2025 AP AP Chemistry Practice Paper with Answers

A. i. A neutral atom of $^{44}\text{Ca}$ contains 24 neutrons, whereas a neutral atom of $^{40}\text{Ca}$ contains 20 neutrons (4 more neutrons). Both atoms contain identical numbers of protons (20) and electrons (20).
ii. The peak with the greatest binding energy corresponds to the electrons closest to the nucleus (core electrons having the greatest nuclear attraction), which is the $1s$ subshell.

B. i. According to Coulomb's law, $F \propto \frac{q_1 q_2}{r^2}$. The $\text{Ca}^{2+}$ ion has a greater magnitude of positive charge ($+2$) than the $\text{K}^+$ ion ($+1$). Furthermore, because $\text{Ca}^{2+}$ has more protons (20 vs 19) attracting the same number of core electrons (18), $\text{Ca}^{2+}$ has a smaller ionic radius than $\text{K}^+$. Both the larger charge and smaller distance lead to a significantly stronger attractive force between $\text{Ca}^{2+}$ and the partial negative oxygen atom of water.
ii. Both $\text{Mg}^{2+}$ and $\text{Ca}^{2+}$ carry a $+2$ charge, but $\text{Mg}^{2+}$ has fewer electron shells (valence shell $n=2$ vs $n=3$), giving $\text{Mg}^{2+}$ a smaller ionic radius. Consequently, the center-to-center distance $r$ to the water molecule is smaller, resulting in a stronger Coulombic force of attraction for $\text{Mg}^{2+}$.

C. Moles of $\text{Ca(NO}_3)_2 = M \times V = (2.50 \times 10^{-3}\text{ mol/L}) \times (0.2500\text{ L}) = 6.25 \times 10^{-4}\text{ mol}$.
$\text{Mass} = (6.25 \times 10^{-4}\text{ mol}) \times (164.10\text{ g/mol}) = 0.10256\text{ g} \approx 0.103\text{ g}$.

D. Total volume $V_{total} = 40.00\text{ mL} + 60.00\text{ mL} = 100.00\text{ mL} = 0.1000\text{ L}$.
i. $[\text{Ca}^{2+}] = \frac{M_1 V_1}{V_{total}} = \frac{(2.50 \times 10^{-3}\text{ M})(40.00\text{ mL})}{100.00\text{ mL}} = 1.00 \times 10^{-3}\text{ M}$.
ii. $[\text{F}^-] = \frac{M_1 V_1}{V_{total}} = \frac{(4.00 \times 10^{-2}\text{ M})(60.00\text{ mL})}{100.00\text{ mL}} = 2.40 \times 10^{-2}\text{ M}$.

E. i. $Q = [\text{Ca}^{2+}][\text{F}^-]^2$.
ii. $Q = (1.00 \times 10^{-3})(2.40 \times 10^{-2})^2 = (1.00 \times 10^{-3})(5.76 \times 10^{-4}) = 5.76 \times 10^{-7}$.
iii. A precipitate of $\text{CaF}_2(s)$ will form because $Q = 5.76 \times 10^{-7} > K_{sp} = 3.90 \times 10^{-11}$. The reaction will proceed in the reverse direction to establish equilibrium, producing solid precipitate.

F. The amount of undissolved $\text{CaF}_2(s)$ will decrease. The added $\text{H}^+(aq)$ reacts with $\text{F}^-(aq)$ ions to produce $\text{HF}(aq)$ (a weak acid): $\text{H}^+(aq) + \text{F}^-(aq) \rightleftharpoons \text{HF}(aq)$. This removal of $\text{F}^-$ lowers $[\text{F}^-]$, causing $Q < K_{sp}$. By Le Châtelier's principle, the dissolution equilibrium shifts to the right (toward products), causing more solid $\text{CaF}_2(s)$ to dissolve.

