Learn Extracted exam questions AP Chemistry 2025 Free Response
2025 Free Response
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Answer the following questions about magnesium.
An incomplete mass spectrum for magnesium is shown in the diagram.
[Bar graph titled "Relative Abundance" vs. "Mass (amu)"; y-axis from 0 to 100 in increments of 20; x-axis shows tick marks at 24, 25, 26; a single thick vertical bar is drawn at mass 24 reaching a relative abundance of about 79.]
The percent abundance of magnesium-24 is 79%. The percent abundances of the other two natural isotopes of magnesium, magnesium-25 and magnesium-26, are approximately equal.
Complete the mass spectrum in part A by drawing thick lines in the appropriate locations to represent the percent abundance of magnesium-25 and magnesium-26.
Describe the difference in atomic structure that accounts for the difference in mass between magnesium-25 and magnesium-26.
A student prepares a $1.85\times10^{-3}\ M$ solution of $\text{Mg(NO}_3)_2(aq)$ in beaker 1 and a $2.80\times10^{-4}\ M$ solution of $\text{NaOH}(aq)$ in beaker 2, as shown.
[Particle diagram: a magnesium ion, $\text{Mg}^{2+}$, at the center surrounded by six water molecules (H2O), each oriented with its oxygen atom pointing toward the central $\text{Mg}^{2+}$ ion and its two hydrogen atoms pointing outward.]
[Beaker 1: labeled "$1.85\times10^{-3}\ M$ $\text{Mg(NO}_3)_2(aq)$", "Beaker 1", "35.00 mL".]
[Beaker 2: labeled "$2.80\times10^{-4}\ M$ $\text{NaOH}(aq)$", "Beaker 2", "50.00 mL".]
The particle diagram shown represents a magnesium ion, $\text{Mg}^{2+}$, in beaker 1. A sodium ion, $\text{Na}^+$, in beaker 2 has a weaker attraction to water than the $\text{Mg}^{2+}$ does. Explain this phenomenon using Coulomb's law and each of the following.
The relative charge of the ions
The relative radii of the ions
Calculate the pH of the solution in beaker 2.
A student combines 35.00 mL of $1.85\times10^{-3}\ M\ \text{Mg(NO}_3)_2(aq)$ with 50.00 mL of $2.80\times10^{-4}\ M\ \text{NaOH}(aq)$, as shown in the diagram. Calculate $\left[\text{Mg}^{2+}\right]$ after the two solutions are combined but before any reaction takes place. (Assume that volumes are additive.)
[Diagram of two conical/Erlenmeyer flasks being poured into an empty Beaker 3. Left flask: "35.00 mL", "$1.85\times10^{-3}\ M$", "$\text{Mg(NO}_3)_2(aq)$", labeled "Beaker 1". Right flask: "50.00 mL", "$2.80\times10^{-4}\ M$", "$\text{NaOH}(aq)$", labeled "Beaker 2". Both flasks have an arrow pointing down into the empty "Beaker 3" between them.]
The dissolution of magnesium hydroxide is represented by the following equation.
Write the expression for the solubility product constant, $K_{sp}$.
After the two solutions are combined in beaker 3 as described in part D, but before any reaction takes place, $\left[\text{OH}^-\right] = 1.65\times10^{-4}\ M$. Using your answer to part D, calculate the value of the reaction quotient, $Q$.
Using the reaction quotient, $Q$, predict whether a precipitate should form as the mixture in beaker 3 approaches equilibrium. Justify your answer.
In a separate experiment, the student adds $\text{HNO}_3(aq)$ to decrease the pH of a saturated solution containing undissolved $\text{Mg(OH)}_2(s)$. Does the amount of undissolved $\text{Mg(OH)}_2(s)$ increase, decrease, or remain the same as the $\text{HNO}_3(aq)$ is added? Justify your answer.
Answer the following questions about ascorbic acid (vitamin C).
A student combusts a sample of ascorbic acid, $\text{C}_x\text{H}_y\text{O}_x$, to determine its chemical composition. The only products of the reaction are 0.2400 mol of $\text{CO}_2$ and 2.883 g of $\text{H}_2\text{O}$.
Calculate the number of moles of $\text{H}_2\text{O}$ produced.
The mole ratio of carbon (C) to oxygen (O) is 1:1 in ascorbic acid. Based on this information and your answer to part A (i), determine the empirical formula of ascorbic acid.
Ascorbic acid, $\text{HAsc}(aq)$, acts as a weak acid, as shown in the equation.
