Kaplan MCAT General Chemistry Review

A stronger antacid can neutralize the same amount of acid while using a smaller quantity of reactant. This means that more reactant will be left over after the neutralization is complete. Often, this leftover reactant will present itself in the precipitate (some antacids, however, will dissolve in the solution; this is why the student’s logic was flawed). (A) is incorrect because the question states that the antacid is present as an excess reagent; increasing the amount of an excess reagent does not increase the amount of product unless more of the limiting reagent becomes available. (B) is incorrect for the same reason.

52. C

Sodium and potassium (and the rest of the alkali metals) form soluble salts, while aluminum, magnesium, and calcium do not; however, you do not need to know this fact in order to answer this question. The increased solubility of the compounds in group B caused the unreacted material to dissolve in solution, so the student was unable to detect any of the starting compound in the precipitate. The compounds in group A are sparingly soluble, so stronger antacids left a larger amount of unreacted solid. The alkalinity of the compound has no bearing on its ability to precipitate (A); also, you should know that strong bases containing sodium and potassium (i.e., NaOH and KOH) are just as alkaline as their counterparts which contain magnesium, calcium, and aluminum. (B) and (D) are accurate statements, but do not explain the student’s results as well.

PRACTICE SECTION 2

ANSWER KEY

1. B

2. D

3. A

4. C

5. B

6. C

7. D

8. B

9. B

10. D

11. A

12. B

13. A

14. B

15. C

16. A

17. C

18. D

19. B

20. A

21. A

22. C

23. A

24. B

25. D

26. C

27. B

28. D

29. A

30. C

31. A

32. A

33. B

34. A

35. B

36. D

37. A

38. C

39. B

40. D

41. B

42. D

43. A

44. C

45. C

46. C

47. A

48. B

49. A

50. D

51. C

52. C

PASSAGE I

1. B

The dissociation reaction tells us that the coefficients for both products are equal to 1. We use this information to write the solubility equation. The solubility equation for the dissociation of CaSO₄ is the following:

4.93 × 10^-5 = [Ca²⁺][SO₄^2-]

Because Ca²⁺ and SO₄^2- have a one-to-one ratio, they can be replaced by the variable x to solve for their concentration. Solving the equation 4.93 × 10^-5 = (x)², we obtain a concentration of 7.02 × 10^-3 M for each ion. Because one mole of ion is equivalent to one mole of salt, this is the concentration of CaSO₄ needed to equal the K_sp, at which point the solution is saturated. The concentration of CaSO₄ ions, 7.02 × 10^-3 M, can be multiplied by the volume, 3.75 × 10⁵ L, to obtain a value of 2.63 × 10³ moles. Converting moles of CaSO₄ to grams is accomplished by multiplying by the molar mass of CaSO₄, 136.14 grams/mole, to give a mass of 3.58 × 10⁵ grams needed to reach saturation. (A) is the number of moles, not grams, of CaSO₄.

2. D

One way to solve this problem is to calculate the Ca²⁺ concentration for one mole of each compound using the K_sp and chemical equilibrium equation. Looking at the answer choices, (C) can be ruled out immediately because of the extremely small K_sp, which indicates that very few of the salt ions will dissolve. The other K_sp values are comparable for (A), (B), and (D). Remember to raise the ion to the power of its coefficient when setting up the equilibrium equation. For example, the equilibrium equation for (D) would be K_sp = 6.47 × 10^-6 = [Ca²⁺][IO₃^-]². The concentration of calcium for each of these answer choices, respectively, equals 6.93 × 10^-5 M, 2.14 × 10^-4 M, and 1.17 × 10^-2 M. (D) has the highest concentration of Ca ²⁺ at 1.17 × 10^-2 M.

3. A

The concentration of a saturated solution of CaSO₄ is equal to approximately 0.015 M, according to the graph. This number was found by determining the corresponding concentration for a conductivity value of 2500 µS/ cm. The concentration of CaSO₄ is equal to the concentration of both Ca²⁺ and SO₄^2- ions in solution according to equation 1 because the coefficients are all 1. Therefore, the K_sp = [Ca²⁺ ][SO₄^2- ] = (.015 M)², which equals 2.25 × 10^-4 M². The concentration of CaSO₄ must be square to obtain the K_sp, so (B) is incorrect. (C) doubles, not squares, the concentration. (D) might be obtained if you had taken the square root, not squared, the concentration.

4. C

As temperature increases in a solution, movement of molecules and ions increase. This will increase the current between the electrodes in the probe and increase the value of conductivity. Temperature’s effect on the solubility product constant cannot be determined without further information. Although many salts have a higher solubility with higher temperatures, not all salts share this property. The only sure way to determine the relationship is by experimentation. An increase in temperature would increase the conductivity, not decrease it, so (B) and (D) are incorrect. (A) is incorrect because it is impossible to determine the relationship between temperature and K_sp.

5. B