Point 01: For stating that $^{44}\text{Ca}$ has 4 more neutrons than $^{40}\text{Ca}$ (or 24 neutrons vs 20 neutrons).
Point 02: For identifying the 1s subshell.
Point 03: For explaining that $\text{Ca}^{2+}$ has a greater ionic charge (+2 vs +1) and smaller ionic radius, producing a greater Coulombic attraction to water.
Point 04: For explaining that $\text{Mg}^{2+}$ has a smaller ionic radius (fewer occupied electron shells) than $\text{Ca}^{2+}$, reducing interparticle distance and increasing Coulombic attraction.
Point 05: For correctly calculating the mass of $\text{Ca(NO}_3)_2$ as 0.103 g (or 0.1026 g).
Point 06: For correctly calculating $[\text{Ca}^{2+}] = 1.00 \times 10^{-3}\text{ M}$.
Point 07: For correctly calculating $[\text{F}^-] = 2.40 \times 10^{-2}\text{ M}$.
Point 08: For correctly calculating $Q = [\text{Ca}^{2+}][\text{F}^-]^2 = 5.76 \times 10^{-7}$ (consistent with parts D(i) and D(ii)).
Point 09: For stating that a precipitate will form and justifying by showing that $Q > K_{sp}$.
Point 10: For answering 'decrease' and explaining that $\text{H}^+$ consumes $\text{F}^-$ (forming HF), lowering $[\text{F}^-]$ and shifting the dissolution equilibrium to the right.

A. i. Moles of $\text{C} = 3.963\text{ g } \text{CO}_2 \times \frac{1\text{ mol } \text{CO}_2}{44.01\text{ g } \text{CO}_2} \times \frac{1\text{ mol C}}{1\text{ mol } \text{CO}_2} = 0.09005\text{ mol C}$.
Moles of $\text{H} = 1.081\text{ g } \text{H}_2\text{O} \times \frac{1\text{ mol } \text{H}_2\text{O}}{18.02\text{ g } \text{H}_2\text{O}} \times \frac{2\text{ mol H}}{1\text{ mol } \text{H}_2\text{O}} = 0.11998\text{ mol H} \approx 0.1200\text{ mol H}$.

ii. Mass of $\text{C} = 0.09005\text{ mol} \times 12.011\text{ g/mol} = 1.0816\text{ g}$.
Mass of $\text{H} = 0.11998\text{ mol} \times 1.008\text{ g/mol} = 0.1209\text{ g}$.
Mass of $\text{O} = 1.682\text{ g} - (1.0816\text{ g} + 0.1209\text{ g}) = 0.4795\text{ g}$.
Moles of $\text{O} = \frac{0.4795\text{ g}}{16.00\text{ g/mol}} = 0.02997\text{ mol O}$.
Mole ratio: $\text{C} : \text{H} : \text{O} = \frac{0.09005}{0.02997} : \frac{0.11998}{0.02997} : \frac{0.02997}{0.02997} = 3.00 : 4.00 : 1.00$.
Empirical formula is $\text{C}_3\text{H}_4\text{O}$.

B. i. At equivalence point: $\text{mol NaOH} = (0.1000\text{ M})(0.02000\text{ L}) = 2.000 \times 10^{-3}\text{ mol}$.
Because $\text{HSor} + \text{OH}^- \rightarrow \text{Sor}^- + \text{H}_2\text{O}$ reacts in a $1:1$ ratio:
$[\text{HSor}] = \frac{2.000 \times 10^{-3}\text{ mol}}{0.02500\text{ L}} = 0.08000\text{ M}$.

ii. At the half-equivalence point, $[\text{HSor}] = [\text{Sor}^-]$, so $\text{pH} = \text{p}K_a = 4.76$.
$K_a = 10^{-\text{p}K_a} = 10^{-4.76} = 1.74 \times 10^{-5}$.

iii. Using the Henderson-Hasselbalch equation:
$$\text{pH} = \text{p}K_a + \log\left(\frac{[\text{Sor}^-]}{[\text{HSor}]}\right)$$
$$5.36 = 4.76 + \log\left(\frac{[\text{Sor}^-]}{[\text{HSor}]}\right)$$
$$\log\left(\frac{[\text{Sor}^-]}{[\text{HSor}]}\right) = 5.36 - 4.76 = 0.60$$
$$\frac{[\text{Sor}^-]}{[\text{HSor}]} = 10^{0.60} = 3.98 \approx 4.0$$

C. i. Comparing Trials 1 and 2: when $[\text{HSor}]$ doubles from $0.050\text{ M}$ to $0.100\text{ M}$ with $[\text{ICl}]$ held constant at $0.020\text{ M}$, the rate doubles from $1.40 \times 10^{-4}$ to $2.80 \times 10^{-4}\text{ M/s}$ (factor of $2^1$). Thus, the reaction is first order with respect to $\text{HSor}$.
Comparing Trials 2 and 3: when $[\text{ICl}]$ triples from $0.020\text{ M}$ to $0.060\text{ M}$ with $[\text{HSor}]$ held constant at $0.100\text{ M}$, the rate triples from $2.80 \times 10^{-4}$ to $8.40 \times 10^{-4}\text{ M/s}$ (factor of $3^1$). Thus, the reaction is first order with respect to $\text{ICl}$.

