2025 AP Chemistry FRQ Answers Explained

Part A(i)1 point

Complete a mass spectrum when Mg-24 is 79% and Mg-25 and Mg-26 are about equal.

The other two isotopes share the remaining 21%, so each gets a bar at about 10.5% (anywhere from 10% to 11% earns the point), at masses 25 and 26.

Part A(ii)1 point

What difference in atomic structure makes Mg-26 heavier than Mg-25?

Mg-26 has one more neutron. Both have 12 protons; isotopes differ only in neutrons. See mass spectrometry practice.

Part B2 points

Use Coulomb’s law to explain why Na⁺ is attracted to water less strongly than Mg²⁺, by charge and by radius.

Charge: Na⁺ has a smaller charge (+1 vs +2), and Coulombic attraction is proportional to the charges, so its attraction to the partial negative oxygen of water is weaker. Radius: Na⁺ is larger than Mg²⁺, so water’s oxygen sits farther from its nucleus, and attraction weakens as distance increases.

Each sub-part needs its own factor tied to Coulomb’s law; naming “charge” or “size” without saying how it changes the attraction does not earn the point.

Part C1 point

Calculate the pH of 2.80 × 10⁻⁴ M NaOH.

NaOH is a strong base, so [OH⁻] = 2.80 × 10⁻⁴ M. pOH = −log(2.80 × 10⁻⁴) = 3.553, and pH = 14 − 3.553 = 10.447. Three significant figures in the concentration give three decimal places in the pH. More in the pH scale.

Part D1 point

Find [Mg²⁺] after 35.00 mL of 1.85 × 10⁻³ M Mg(NO₃)₂ is mixed with 50.00 mL of the NaOH, before any reaction.

This is a dilution into 85.00 mL: M₂ = (1.85 × 10⁻³ M)(35.00 mL) ÷ 85.00 mL = 7.62 × 10⁻⁴ M. The dilution guide covers M₁V₁ = M₂V₂.

Part E3 points

Write the Ksp expression for Mg(OH)₂, calculate Q using [OH⁻] = 1.65 × 10⁻⁴ M, and predict whether a precipitate forms (Ksp = 5.61 × 10⁻¹²).

(i) K_sp = [Mg²⁺][OH⁻]². (ii) Q = (7.62 × 10⁻⁴)(1.65 × 10⁻⁴)² = 2.07 × 10⁻¹¹. (iii) Q > K_sp, so the solution holds more ions than a saturated solution can, and a precipitate forms.

Squaring [OH⁻] is the step most often missed. The comparison is explained in the reaction quotient and Ksp and molar solubility.

Part F1 point

Adding HNO₃ to saturated Mg(OH)₂ with solid present: does the amount of solid increase, decrease, or stay the same?

Decrease. H⁺ reacts with OH⁻, lowering [OH⁻] so that Q < K_sp; more Mg(OH)₂ dissolves to restore equilibrium.

Part A3 points

Combustion gives 0.2400 mol CO₂ and 2.883 g H₂O. Find the moles of water, then the empirical formula if C:O is 1:1.

(i) 2.883 g ÷ 18.02 g/mol = 0.1600 mol H₂O. (ii) Moles of H = 2 × 0.1600 = 0.3200 mol; moles of C = 0.2400 mol (one C per CO₂); moles of O = 0.2400 mol (C:O = 1:1). The ratio 0.2400 : 0.3200 : 0.2400 = 3 : 4 : 3, so the empirical formula is C₃H₄O₃. One point is for the moles of H and one for the formula.

The same steps, with the oxygen found by difference, run in the empirical formula calculator.

Part B3 points

From a titration curve for 10.0 mL of ascorbic acid with 0.0550 M NaOH, find the acid concentration, the pKa, and [Asc⁻]/[HAsc] at pH 4.7.

(i) The curve’s equivalence point is at 16.0 mL. Moles NaOH = (0.0550 M)(0.0160 L) = 8.80 × 10⁻⁴ mol = moles HAsc, so [HAsc] = 8.80 × 10⁻⁴ mol ÷ 0.0100 L = 0.0880 M. (ii) At the half-equivalence point (8.0 mL), pH = pKa ≈ 4.1 (4.0 to 4.3 accepted). (iii) Henderson-Hasselbalch: 4.7 = 4.1 + log([Asc⁻]/[HAsc]), so the ratio = 10^0.6 ≈ 4.0.

Read the pKa at the half-equivalence volume, not at the equivalence point. See titration curve pH calculations and the Henderson-Hasselbalch equation.

Part C3 points

Given rate = k[HAsc][I₃⁻] and three trials, justify first order in HAsc and find k with units.

(i) Between trials 1 and 3, [HAsc] doubles (0.450 to 0.900 M) with [I₃⁻] fixed at 1.200 M, and the rate doubles (2.457 to 4.914 × 10⁻⁴ M/s), so the order in HAsc is 1. (ii) From trial 1: k = (2.457 × 10⁻⁴ M/s) ÷ [(0.450 M)(1.200 M)] = 4.55 × 10⁻⁴ M⁻¹ s⁻¹. The value and the units are separate points. Method in the rate law guide.

Part D1 point

Name an intermolecular force between I₃⁻ and water that I₂ and water do not have.

Ion-dipole forces, because I₃⁻ is an ion and I₂ is a neutral nonpolar molecule. See intermolecular forces.

Part A1 point

Add the nonbonding electrons to the Lewis diagram of tetrahedral P₄.

Each P makes three single bonds, so each needs one lone pair to complete its octet: four lone pairs in total. See Lewis structures.

