targets SP 2 (method), SP 3 (graphing data), SP 5 (mathematical routine), SP 6 (argument from evidence). Setup (5 min): Calibrate one probe per group using pH 4.00 and pH 7.00 buffers. Distribute 50 mL of each stock. Warn: acids into water, never water into acid. Rinse probe with DI between samples and blot with a Kimwipe — do not wipe the glass bulb. Run (18 min): Students prepare dilutions by pipeting from the previous solution into the next beaker (serial dilution) with DI water to a 25.0 mL total volume. They measure pH of each and record. Then measure the 0.050 M H₂SO₄ sample. Debrief (7 min): Groups plot pH (y) vs. −log[HCl] (x). For a pure strong monoprotic acid the plot should be a straight line of slope 1 through the origin. They mark the H₂SO₄ point on the same axes using [H₂SO₄] as the x-coordinate — it should sit BELOW the line (more acidic than a monoprotic acid of the same formal concentration). They write a one-sentence claim + evidence answering: 'Does H₂SO₄ behave as a monoprotic or diprotic strong acid at 0.050 M?' Walk around and check: (a) Are students using log correctly on the calculator? (b) Are they reporting pH to 2 decimal places? (c) Did they forget to rinse the probe between samples (drift downward)? --- Student handout: Part 1 — Prepare the HCl dilution series (all volumes total 25.0 mL) - Beaker A: 25.0 mL of 0.10 M HCl stock (no dilution) - Beaker B: 2.5 mL of Beaker A + 22.5 mL DI water - Beaker C: 2.5 mL of Beaker B + 22.5 mL DI water - Beaker D: 2.5 mL of Beaker C + 22.5 mL DI water - Beaker E: 2.5 mL of Beaker D + 22.5 mL DI water Safety: acid into water, never water into acid. Wear goggles. Part 2 — Data table (HCl) For each beaker, compute [HCl] from the dilution, calculate expected pH, then measure pH with the probe. - Beaker A: [HCl] = 0.10 M · calc pH = ______ · measured pH = ______ - Beaker B: [HCl] = ______ M · calc pH = ______ · measured pH = ______ - Beaker C: [HCl] = ______ M · calc pH = ______ · measured pH = ______ - Beaker D: [HCl] = ______ M · calc pH = ______ · measured pH = ______ - Beaker E: [HCl] = ______ M · calc pH = ______ · measured pH = ______ Part 3 — H₂SO₄ trial Prepare Beaker F: 12.5 mL of 0.10 M H₂SO₄ stock + 12.5 mL DI water → [H₂SO₄] = 0.050 M. - Predicted pH assuming ONLY the first proton dissociates: ______ - Predicted pH assuming BOTH protons fully dissociate: ______ - Measured pH of Beaker F: ______ Part 4 — Graph On the graph provided (or graph paper), plot measured pH (y-axis) vs. −log[acid] (x-axis) for beakers A–E. Draw the best-fit line. Plot the H₂SO₄ point (F) on the same axes as a differently-shaped marker. Part 5 — Claim, Evidence, Reasoning (write 2–3 sentences) Does 0.050 M H₂SO₄ behave more like a monoprotic or a diprotic strong acid? Cite your measured pH and your two predicted pHs as evidence. Explain in terms of Ka₂ of HSO₄⁻ ≈ 1.2 × 10⁻². Part 6 — pOH check For Beaker A, calculate [OH⁻] using Kw and report pOH. Confirm pH + pOH = 14.00 ± 0.05. - [OH⁻] = ______ M - pOH = ______ - pH + pOH = ______

   Materials

   • 0.10 M HCl stock (50 mL per group)
   • 0.10 M H₂SO₄ stock (50 mL per group)
   • Distilled water squirt bottle
   • Six 50-mL beakers per group
   • 10-mL and 25-mL volumetric pipets (or graduated pipets) with bulbs
   • Calibrated pH probe (LabQuest, Vernier Go Direct, or equivalent) with pH 4 and pH 7 buffer for calibration
   • Waste beaker for acid disposal
   • Kimwipes for probe
   • Calculator with log button
   • Student handout (printed from description)

