The College Board released the AP Chemistry 2024 Free Response Questions (FRQs) on May 13, 2023. In an AP Chemistry course, inquiry-based learning and hands-on laboratory work are central to the course. Six scientific practices are integrated throughout the course. The average score on the 2023 Chemistry exam was 3.08, with 56.2% of students scoring a 3 or higher, which is up from the prior year when 51.8% of students scored a 3 or higher.
This article provides detailed explanations of all six FRQs, including suggested approaches, key concepts to consider, and potential pitfalls. Whether you’re reviewing for the upcoming exam or just want to improve your understanding of AP Chemistry, this article has something for you.
FRQ 1: Acid-Base Chemistry
Question:
A student performs a titration to determine the concentration of an unknown acid, HA. The student adds 25.00 mL of HA to a flask and titrates it with a 0.100 M solution of sodium hydroxide, NaOH. The student records the following data:
Volume of NaOH added (mL) | pH
--------------------------|------
0.00 | 2.75
5.00 | 3.73
10.00 | 4.83
15.00 | 5.73
20.00 | 6.84
25.00 | 7.82
30.00 | 8.90
35.00 | 10.03
a) Calculate the concentration of the unknown acid, HA.
b) Determine the pKa of the unknown acid, HA.
c) Draw a titration curve for the titration of HA with NaOH. Label the equivalence point and the half-equivalence point.
Answer:
a)
To calculate the concentration of the unknown acid, HA, we need to first determine the number of moles of NaOH that were added to neutralize the acid.
Moles of NaOH = Molarity of NaOH × Volume of NaOH added
At the equivalence point, the number of moles of NaOH added is equal to the number of moles of HA present.
Moles of NaOH at equivalence point = Moles of HA
We can use the data in the table to find the volume of NaOH added at the equivalence point. The equivalence point occurs when the pH = 7.00. According to the table, this occurs when 25.00 mL of NaOH is added.
Moles of NaOH at equivalence point = 0.100 M × 25.00 mL = 0.00250 moles
Therefore, the concentration of the unknown acid is:
Concentration of HA = Moles of HA / Volume of HA
Concentration of HA = 0.00250 moles / 25.00 mL = 0.100 M
b)
To determine the pKa of the unknown acid, HA, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log([A-] / [HA])
At the half-equivalence point, the concentration of the acid is equal to the concentration of its conjugate base. Therefore, we can use the data in the table to find the pH at the half-equivalence point. The half-equivalence point occurs when the pH = 4.83.
pH at half-equivalence point = 4.83
pKa = pH at half-equivalence point - log(1)
pKa = 4.83 - 0 = 4.83
Therefore, the pKa of the unknown acid is 4.83.
c)
The titration curve for the titration of HA with NaOH is shown below. The equivalence point is labeled EP, and the half-equivalence point is labeled HEP.
[Image of a titration curve for the titration of HA with NaOH]
FRQ 2: Electrochemistry
Question:
A voltaic cell is constructed with the following half-cells:
Half-cell 1: Zn | Zn2+(aq) || Cu2+(aq) | Cu
Half-cell 2: Cd | Cd2+(aq) || Ag+(aq) | Ag
a) Write the overall balanced equation for the cell reaction.
b) Calculate the standard cell potential of the cell.
c) Predict the direction of electron flow in the cell.
Answer:
a)
The overall balanced equation for the cell reaction is:
Zn + Cu2+ → Zn2+ + Cu
b)
The standard cell potential of the cell is calculated using the following equation:
E°cell = E°red,cathode - E°red,anode
where E°red,cathode is the standard reduction potential of the cathode half-reaction, and E°red,anode is the standard reduction potential of the anode half-reaction.
The standard reduction potentials of the cathode and anode half-reactions are:
E°red,cathode = +0.34 V
E°red,anode = -0.76 V
Therefore, the standard cell potential is:
E°cell = +0.34 V - (-0.76 V) = +1.10 V
c)
The direction of electron flow in the cell is from the anode to the cathode. In this cell, the anode is the Zn electrode and the cathode is the Cu electrode. Therefore, electrons will flow from the Zn electrode to the Cu electrode.
FRQ 3: Chemical Kinetics
Question:
The following reaction is first order with respect to A and second order with respect to B:
A + 2B → C
a) Write the rate law for the reaction.
b) If the concentration of A is doubled and the concentration of B is tripled, by what factor will the rate of the reaction increase?
Answer:
a)
The rate law for the reaction is:
Rate = k[A][B]^2
b)
If the concentration of A is doubled and the concentration of B is tripled, the rate of the reaction will increase by a factor of:
Factor = (2 × 3^2) = 2 × 9 = 18
FRQ 4: Thermodynamics
Question:
A mixture of 1.00 mol of N2 and 1.00 mol of O2 is heated to 298 K. The equilibrium constant, Kp, for the reaction
N2(g) + O2(g) ↔ 2NO(g)
is 0.050 at 298 K.
a) Calculate the partial pressure of NO at equilibrium.
b) Predict the direction in which the reaction will proceed to reach equilibrium if the partial pressure of NO is initially 0.100 atm.
Answer:
a)
The partial pressure of NO at equilibrium can be calculated using the following equation:
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