AP Chemistry Exam 2024 FRQ #1 Walkthrough

The question presents a balanced equation between lactic acid (C3H6O3) and sodium hydroxide (NaOH).
It asks to circle the hydrogen atom in the structural formula of lactic acid that most readily participates in the reaction with NaOH.
The correct answer is the terminal hydrogen atom of the carboxylic acid group (-COOH), as it is the most acidic and easily ionized.

The student dissolves 10.22 grams of NaOH in enough water to produce 500 mL of solution.
We need to calculate the molarity of the NaOH solution.
First, convert grams of NaOH to moles using its molar mass (approximately 40 g/mol):

$Moles \space NaOH = \frac{10.22 \space g}{40 \space g/mol} = 0.2555 \space mol$

Next, divide the moles of NaOH by the volume of the solution in liters (500 mL = 0.5 L) to find the molarity:

$Molarity = \frac{0.2555 \space mol}{0.5 \space L} = 0.511 \space M$

A titration curve is provided, representing the titration of lactic acid with the NaOH solution.
The question asks to use the graph to determine the approximate pKa of lactic acid.
Recall that at the half-equivalence point, the pH is equal to the pKa of the weak acid.
From the titration curve, the equivalence point is at approximately 16 mL.
Therefore, the half-equivalence point is at approximately 8 mL.
The pH at 8 mL is approximately 3.9, so the pKa of lactic acid is approximately 3.9.

A diagram represents the relative amounts of major species (conjugate base and conjugate acid) in the solution at one point during the titration.
The question asks to draw an X on the titration curve at a point where the reaction mixture would be represented by the diagram and justify the placement.
The diagram shows a ratio of 2 ions of conjugate base for every 4 molecules of conjugate acid.
Use the Henderson-Hasselbalch equation to determine the pH:

$pH = pKa + log \frac{[conjugate \space base]}{[weak \space acid]}$

$pH = 3.9 + log \frac{2}{4}$

$pH = 3.9 + log(0.5)$

$pH \approx 3.6$

The student repeats the experiment with a NaOH solution that has twice the concentration.
The question asks to draw the expected titration curve for the second experiment on the same graph as the first experiment.
Key considerations:- The initial pH should be the same because the acid being titrated hasn't changed.
- Since the concentration of NaOH is doubled, it will require half the volume to reach the equivalence point (8 mL instead of 16 mL).
- At the new half-equivalence point (4 mL), the pH should equal the pKa (3.9).
The new titration curve should start at the same initial pH, reach a pH of 3.9 at 4 mL, have an inflection point at 8 mL, and then continue to rise.

The student combines 100 mL of 0.5 M lactic acid at 20°C with 100 mL of 0.5 M NaOH at 20°C in a calorimeter.
The final temperature of the combined solution is 23.2°C.
Assume the density of each solution is 1 g/mL and the specific heat capacity of the combined solution is 4.2 J/g°C.
Calculate the quantity of heat produced (released) in the reaction using $q = mc \Delta T$ :- Total volume = 100 mL + 100 mL = 200 mL
- Total mass = 200 mL * 1 g/mL = 200 g
- $\Delta T$ = 23.2°C - 20°C = 3.2°C
- $q = (200 \space g)(4.2 \space J/g°C)(3.2°C) = 2688 \space J \approx 2700 \space J$

Calculate the molar enthalpy of reaction in kJ/mol, including the sign.
First, determine the number of moles of lactic acid and NaOH:- Moles of lactic acid = (0.1 L)(0.5 M) = 0.05 mol
- Moles of NaOH = (0.1 L)(0.5 M) = 0.05 mol
Since the stoichiometry is 1:1, 0.05 moles of each reactant are consumed.
Convert the heat released to kilojoules: 2700 J = 2.7 kJ
Since the temperature increased, the reaction is exothermic, so the sign is negative: -2.7 kJ
Calculate the molar enthalpy of reaction:

$\Delta H = \frac{-2.7 \space kJ}{0.05 \space mol} = -54 \space kJ/mol$

The student claims that if heat is lost from the calorimeter to the surroundings, the experimental value of the molar enthalpy of reaction will be smaller in magnitude than the actual value.
Agree with the student's claim.
Justification: If heat is lost to the surroundings, the measured $\Delta T$ will be lower than the actual $\Delta T$ . This leads to a smaller calculated value for q. Consequently, the magnitude of $\Delta H$ will also be smaller than the actual value.