The following titration curve was produced when a 10.0 mL sample of $\text{HAsc}(aq)$ was titrated using $0.0550\ M\ \text{NaOH}(aq)$.
[Titration curve graph: x-axis "Volume of 0.0550 M NaOH Added (mL)" from 0.0 to 20.0 in increments of 2.0; y-axis "pH" from 0 to 14 in increments of 2. The curve starts at about (0.0, 2.6), rises gradually and levels into a broad buffer region around pH 4 from about 2 mL to 14 mL, has a small inflection/bump near (14.0, 4.5)-(15.0, 5.2), then rises steeply (equivalence point) through the region 15.0-16.5 mL up to about pH 10.5, then levels off gradually approaching about (20.0, 11.3).]
Calculate the molar concentration of the ascorbic acid solution.
From the titration curve, determine the approximate $\text{p}K_a$ of ascorbic acid.
What is the value of the ratio $\dfrac{[\text{Asc}^-]}{[\text{HAsc}]}$ when the pH of the solution is 4.7?
Dehydroascorbic acid (DHAsc) can be produced by reacting ascorbic acid with the triiodide ion, $\text{I}_3^-$, as represented by the following equation.
The student runs three trials of the reaction with different initial concentrations of HAsc and $\text{I}_3^-$, producing the following data.
| Trial | $[\text{HAsc}]\ (M)$ | $[\text{I}_3^-]\ (M)$ | Initial Rate of DHAsc Formation $(M/\text{s})$ |
|---|---|---|---|
| 1 | 0.450 | 1.200 | $2.457\times10^{-4}$ |
| 2 | 0.450 | 0.600 | $1.229\times10^{-4}$ |
| 3 | 0.900 | 1.200 | $4.914\times10^{-4}$ |
The rate law for the reaction is $rate = k[\text{HAsc}][\text{I}_3^-]$. Explain how the data in the table support the conclusion that the reaction is first order with respect to $[\text{HAsc}]$.
Calculate the value of the rate constant, $k$, for the reaction. Include units with your answer.
The triiodide ion, $\text{I}_3^-$, is significantly more soluble in water than elemental iodine, $\text{I}_2$, is. Identify an intermolecular force between $\text{I}_3^-$ and water that is not present between $\text{I}_2$ and water, which could explain the difference in solubility. Lewis diagrams for $\text{I}_2$ and $\text{I}_3^-$ are provided.
[Lewis structures: $\text{I}_2$ shown as two iodine atoms singly bonded together, each with three lone pairs. $\text{I}_3^-$ shown in brackets with an overall negative charge, as three iodine atoms in a row connected by single bonds (I—I—I), each terminal iodine with three lone pairs and the central iodine with two lone pairs.]
White phosphorus is composed of $\text{P}_4$ molecules with a tetrahedral structure, as shown in the diagram on the left. Each P atom is bonded to the other three P atoms by single bonds, as shown in the incomplete Lewis diagram on the right.
[Left: a ball-and-stick tetrahedral structure of $\text{P}_4$, showing four P atoms (spheres) at the vertices of a tetrahedron connected by bonds (sticks).]
[Right: an incomplete Lewis diagram showing four P atoms arranged in a tetrahedral projection (one P at top, one P in the middle/front, two P at the bottom left and right), connected by single bonds forming the P4 skeleton, with no lone pairs drawn yet.]
In the box in part A, complete the Lewis diagram for $\text{P}_4$ by drawing the nonbonding electrons.
The reaction of white phosphorus with oxygen to form $\text{P}_4\text{O}_{10}(s)$ is thermodynamically favorable at 298 K. The reaction is represented by equation 1.
Equation 1: $\text{P}_4(s) + 5\,\text{O}_2(g) \rightarrow \text{P}_4\text{O}_{10}(s)$
The entropy change of the reaction, $\Delta S^\circ$, is negative. Using particle-level reasoning, explain why the entropy decreases as the reaction progresses.
The enthalpy change of the reaction, $\Delta H^\circ$, is also negative. A student claims that the favorability of the reaction is driven by enthalpy and not by entropy. Is the student's claim correct? Justify your answer by using the relationship between $\Delta G^\circ$, $\Delta H^\circ$, and $\Delta S^\circ$.
$\text{P}_4\text{O}_{10}(s)$ reacts exothermically with water to form phosphoric acid, as represented by equation 2.
