# AP Chemistry: Free Response Practice Questions

The free-response section of the AP Chemistry test can be daunting. It consists of seven questions–three long and four short–that you will have 105 minutes to answer. It is suggested that you spend 7 minutes on each short questions and 25 minutes on each long question. You will be able to use a calculator and the formula sheets provided to you in the test. Brush up on the free-response strategies, then try out the following two practice questions. Question 1 is a short question and Question 2 is a long question.

Begin by writing the balanced chemical equation:

PbI2 Pb2++ 2I

(a) We need to explain why this solution is not saturated by showing that Q < K. The solubility product constant, Ksp of PbI2 at this temperature, is 1.4 × 10−8 and needs to be compared with the ion product, Q (calculated from initial concentrations of Pb2+ and I). According to the balanced chemical equation above, we see that dissolution of 3.0 × 10−4 moles of PbI2 will produce 3.0 × 10−4 moles of Pb2+ and 6.0 × 10−4 moles of pb2+. Since the volume of solution is 500 mL: Since Q < K, we have shown that the solution is NOT saturated. This is because the ion product is smaller than the solubility constant.

(b) We need to assess how much Ineeds to be added to the solution in order to begin precipitation of PbI2. Precipitation will occur when Q > K, so we need to find the point at which Q = K to find the concentration of Iat which precipitation will begin. Remember here that the principle we are using is the common ion effect. So:

Q=[I]0 2[Pb2+]0 =Ksp =1.4×10−8

We know that the concentration of Pb2+ has not changed. So:

[I]02 [6×104]=1.4×10−8 [I]0 = 4.8 × 10−3 M

This is the total concentration of Irequired to begin concentration of PbI2. The corresponding number of moles in our 500 mL solution is given by:

# moles I= (4.8 × 10−3 moles/L) (0.500 L) = 2.4 × 10−3 moles I

We already had 6.0 × 10−4 moles of Idue to dissolution of PbI2, so the number of moles of NaI that need to be added is:

(2.4 × 10−3) − (0.6 × 10−3) = 1.8 × 10−3 moles of NaI

a. The general situation described in this question is cathodic protection of iron.

(b) A more active metal (magnesium) is electrically connected to the iron pipe and will corrode (oxidize) in place of the iron. The magnesium serves as a sacrificial anode.

For a metal to function as a sacrificial anode, it must be more active (a stronger reducer) than the metal that is protected and also be stable in the conditions present underground (sodium metal would make for a poor sacrificial anode).

c. i. The number of moles of iron corroded per meter per year is found by dividing (1g / 55.85 g/mol) = 0.0179 mol. Assume that the rates of corrosion of Fe and Mg are equal, so 0.0179 mol Mg per meter per year would be consumed, or 0.0179 mol × 24.30 g/mol Mg = 0.435 g Mg. Because one block is attached to a 200 m pipe, multiply by 200 to get 87 g per 200 m per year.

Finally, take (1,000 g block / 87 g yr-1) = 11.5 yr. For each 0.5 mole of magnesium oxidized, one mole of electrons must be released into the wire. We know that the rate of magnesium loss is 0.0179 mol m−1 yr−1 and each block is 200 m apart, so we need to multiply: 0.0179 mol m−1 yr−1 × 200 m = 3.58 mol (Mg) yr−1.

Finally, calculate (0.5 mol (MG) / 3.58 mol (Mg) yr-1) = 0.14 yr or 51 days.

d. To calculate the equilibrium constants
for the reactions, we need to calculate the voltage potential for both reactions. This is found in the standard reduction potentials table. For iron, this produces E °
red = 1.23 V + E °ox, and since −E °red = − (−0.44 V), we can substitute to make a simple equation: 1.23 V + 0.44 V = 1.67 V

For magnesium, this produces E °red = 1.23 V + E °ox, and since– E °red = –(–2.37 V), we again can substitute: 1.23 V + 2.37 V = 3.60 V.

Finally, we plug these values into the Nernst equation where log K = (nE ° / 0.0592).

The value for n is equal to the number of electrons transferred in the reaction (2 for this reaction). Calculating the values produces KFe = 1056.4 for iron and KMg = 10121 for magnesium.

e. We need to recalculate the voltage potential

again for zinc: E° red = 1.23 V + E° ox = 1.23 V − E°red = 1.23 V + 0.76 V. = 1.99 V. Then, ∆G ° = − nFE ° = − 2 mol × 96,500 C/mol × 1.99 V = 3.84 × 106J = 3.84 × 103 kJ.

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