ii. Rate law: $\text{Rate} = k[\text{HSor}][\text{ICl}]$.
Using Trial 1 data: $1.40 \times 10^{-4}\text{ M}\cdot\text{s}^{-1} = k (0.050\text{ M})(0.020\text{ M})$.
$k = \frac{1.40 \times 10^{-4}\text{ M/s}}{1.00 \times 10^{-3}\text{ M}^2} = 0.14\text{ M}^{-1}\text{s}^{-1}$ (or $\text{L}\cdot\text{mol}^{-1}\cdot\text{s}^{-1}$).

D. Potassium sorbate is an ionic compound that dissociates in water into $\text{K}^+$ and $\text{Sor}^-$ ions, which interact with polar water molecules via strong ion-dipole forces. Neutral sorbic acid ($\text{HSor}$) contains a nonpolar 5-carbon hydrocarbon tail, and while its $-\text{COOH}$ group can form hydrogen bonds, the nonpolar portion makes hydration less favorable than the strong ion-dipole attractions in $\text{KSor}$.

Point 01: For calculating the correct moles of carbon (0.09005 mol) and hydrogen (0.1200 mol).
Point 02: For calculating the mass and moles of oxygen (0.02997 mol).
Point 03: For stating the correct empirical formula $\text{C}_3\text{H}_4\text{O}$ supported by mole ratio work.
Point 04: For correctly calculating the initial concentration of sorbic acid as $0.08000\text{ M}$.
Point 05: For identifying $\text{p}K_a = 4.76$ and calculating $K_a = 1.74 \times 10^{-5}$ (accept range $1.7 \times 10^{-5}$ to $1.8 \times 10^{-5}$).
Point 06: For correctly calculating the ratio $\frac{[\text{Sor}^-]}{[\text{HSor}]} = 3.98$ or $4.0$.
Point 07: For determining that the reaction is first order with respect to $\text{HSor}$ and first order with respect to $\text{ICl}$ with correct justification referencing trial ratios.
Point 08: For calculating the correct numerical value of $k = 0.14$.
Point 09: For reporting the correct units of $k$ as $\text{M}^{-1}\text{s}^{-1}$ or $\text{L}\cdot\text{mol}^{-1}\cdot\text{s}^{-1}$.
Point 10: For explaining that $\text{KSor}$ forms strong ion-dipole interactions with water, which are much stronger and more favorable than the interactions formed by neutral $\text{HSor}$.

A. Sulfur has 6 valence electrons and each oxygen has 6 valence electrons, total 18 valence electrons. To minimize formal charges, sulfur forms double bonds to both oxygen atoms: $\ddot{\text{O}}=\ddot{\text{S}}=\ddot{\text{O}}$, with 2 lone pairs on each oxygen and 1 lone pair on sulfur. All atoms have formal charge 0.

B. i. $\Delta S^\circ$ is negative ($\Delta S^\circ < 0$). In the forward reaction, 3 moles of gas particles ($2\,\text{SO}_2 + 1\,\text{O}_2$) are converted into 2 moles of gas particles ($2\,\text{SO}_3$). A decrease in the number of gaseous particles results in a decrease in the number of possible microstates and spatial dispersal of matter.
ii. The reaction is favorable only at low temperatures. In $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$, since $\Delta H^\circ < 0$ (exothermic) and $\Delta S^\circ < 0$, the term $-T\Delta S^\circ$ is positive. For $\Delta G^\circ$ to be negative, the magnitude of $\Delta H^\circ$ must exceed $T|\Delta S^\circ|$, which occurs only at sufficiently low temperatures ($T < \frac{\Delta H^\circ}{\Delta S^\circ}$).

