1972 Ap — Chemistry Free Repack Response Answers

The Crucible of 1972: Deconstructing the AP Chemistry Free Response

By: RetroChem Analyst

In the spring of 1972, Richard Nixon was in the White House, the Vietnam War raged on, and “America’s Top 40” featured Don McLean’s “American Pie.” But in high school gymnasiums converted into exam halls, a different kind of tension crackled. The future STEM elite were hunched over booklets labeled AP Chemistry.

For modern students, the 1972 free response section is a fascinating fossil. Before the era of calculators, data tables, or multiple-choice-heavy scoring, the 1972 exam was a raw, computational beast. It didn't test conceptual understanding as much as it tested mathematical endurance.

Here is a look back at what those students faced—and the answers that separated the 5s from the 1s.

Problem 3: Thermodynamics – Hess’s Law & ΔH

Typical Prompt: Given the following thermochemical equations at 25°C: 1972 ap chemistry free response answers

1. C(s) + O2(g) → CO2(g) ΔH = -393.5 kJ
2. H2(g) + ½O2(g) → H2O(l) ΔH = -285.8 kJ
3. C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) ΔH = -1367 kJ

Calculate the standard enthalpy of formation (ΔH°f) of C2H5OH(l).

Problem 1: The Ksp and Stoichiometry Classic (Molar Solubility)

Question Summary:
The solubility product of ( PbF_2 ) is ( 3.7 \times 10^-8 ). Calculate:
(a) The molar solubility of ( PbF_2 ) in pure water.
(b) The molar solubility of ( PbF_2 ) in a 0.10 M ( NaF ) solution.

1972 Answer Key (Validated):

• (a) In pure water:
  Let ( s ) = molar solubility.
  ( K_sp = [Pb^2+][F^-]^2 = (s)(2s)^2 = 4s^3 )
  ( 4s^3 = 3.7 \times 10^-8 )
  ( s^3 = 9.25 \times 10^-9 )
  ( s = \sqrt[3]9.25 \times 10^-9 )
  ( s \approx 2.10 \times 10^-3 , \textM ) The Crucible of 1972: Deconstructing the AP Chemistry

• (b) In 0.10 M NaF:
  Initial ([F^-] = 0.10 , \textM) from NaF (common ion).
  Let ( x ) = additional solubility from PbF₂.
  ( K_sp = [Pb^2+][F^-]^2 = (x)(0.10 + 2x)^2 ). Assume ( 2x \ll 0.10 ), so ( 0.10 + 2x \approx 0.10 ).
  ( 3.7 \times 10^-8 = x (0.10)^2 )
  ( 3.7 \times 10^-8 = x (1.0 \times 10^-2) )
  ( x = 3.7 \times 10^-6 , \textM )

Score Rubric (1972): 5 points for part (a) – setup (2), cube root (2), units (1). 5 points for part (b) – common ion effect (2), approximation (2), final (1).

Sample Question Breakdown: The "1972" Vibe

Let’s reconstruct a typical question from that year (paraphrased from actual historical prompts):

Question 3: "A 0.500 gram sample of a pure metal, X, reacts completely with excess hydrochloric acid to produce 280. mL of hydrogen gas collected over water at 25.0°C and a total pressure of 740. mm Hg. The vapor pressure of water at 25.0°C is 23.8 mm Hg. Determine the equivalent weight of the metal." C(s) + O2(g) → CO2(g) ΔH = -393

Why this question was cruel: Modern students would use a P-table and an ICE chart. 1972 students had to:

1. Correct for water vapor pressure (Dalton’s Law).
2. Convert mL to L manually.
3. Convert mm Hg to atm (mentally: 760 mm Hg = 1 atm).
4. Use the ideal gas law to find moles of $H_2$.
5. Realize that moles of metal = 2 * moles of $H_2$ (for a +2 oxidation state).
6. Divide the mass by the moles.

The Answer (1972 Style):

• Pressure of $H_2$: $740.0 - 23.8 = 716.2 \text mm Hg = 0.942 \text atm$
• Moles $H_2$: $n = \fracPVRT = \frac(0.942)(0.280)(0.0821)(298) \approx 0.0108 \text mol$
• Moles Metal (assuming divalent): $0.0108 \times 2 = 0.0216 \text mol$
• Atomic Mass: $0.500 / 0.0216 \approx 23.1 \text g/mol$ (Sodium, or near it).
• Partial credit was given if you remembered to subtract the water vapor.

Section II: Reaction Prediction & Theory

(These questions often required writing equations or explaining concepts without calculation).

Core formulas & reminders (memorize)

• PV = nRT (R = 0.08206 L·atm·mol−1·K−1)
• n = m/M (moles = mass / molar mass)
• Molarity: M = moles solute / liters solution
• q = mcΔT ; ΔHrxn = −qsolution / moles reacted (for calorimetry)
• ΔH°rxn = ΣΔH°f(products) − ΣΔH°f(reactants)
• K = [products]coeff / [reactants]coeff (heterogeneous phases omitted)
• Q vs K: Q<K → shift right; Q>K → shift left
• pH = −log[H+]; pOH = −log[OH−]; pH + pOH = 14 (at 25°C)
• Henderson–Hasselbalch: pH = pKa + log([A−]/[HA])
• E°cell = E°cathode − E°anode; ΔG° = −nFE°cell
• Rate law: rate = k[A]^m[B]^n ; ln(k) vs 1/T for activation energy (Arrhenius)
• Ideal gas rms speed: urms = sqrt(3RT/M) (R in J·mol−1·K−1, M in kg·mol−1)

Problem 4: The Quantitative Titration (Acid-Base)

Question Summary:
A 0.500 g sample of an unknown monoprotic weak acid (HA) is dissolved in water and titrated with 0.100 M NaOH. It requires 40.0 mL of NaOH to reach the phenolphthalein endpoint. Calculate the molar mass of HA.

1972 Answer Key:

• Moles of NaOH = ( 0.100 , \textmol/L \times 0.0400 , \textL = 0.00400 , \textmol )
• 1:1 reaction: moles HA = 0.00400 mol
• Molar mass = ( \frac0.500 , \textg0.00400 , \textmol = 125 , \textg/mol )

Follow-up (part b): If the pH at half-neutralization was 4.80, find ( K_a ).

• At half-neutralization, pH = ( pK_a ) → ( pK_a = 4.80 )
• ( K_a = 10^-4.80 = 1.58 \times 10^-5 )

One-week mini study plan

• Day 1: Stoichiometry, empirical/molecular formula (1 practice).
• Day 2: Gases and kinetic theory (1 practice).
• Day 3: Thermochemistry & calorimetry (1 practice).
• Day 4: Equilibrium & acid–base (2 practices).
• Day 5: Electrochemistry & kinetics (2 practices).
• Day 6: Lab techniques, titration, qualitative analysis (1 practice).
• Day 7: Mixed timed 45-min free-response: pick 3 problems across topics.