Equation 2: $\text{P}_4\text{O}_{10}(s) + 6\,\text{H}_2\text{O}(l) \rightarrow 4\,\text{H}_3\text{PO}_4(aq)$
A chemist uses a calorimetry experiment to determine the enthalpy change for the reaction, as represented by the following diagram.
[Calorimeter diagram: a coffee-cup-style calorimeter labeled "Thermometer" (inserted through the insulating cover), "Stirrer" (inserted alongside the thermometer), "Insulating Cover" (the lid), "Inner Vessel" (the cup holding the solution, shaded), and "Outer Vessel" (the outer supporting cup).]
The chemist carries out the calorimetry experiment and records the following information.
| Quantity | Value |
|---|---|
| Mass of $\text{P}_4\text{O}_{10}$ | 0.100 g |
| Mass of $\text{H}_2\text{O}$ | 100.0 g |
| Initial temperature | $22.00^\circ\text{C}$ |
| Final temperature | $22.38^\circ\text{C}$ |
| Molar mass of $\text{P}_4\text{O}_{10}$ | 283.9 g/mol |
| Specific heat of $\text{H}_2\text{O}$ | $4.18\ \text{J}/(\text{g}\cdot^\circ\text{C})$ |
Calculate the amount of heat, $q$, released during the experiment, in kJ. Assume that the specific heat of the solution is the same as that of water.
Calculate the value of $\Delta H^\circ_{rxn}$ for equation 2 in $\text{kJ/mol}_{rxn}$. Include the sign in your answer.
The chemist weighed out 0.100 g $\text{P}_4\text{O}_{10}$ of and 100.0 g of $\text{H}_2\text{O}$ to perform a second trial. In the second trial, some of the solid $\text{P}_4\text{O}_{10}$ stuck to the weighing paper and was not transferred to the calorimeter. Given that $\text{P}_4\text{O}_{10}$ is the limiting reactant, would $\Delta T$ for the second trial be greater than, less than, or equal to the value in the first trial? Justify your answer.
$\text{P}_4(s)$ also reacts readily with $\text{Cl}_2(g)$ to produce phosphorus trichloride, $\text{PCl}_3(g)$, which in turn reacts with $\text{Cl}_2(g)$ in an equilibrium process to produce $\text{PCl}_5(g)$. The reactions are represented by equations 3 and 4.
Equation 3: $\text{P}_4(s) + 6\,\text{Cl}_2(g) \rightarrow 4\,\text{PCl}_3(g) \qquad \Delta H_1^\circ = -1148\ \text{kJ/mol}_{rxn}$
Equation 4: $\text{PCl}_3(g) + \text{Cl}_2(g) \rightleftharpoons \text{PCl}_5(g) \qquad \Delta H_2^\circ = -88\ \text{kJ/mol}_{rxn}$
Calculate the standard enthalpy of formation of $\text{PCl}_5(g)$ represented by equation 5.
Equation 5: $\dfrac{1}{4}\text{P}_4(s) + \dfrac{5}{2}\text{Cl}_2(g) \rightarrow \text{PCl}_5(g) \qquad \Delta H_f^\circ = ?$
The following particle-level diagram represents the contents of the vessel in an equilibrium mixture at 546 K involving equation 4.
[Particle diagram: a box containing multiple particles at random positions representing an equilibrium mixture; a key indicates two-atom particles (grey-grey) = $\text{Cl}_2$, four-atom tetrahedral-looking particles (grey with three white) = $\text{PCl}_3$, and six-atom particles (grey with five white) = $\text{PCl}_5$. The box shows several $\text{Cl}_2$ particles (pairs of grey circles), several $\text{PCl}_3$ particles (grey center with three white circles), and several $\text{PCl}_5$ particles (grey center with five white circles) scattered throughout.]
Equation 4 for the reaction that occurs is shown.
Equation 4: $\text{PCl}_3(g) + \text{Cl}_2(g) \rightleftharpoons \text{PCl}_5(g) \qquad \Delta H_2^\circ = -88\ \text{kJ/mol}_{rxn}$
If each particle in the diagram represents a partial pressure of 1.00 atm, what is the value of $K_p$ for the equilibrium mixture at 546 K?
Does the value of $K_p$ increase, decrease, or remain the same when the temperature is increased to 596 K? Justify your answer based on $\Delta H_2^\circ$.
A scientist is investigating the properties of a mixture of $\text{CH}_3\text{OH}$ and $\text{H}_2\text{CO}$. The scientist generates the following diagram to represent the mixture.