C. i. $\Delta T = 24.70^\circ\text{C} - 21.50^\circ\text{C} = 3.20^\circ\text{C}$.
$q_{surr} = m \cdot c \cdot \Delta T = (102.5\text{ g}) \times (4.18\text{ J}/(\text{g}\cdot^\circ\text{C})) \times (3.20^\circ\text{C}) = 1371\text{ J} = 1.37\text{ kJ}$.

ii. $\text{Moles of } \text{Na}_2\text{SO}_3 = \frac{2.52\text{ g}}{126.04\text{ g/mol}} = 0.01999\text{ mol} \approx 0.0200\text{ mol}$.
$q_{rxn} = -q_{surr} = -1.371\text{ kJ}$.
$\Delta H_{rxn}^\circ = \frac{-1.371\text{ kJ}}{0.01999\text{ mol}} = -68.6\text{ kJ/mol}_{rxn}$ (or $-68.5\text{ kJ/mol}_{rxn}$ using $0.0200\text{ mol}$).

D. Less than. Because heat escaped to the surroundings, the measured final temperature and thus $\Delta T$ would be lower than the true insulated value. Consequently, the calculated $q_{surr}$ and the resulting magnitude of $\Delta H_{rxn}^\circ$ would be smaller (less than the actual value).

E. Using Hess's law:
Reverse Reaction (a) and multiply by 2: $2\,\text{SO}_2(g) \rightarrow 2\,\text{S}(s) + 2\,\text{O}_2(g) \quad \Delta H_1 = -2(-296.8\text{ kJ}) = +593.6\text{ kJ}$.
Keep Reaction (b): $2\,\text{S}(s) + 3\,\text{O}_2(g) \rightarrow 2\,\text{SO}_3(g) \quad \Delta H_2 = -791.4\text{ kJ}$.
Sum of reactions: $2\,\text{SO}_2(g) + \text{O}_2(g) \rightarrow 2\,\text{SO}_3(g)$.
$\Delta H^\circ = \Delta H_1 + \Delta H_2 = +593.6\text{ kJ} + (-791.4\text{ kJ}) = -197.8\text{ kJ/mol}_{rxn}$.

F. i. $K_p = \frac{(P_{\text{SO}_3})^2}{(P_{\text{SO}_2})^2(P_{\text{O}_2})} = \frac{(1.20)^2}{(0.20)^2(0.10)} = \frac{1.44}{(0.040)(0.10)} = \frac{1.44}{0.0040} = 360$.
ii. The value of $K_p$ will decrease. The forward reaction is exothermic ($\Delta H^\circ < 0$). According to Le Châtelier's principle, an increase in temperature shifts the equilibrium in the endothermic (reverse) direction to absorb added heat, reducing the partial pressure of product $\text{SO}_3$ and increasing the partial pressures of reactants $\text{SO}_2$ and $\text{O}_2$, thereby decreasing $K_p$.

Point 01: For drawing a valid Lewis diagram of $\text{SO}_2$ with an expanded octet (two double bonds, one lone pair on S, two lone pairs on each O) that minimizes formal charge.
Point 02: For predicting that $\Delta S^\circ < 0$ and explaining that fewer moles of gas in products (2 mol vs 3 mol) means fewer microstates/lower disorder.
Point 03: For choosing 'low temperatures' and justifying with $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ showing that negative $\Delta H^\circ$ dominates when $T$ is small.
Point 04: For correctly calculating $q = 1.37\text{ kJ}$ (or $1370\text{ J}$) with correct significant figures (3 sig figs).
Point 05: For calculating $\Delta H_{rxn}^\circ = -68.5\text{ kJ/mol}_{rxn}$ (or $-68.6\text{ kJ/mol}_{rxn}$).
Point 06: For including the negative sign for $\Delta H_{rxn}^\circ$ consistent with an exothermic reaction.
Point 07: For predicting 'less than' and explaining that heat loss lowers the observed $\Delta T$, lowering calculated $q$.
Point 08: For correctly calculating $\Delta H^\circ = -197.8\text{ kJ/mol}_{rxn}$ using Hess's Law.
Point 09: For correctly setting up and calculating $K_p = 360$.
Point 10: For stating that $K_p$ decreases and justifying using Le Châtelier's principle and the exothermic nature of the forward reaction.

A. The central carbon atom in formamide is bonded to three electron domains (one single bond to H, one double bond to O, and one single bond to N) with zero lone pairs. Three electron domains corresponds to \(sp^2\) hybridization.