Part B2 points

For P₄(s) + 5 O₂(g) → P₄O₁₀(s), explain why entropy decreases, and decide whether favorability is driven by enthalpy rather than entropy.

(i) Gas reactants become a solid product. Gas particles are far more dispersed, with many more possible arrangements (microstates), so entropy decreases. (ii) Yes. In ΔG° = ΔH° − TΔS°, a negative ΔH° makes ΔG° more negative while a negative ΔS° makes it less negative, so only enthalpy favors the reaction. See Gibbs free energy.

Part C3 points

0.100 g P₄O₁₀ reacts with 100.0 g water; the temperature rises from 22.00 °C to 22.38 °C. Find q in kJ and ΔH for P₄O₁₀ + 6 H₂O → 4 H₃PO₄.

(i) q = mcΔT = (100.1 g)(4.18 J/g·°C)(0.38 °C) = 159 J = 0.16 kJ (two significant figures, from ΔT = 0.38 °C). Using 100.0 g gives the same answer. (ii) Moles P₄O₁₀ = 0.100 g ÷ 283.9 g/mol = 3.52 × 10⁻⁴ mol. The reaction released the heat, so ΔH = −0.16 kJ ÷ 3.52 × 10⁻⁴ mol = −450 kJ/mol_rxn. The sign is its own point.

The water gained heat, so q for the reaction is negative. Practice this in the calorimetry guide.

Part D1 point

If some P₄O₁₀ sticks to the weighing paper (it is limiting), is ΔT greater, less, or the same?

Less than. Less of the limiting reactant reacts, so less heat is released into the same mass of water.

Part E1 point

Use P₄ + 6 Cl₂ → 4 PCl₃ (ΔH = −1148 kJ) and PCl₃ + Cl₂ → PCl₅ (ΔH = −88 kJ) to find ΔH_f° of PCl₅(g).

Formation of one PCl₅ is ¼ of the first equation plus the second: ΔH_f° = ¼(−1148) + (−88) = −287 − 88 = −375 kJ/mol. See Hess’s law.

Part F2 points

A particle diagram at 546 K (1 particle = 1.00 atm) shows 4 PCl₅, 2 PCl₃, and 6 Cl₂. Find Kp, then predict how Kp changes at 596 K.

(i) K_p = P_PCl5 ÷ (P_PCl3P_Cl2) = 4.00 ÷ (2.00 × 6.00) = 0.333. (ii) Decreases. The forward reaction is exothermic (ΔH = −88 kJ/mol), so raising the temperature favors the reactants and lowers K_p. See Kc and Kp.

Parts A–C4 points, 1 each

A. The carbon in H₂CO has three electron domains (one double bond, two single bonds), so it is sp² hybridized. See hybridization.

B. The hydrogen bond runs from the H on the O–H of methanol to the O of formaldehyde. Formaldehyde’s H atoms are bonded to carbon, so they cannot hydrogen bond. See hydrogen bonding.

C(i). Both are liquid between H₂CO’s melting point (181 K) and its boiling point (254 K), so any temperature from 181 K to 254 K, such as 200 K, works.

C(ii). (8.59 g ÷ 32.04 g/mol)(37.6 kJ/mol) = 10.1 kJ removed. Condensing releases this heat, which is why it is the energy removed from the sample.

Parts A–D4 points, 1 each

A. Si has four bonding domains and no lone pairs, so its geometry is tetrahedral. See the VSEPR chart.

B. Agree. Si has more occupied electron shells than C, so Y has a larger, more polarizable electron cloud and stronger London dispersion forces, which raises its boiling point. See London dispersion forces.

C. Compound X. At 82 °C it is at its boiling point while Y is not, because X has weaker intermolecular forces, so more X molecules are in the vapor. See vapor pressure.

D. n = PV/RT = (2.30 atm)(12.5 L) ÷ [(0.08206 L·atm/mol·K)(471 K)] = 0.744 mol. Convert 198 °C to kelvin first. The gas law calculator does this step by step.

Parts A–D4 points, 1 each

A. Aluminum is oxidized at the anode: Al(s) → Al³⁺(aq) + 3 e⁻.

B. Multiply to share 6 electrons: 2 Al(s) + 3 Zn²⁺(aq) → 2 Al³⁺(aq) + 3 Zn(s). Check it with the net ionic equation calculator.

C. The zinc electrode. Per mole of reaction, 3 mol Zn (196.1 g) is deposited while 2 mol Al (53.96 g) dissolves, so the zinc mass changes more.

D. Pair the zinc half-cell with the half-reaction farthest above it, Au³⁺/Au at +1.50 V, with Zn as the anode: E°_cell = 1.50 V + 0.76 V = 2.26 V. See galvanic cells and cell potential.

Do not multiply E° values by the coefficients; cell potential does not depend on the amount of reaction.

Parts A–C4 points, 1 each

A. The proton is accepted by one of the negatively charged oxygen atoms of the carboxylate (COO⁻) group; the two C–O bonds there are equivalent by resonance.

B(i). K_b = (1.3 × 10⁻⁵)² ÷ 2.5 = 6.8 × 10⁻¹¹, because [HC₂H₃O₃] = [OH⁻] and the change in 2.5 M is negligible.

B(ii). K_a = K_w ÷ K_b = 1.0 × 10⁻¹⁴ ÷ 6.8 × 10⁻¹¹ = 1.5 × 10⁻⁴. See weak acids, Ka, and pKa.

C. Agree. H₃O⁺ is consumed in step 1 and regenerated in step 2, so it speeds the reaction without being used up, which is how a catalyst behaves. See catalysts and reaction mechanisms.