[Diagram in a large box showing six Lewis-structure molecules scattered at different positions and orientations, representing a mixture of $\text{CH}_3\text{OH}$ and $\text{H}_2\text{CO}$ molecules:
- Top-left: an $\text{H}_2\text{CO}$ molecule, $\ddot{\text{O}}=\text{C}$ with two H atoms attached, two lone pairs on O.
- Top-right: an $\text{H}_2\text{CO}$ molecule (different orientation), H atoms attached to C, C double-bonded to O with two lone pairs.
- Middle: a $\text{CH}_3\text{OH}$ molecule, H—C(—H)(—H)—O(—H) with two lone pairs on O, drawn with C in the center bonded to three H atoms and one O, and the O bonded to one more H.
- Middle-left: a $\text{CH}_3\text{OH}$ molecule (different orientation) with O bonded to C and H, two lone pairs on O, C bonded to three H atoms.
- Bottom-middle: an $\text{H}_2\text{CO}$ molecule, C double-bonded to O (two lone pairs), C bonded to two H atoms below.
- Bottom-right: an $\text{H}_2\text{CO}$ molecule (different orientation), similar structure with O double-bonded to C bonded to two H atoms.]
Identify the hybridization of the valence orbitals of the C atom in the $\text{H}_2\text{CO}$ molecule.
In the diagram provided, draw a SINGLE dashed line (----) to represent a strong hydrogen-bonding attraction between one $\text{CH}_3\text{OH}$ molecule and one $\text{H}_2\text{CO}$ molecule in the mixture.
The scientist plans to cool a gaseous mixture of $\text{CH}_3\text{OH}$ and $\text{H}_2\text{CO}$ to form a liquid mixture and finds data on the two compounds. The data are summarized in the table.
| Substance | Melting Point (K) | Boiling Point (K) | Enthalpy of Vaporization (kJ/mol) |
|---|---|---|---|
| $\text{CH}_3\text{OH}$ | 176 | 338 | 37.6 |
| $\text{H}_2\text{CO}$ | 181 | 254 | 24.2 |
Propose a temperature to which the mixture should be cooled such that $\text{CH}_3\text{OH}$ and $\text{H}_2\text{CO}$ will both be liquids.
The scientist analyzes the mixture after it is cooled and determines that 8.59 g of $\text{CH}_3\text{OH}(l)$ is present. Calculate the amount of thermal energy, in kJ, that was removed to condense the 8.59 g of $\text{CH}_3\text{OH}$ (molar mass 32.04 g/mol) at its boiling point.
Complete Lewis diagrams and some physical properties for compounds X and Y are given.
| Compound | X | Y |
|---|---|---|
| Lewis diagram | [Lewis structure of compound X: a central chain of three C atoms; the left C is bonded to three H atoms and to the middle C; the middle C is bonded to an O atom above it (O has two lone pairs, drawn as $:\!\text{O}\!:$) and to the right C; the right C is bonded to three H atoms and to the middle C; the middle C is also bonded to a fourth C below it, which is bonded to two H atoms.] | [Lewis structure of compound Y: a central chain analogous to X but with a Si atom in place of the middle C; the left C is bonded to three H atoms and to Si; Si is bonded to an O atom above it (O has two lone pairs, drawn as $:\!\text{O}\!:$) and to the right C; the right C is bonded to three H atoms and to Si; Si is also bonded to a C below it, which is bonded to two H atoms.] |
| Molar mass | 74.1 g/mol | 90.2 g/mol |
| Boiling point | $82^\circ\text{C}$ | $98^\circ\text{C}$ |
Based on VSEPR theory, predict the geometry around the Si atom in compound Y.
A student claims that compound Y has a higher boiling point than that of compound X because compound Y has stronger London dispersion forces. Do you agree or disagree? Justify your answer.
An equimolar mixture of the two compounds is heated. When the mixture reaches $82^\circ\text{C}$, which compound will have the higher vapor pressure? Justify your answer.
The mixture is heated to $198^\circ\text{C}$ in a sealed, rigid 12.5 L container, at which point both substances are gases and the total pressure in the container is 2.30 atm. Calculate the number of moles of gas particles in the container.
A scientist constructs a galvanic cell as shown in the diagram. As the cell operates, the $\text{Zn}(s)$ electrode increases in mass and the $\text{Al}(s)$ electrode decreases in mass. A data table with the standard reduction potentials for the substances follows the diagram.