   Example outputs

   • Beaker C: [HCl] = 0.0010 M, calc pH = 3.00, measured pH = 3.06. Beaker E: [HCl] = 1.0 × 10⁻⁵ M, calc pH = 5.00, measured pH ≈ 5.4 (drift toward 7 from DI water). Plot of A–E is linear with slope ≈ 1.
   • H₂SO₄ Part 3: monoprotic prediction pH = 1.30; diprotic prediction pH = 1.00; measured pH ≈ 1.05. CER: 'H₂SO₄ at 0.050 M behaves essentially as a diprotic strong acid because the measured pH (1.05) matches the diprotic prediction (1.00) far better than the monoprotic prediction (1.30). At this concentration, Ka₂ ≈ 0.012 is large enough that HSO₄⁻ dissociates almost completely.'
   • Beaker A pOH check: [OH⁻] = 1.0 × 10⁻¹⁴ / 0.10 = 1.0 × 10⁻¹³ M; pOH = 13.00; pH + pOH = 1.00 + 13.00 = 14.00 ✓

1. A 0.020 M solution of Sr(OH)₂ is prepared at 25 °C. Calculate the pH of the solution, reporting the answer with the correct number of significant figures.

   calculationSr(OH)₂ is a strong base that dissociates completely into 1 Sr²⁺ and 2 OH⁻. [OH⁻] = 2 × 0.020 M = 0.040 M pOH = −log(0.040) = 1.40 pH = 14.00 − 1.40 = 12.60 Answer: pH = 12.60 (2 decimal places because 0.040 has 2 sig figs). Targets SP 5. Common wrong answer 12.30 comes from forgetting the factor of 2.

2. Which statement best explains why pure water at 50 °C has a pH of about 6.63? (A) The water has become slightly acidic. (B) The water is still neutral because [H₃O⁺] = [OH⁻]; Kw has increased with temperature. (C) The pH scale is invalid above 25 °C. (D) An impurity in the water has lowered the pH.

   multiple choiceB. Autoionization is endothermic, so heating shifts the equilibrium to produce more H₃O⁺ and more OH⁻ equally. The water is still neutral — [H₃O⁺] still equals [OH⁻] — but both concentrations are larger than 10⁻⁷ M, so pH < 7. Neutral is defined by equal ion concentrations, not by pH = 7. Targets SP 4 and SP 6.

3. A student writes: 'The pH of 3.0 M HCl is 0 because pH cannot be negative.' Evaluate this claim. Calculate the actual pH and justify whether the student is correct.

   short answerThe student is incorrect. HCl is a strong monoprotic acid, so [H₃O⁺] = 3.0 M. pH = −log(3.0) = −0.48. Because pH is defined as −log[H₃O⁺], any [H₃O⁺] greater than 1 M gives a negative pH; the 0–14 window is a convenience for dilute solutions, not a hard limit. Answer: pH ≈ −0.48. Targets SP 5 and SP 6.

4. For 0.10 M H₂SO₄ at 25 °C, which is the best estimate of [OH⁻]? (A) 1.0 × 10⁻¹³ M (B) 5.0 × 10⁻¹⁴ M (C) 1.0 × 10⁻⁷ M (D) 0.20 M

   multiple choiceB. Assuming both protons fully dissociate (justified because Ka₂ ≈ 1.2 × 10⁻² is large at this concentration), [H₃O⁺] ≈ 0.20 M. Then [OH⁻] = Kw / [H₃O⁺] = 1.0 × 10⁻¹⁴ / 0.20 = 5.0 × 10⁻¹⁴ M. Choice A forgets the diprotic factor; C is neutral water; D confuses [OH⁻] with [H₃O⁺]. Targets SP 5.

• ☐0.10 M HCl stock (50 mL/group)
• ☐0.10 M H₂SO₄ stock (50 mL/group)
• ☐Distilled water in squirt bottles
• ☐Six 50-mL beakers per group
• ☐10-mL and 25-mL pipets with bulbs
• ☐Vernier or LabQuest pH probes, one per group, pre-calibrated
• ☐pH 4.00 and pH 7.00 buffer solutions for calibration
• ☐Kimwipes
• ☐Waste beaker for acid disposal (neutralize with NaHCO₃ before sink)
• ☐Safety goggles for every student
• ☐Printed student handout (Parts 1–6)
• ☐Calculators with log function
• ☐Graph paper or preprinted axes for pH vs. −log[acid]