B. Hydrogen bonding occurs when a hydrogen atom covalently bonded to an electronegative atom (N, O, or F) is attracted to a lone pair on an electronegative atom of another molecule. Acceptable interactions include:
- The \(\text{H}\) atom of the \(-\text{OH}\) group on methanol interacts with a lone pair on the carbonyl oxygen atom of formamide.
- An \(\text{H}\) atom of the \(-\text{NH}_2\) group on formamide interacts with a lone pair on the oxygen atom of methanol.

C. i. For both substances to exist in the liquid state, the temperature must be above the higher melting point (\(276\text{ K}\)) and below the lower boiling point (\(338\text{ K}\)). Any temperature strictly between \(276\text{ K}\) and \(338\text{ K}\) is correct (for example, \(298\text{ K}\) or \(300\text{ K}\)).

ii.
$$\text{Moles of } \text{CH}_3\text{OH} = 6.40\text{ g} \times \frac{1\text{ mol}}{32.04\text{ g}} = 0.1998\text{ mol}$$
$$\text{Thermal energy released } q = 0.1998\text{ mol} \times 37.6\text{ kJ/mol} = 7.51\text{ kJ}$$

Part A (1 point):
- 1 point for the correct hybridization: \(sp^2\).

Part B (1 point):
- 1 point for correctly identifying an H atom covalently bonded to O or N on one molecule and an O or N atom with lone pairs on the other molecule (e.g., hydroxyl H of methanol and carbonyl O of formamide, or amino H of formamide and hydroxyl O of methanol).

Part C(i) (1 point):
- 1 point for proposing any temperature strictly between \(276\text{ K}\) and \(338\text{ K}\) (or equivalent in °C, \(3^\circ\text{C}\) to \(65^\circ\text{C}\)).

Part C(ii) (1 point):
- 1 point for the correct calculation of energy: \(7.51\text{ kJ}\) (accept \(-7.51\text{ kJ}\)). Correct calculation showing conversion of grams to moles using \(32.04\text{ g/mol}\) and multiplication by \(37.6\text{ kJ/mol}\).

A. The central carbon atom in 2,2-dimethylpropane is bonded to 4 methyl (\(-\text{CH}_3\)) groups with zero lone pairs. Four bonding pairs with zero lone pairs gives a tetrahedral geometry.

B. Agree. Both molecules are nonpolar hydrocarbons with the same molecular formula and molar mass. However, pentane is a linear/elongated chain molecule, which gives it a larger surface area of contact between adjacent molecules compared to the compact, spherical shape of 2,2-dimethylpropane. This greater surface area allows pentane's electron cloud to be more easily distorted (more polarizable), leading to stronger London dispersion forces and a higher boiling point.

C. Compound X (2,2-dimethylpropane) will have the higher vapor pressure. At \(20.0^\circ\text{C}\), 2,2-dimethylpropane is above its boiling point (\(9.5^\circ\text{C}\)) and has weaker intermolecular attractions than pentane, meaning molecules more readily enter the vapor phase.

D. Using the ideal gas law \(PV = nRT\):
$$T = 100.0 + 273.15 = 373.15\text{ K}$$
$$P = \frac{nRT}{V} = \frac{(0.294\text{ mol})(0.08206\text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K}))(373.15\text{ K})}{5.00\text{ L}} = 1.80\text{ atm}$$

Part A (1 point):
- 1 point for predicting tetrahedral geometry.

Part B (1 point):
- 1 point for agreeing and providing a valid explanation stating that pentane's linear/elongated shape provides more surface contact / greater polarizability of its electron cloud, resulting in stronger London dispersion forces.

Part C (1 point):
- 1 point for selecting Compound X with a valid justification relating lower boiling point / weaker intermolecular forces to higher vapor pressure.

Part D (1 point):
- 1 point for the correct calculated pressure of \(1.80\text{ atm}\) (accept \(1.8\text{ atm}\)), with appropriate substitution into \(PV = nRT\) and temperature in kelvins (\(373\text{ K}\)).

