Kaplan MCAT General Chemistry Review

We are able to tell that HCO₃^- is a buffer because it can either donate or accept a proton. Being able to donate its remaining H⁺ allows it to act as an acid, while its overall negative charge allows it to act as base and accept an H⁺. (A) is correct because HCO₃^- will accept a proton from the acid introduced to the blood stream, and act as a buffer. (B) is incorrect because while it will act as a buffer, it will accept an H⁺ ion, not donate an H⁺ ion. (C) is incorrect because HCO₃^- does act as a buffer. (D) is incorrect because there is enough information.

5. B

Write down the steps as you go. First, we need to figure out what the original [H⁺] is in the rain. Because H₂SO₄ is bivalent, we double its concentration to find its contribution to the total [H⁺], which is 4.0 × 10^-3 M. We then add this to 3.2 × 10^-3 M to get [H⁺]_rain= 7.2 × 10^-3 M. If we divide this number by 4 (the factor by which the volume decreased, use V₁C₁ = V₂C₂), we get [H⁺]_rain+pure = 1.8 × 10^-3 M. Finally, we can use Kaplan’s logarithm estimation strategy to find that the pH is just below 3, so 2.8 is the answer.

6. D

A conjugate base is an acid that loses a proton. (D) shows an acid and a molecule that has lost a molecule of H₂O, so it is not an acid/conjugate base pair. The other choices show an acid and conjugate base that has one less proton.

7. C

This question tests your ability to actively read the passage and synthesize the information presented. The student would likely agree with (A) because pollution is a major cause of acid rain, and industrialization over the past 150 years has greatly increased pollution levels. The student would likely agree with (B) because, as shown in the passage, radicals are integral in the formation of acid rain. The student would likely disagree with (C) because more acid content in water would increase the conductive capacity of water. The student would likely agree with (D) because while rain is naturally acidic, acid rain contains dangerous levels of acid.

8. C

The first pK_a in this curve can be estimated by eye. It is located between the starting point (when no base had been added yet), and the first equivalence point. This point is at pH of approximately 6.3. The value of the second pK_a is notable because it is found in a slightly different way than the first pK_a. It is located at the midpoint between the first and second equivalence points. In this curve, that corresponds to pH = 10.3

9. B

Equivalence points are at the midpoint of the quickly escalating slope range. In this titration curve, the value of the first equivalence point is 7.8 and the value of the second is 12.0.

PASSAGE II

10. B

Many MCAT Physical Sciences questions include dimensional analysis calculations. If you do not know the specific heat of water from memory, you’ll want to know this constant on Test Day! (It is reported in paragraph 1.) The correct answer is (B). (A) is tempting, as it has the same magnitude as the specific heat of water in Jg^-1K^-1. You will need to convert (A) from grams to moles using the molecular mass of pure water, 18 g/mol. (C) and (D) differ from (B) and (D) by factors of 1,000, a frequent source of error in thermochemistry problems (using kJ rather than J).

11. D

This is a discrete question which draws on information from the passage. The passage does not provide an explicit comparison of intensive and extensive physical properties. You can infer from paragraph 1, however, that intensive properties do not depend on the amount of substance present in the measurement, and that extensive properties do. While viscosity is irrelevant to this specific experiment, it is an intensive property of a liquid. Mass, heat, and enthalpy are all extensive properties. (Heat is an extensive property; temperature is an intensive property.) The correct answer is (D).

12. B

Use a calorimetry equation to determine the heat capacity of the calorimeter. This setup requires referring to table 1, which describes how the student calibrated the instrument. The equation must account for all heat inputs and outputs in the closed system. In short, the heat lost by one part of the system must be gained by another part of the system. Here, the cold water heats up, and the hot water cools.

That transition is summarized by the equation m_hotC_H20(T_f -T _{i,, hot}) = m_cold C_H20(T_f - T_{i, cold}) + K_cal (T_f - T_{i, cold}). Use the density of pure liquid water, 1 g/mL, to convert the volumes of hot and cold water to mass quantities. (B) is the answer. (A) is the specific heat of water, not the heat capacity of the calorimeter. (C) results from reversing the sign of the hot water temperature change, which obtains a negative heat capacity value. (D) is off by a factor of 1,000, using water’s specific heat in kg.

13. D