[Galvanic cell diagram: a voltmeter reading "0.90 V" connects two electrodes across the top. The left electrode is labeled "$\text{Zn}(s)$" (with an arrow pointing to it), immersed in a beaker of "$1.0\ M\ \text{Zn(NO}_3)_2(aq)$". The right electrode is labeled "$\text{Al}(s)$" (with an arrow pointing to it), immersed in a beaker of "$1.0\ M\ \text{Al(NO}_3)_3(aq)$". The two beakers are connected by a U-shaped "Salt Bridge" with cotton plugs at each end, shown dipping into each solution.]
| Half-Reaction | $E^\circ\ (\text{V})$ |
|---|---|
| $\text{Zn}^{2+}(aq) + 2\,e^- \rightarrow \text{Zn}(s)$ | $-0.76$ |
| $\text{Al}^{3+}(aq) + 3\,e^- \rightarrow \text{Al}(s)$ | $-1.66$ |
Write the half-reaction for the oxidation that occurs at the anode.
Write the balanced net ionic equation for the overall reaction that occurs in the galvanic cell.
Initially, each electrode has a mass of 50.0 g. The cell is allowed to run for a period of time and is then stopped. Which electrode's mass changed the most? Justify your answer with a calculation.
| Reduction Half-Reaction | $E^\circ\ (\text{V})$ |
|---|---|
| $\text{Au}^{3+}(aq) + 3\,e^- \rightarrow \text{Au}(s)$ | $+1.50$ |
| $\text{Zn}^{2+}(aq) + 2\,e^- \rightarrow \text{Zn}(s)$ | $-0.76$ |
| $\text{Mn}^{2+}(aq) + 2\,e^- \rightarrow \text{Mn}(s)$ | $-1.19$ |
| $\text{Al}^{3+}(aq) + 3\,e^- \rightarrow \text{Al}(s)$ | $-1.66$ |
| $\text{Be}^{2+}(aq) + 2\,e^- \rightarrow \text{Be}(s)$ | $-1.85$ |
The standard Zn/Al cell has a value of $E^\circ_{cell}$ equal to 0.90 V. The scientist needs a galvanic cell that produces a greater voltage. The scientist has access to the chemical systems in the table. If the scientist uses the Zn half-cell and one of the other options from the table, what is the MAXIMUM voltage that could be generated at standard conditions?
Answer the following questions about the glycolate ion, $\text{C}_2\text{H}_3\text{O}_3^-$, which acts as a base in aqueous solution. A Lewis diagram for the ion is provided.
[Lewis structure of the glycolate ion in brackets with an overall negative charge: H—O—C(—H)(—H)—C(=O)—O:, i.e., an H atom bonded to an O atom (with two lone pairs) which is bonded to a C atom; that C atom is bonded to two H atoms and to a second C atom; the second C atom is double-bonded to an O atom (with two lone pairs) above it and singly bonded to another O atom (with three lone pairs) on the right.]
On the Lewis diagram in part A, circle the atom that accepts the proton when the glycolate ion reacts with water.
When the glycolate ion reacts with water, it forms glycolic acid, $\text{HC}_2\text{H}_3\text{O}_3$, according to the following equation. The $K_b$ expression for the reaction is provided.
At $25^\circ\text{C}$, a 2.5 $M$ solution of glycolate is found to have $[\text{OH}^-] = 1.3\times10^{-5}\ M$.
Calculate the value of $K_b$ for the glycolate ion.
Using your answer to part B (i), calculate the value of $K_a$ for glycolic acid at $25^\circ\text{C}$.
Glycolic acid can be produced from the hydrolysis of methyl glycolate, $\text{C}_3\text{H}_6\text{O}_3$. A proposed mechanism for the reaction is given.
Step 1: $\text{C}_3\text{H}_6\text{O}_3 + \text{H}_3\text{O}^+ \rightleftharpoons \text{C}_2\text{H}_5\text{O}_3^+ + \text{CH}_3\text{OH}$
Step 2: $\text{C}_2\text{H}_5\text{O}_3^+ + \text{H}_2\text{O} \rightleftharpoons \text{HC}_2\text{H}_3\text{O}_3 + \text{H}_3\text{O}^+$
Overall: $\text{C}_3\text{H}_6\text{O}_3 + \text{H}_2\text{O} \rightleftharpoons \text{HC}_2\text{H}_3\text{O}_3 + \text{CH}_3\text{OH}$
A student claims that $\text{H}_3\text{O}^+$ is a catalyst for the reaction. Do you agree or disagree? Justify your answer based on the mechanism given.