A. The nickel electrode decreases in mass, indicating oxidation occurs at the nickel anode:
$$\text{Ni}(s) \rightarrow \text{Ni}^{2+}(aq) + 2\,e^-$$

B. The reduction half-reaction is \(\text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s)\). Multiplying by 2 to balance electrons and adding to the oxidation half-reaction:
$$\text{Ni}(s) + 2\,\text{Ag}^+(aq) \rightarrow \text{Ni}^{2+}(aq) + 2\,\text{Ag}(s)$$

C. When \(0.0500\text{ mol}\) of electrons transfer:
- Nickel lost:
$$\Delta m_{\text{Ni}} = 0.0500\text{ mol } e^- \times \frac{1\text{ mol Ni}}{2\text{ mol } e^-} \times \frac{58.69\text{ g Ni}}{1\text{ mol Ni}} = 1.47\text{ g}$$
- Silver gained:
$$\Delta m_{\text{Ag}} = 0.0500\text{ mol } e^- \times \frac{1\text{ mol Ag}}{1\text{ mol } e^-} \times \frac{107.87\text{ g Ag}}{1\text{ mol Ag}} = 5.39\text{ g}$$
Therefore, the silver electrode experienced the greater change in mass.

D. To maximize \(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\) involving the nickel half-cell (\(E^\circ = -0.25\text{ V}\)):
- If Ni is the anode (\(E^\circ_{\text{ox}} = +0.25\text{ V}\)), pairing with the highest reduction potential (\(\text{Ag}^+\), \(+0.80\text{ V}\)) yields \(E^\circ_{\text{cell}} = 0.80 - (-0.25) = 1.05\text{ V}\).
- If Ni is the cathode (\(E^\circ_{\text{red}} = -0.25\text{ V}\)), pairing with the lowest reduction potential (\(\text{Mg}\), \(-2.37\text{ V}\)) as the anode yields:
$$E^\circ_{\text{cell}} = -0.25\text{ V} - (-2.37\text{ V}) = +2.12\text{ V}$$
The maximum voltage that can be generated is \(2.12\text{ V}\).

Part A (1 point):
- 1 point for the correct oxidation half-reaction: \(\text{Ni}(s) \rightarrow \text{Ni}^{2+}(aq) + 2\,e^-\) (states not required).

Part B (1 point):
- 1 point for the correct balanced net ionic equation: \(\text{Ni}(s) + 2\,\text{Ag}^+(aq) \rightarrow \text{Ni}^{2+}(aq) + 2\,\text{Ag}(s)\).

Part C (1 point):
- 1 point for identifying the silver (\(\text{Ag}\)) electrode and showing a correct quantitative comparison based on stoichiometry and molar masses (\(5.39\text{ g}\) of Ag vs \(1.47\text{ g}\) of Ni, or \(215.7\text{ g Ag}\) per \(58.69\text{ g Ni}\)).

Part D (1 point):
- 1 point for the correct calculated maximum standard cell potential: \(2.12\text{ V}\) (using Mg as anode and Ni as cathode).

A. The proton (\(\text{H}^+\)) is accepted by an oxygen atom in the carboxylate group (\(-\text{COO}^-\)), because the negative charge and lone pairs are localized/delocalized on the two carboxylate oxygen atoms.

B. i. From the stoichiometry of the reaction, \([\text{C}_2\text{H}_5\text{COOH}] = [\text{OH}^-] = 1.8 \times 10^{-5}\text{ M}\).
$$K_b = \frac{[\text{C}_2\text{H}_5\text{COOH}][\text{OH}^-]}{[\text{C}_2\text{H}_5\text{COO}^-]} = \frac{(1.8 \times 10^{-5})(1.8 \times 10^{-5})}{0.40 - 1.8 \times 10^{-5}} \approx \frac{(1.8 \times 10^{-5})^2}{0.40} = 8.1 \times 10^{-10}$$

ii. Using \(K_w = K_a \times K_b = 1.0 \times 10^{-14}\):
$$K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{8.1 \times 10^{-10}} = 1.2 \times 10^{-5}$$

C. Agree. \(\text{H}_3\text{O}^+\) is consumed as a reactant in Step 1 and regenerated as a product in Step 2 without being permanently consumed in the overall reaction, which is the defining behavior of a catalyst.

Part A (1 point):
- 1 point for stating either of the oxygen atoms in the carboxylate group (\(-\text{COO}^-\)).

Part B(i) (1 point):
- 1 point for the correct calculated value: \(K_b = 8.1 \times 10^{-10}\) (accept \(8.10 \times 10^{-10}\)).

Part B(ii) (1 point):
- 1 point for the correct calculated value of \(K_a\) consistent with part B(i): \(K_a = \frac{1.0 \times 10^{-14}}{K_b} = 1.2 \times 10^{-5}\).

Part C (1 point):
- 1 point for agreeing and providing a valid justification stating that \(\text{H}_3\text{O}^+\) is consumed in Step 1 and regenerated in Step 2.