Kaplan MCAT General Chemistry Review

The common ion effect is when an ion is already present in the solution and affects the dissociation of a compound that contains the same ion. In this case, Ca(OCl)₂ contains calcium and will cause the reaction in equation 1 to shift to the left due to Le Châtelier’s principle. If there is no Ca(OCl)₂ and therefore no extra Ca²⁺ present in solution, more CaSO₄ will dissociate. (C) is the opposite. (A) is misleading because chlorine gas will not react with SO₄^2-. (D) is incorrect because changing from Ca(OCl)₂ to chlorine gas will remove the common ion effect and cause more dissociation of CaSO₄.

6. C

An increase in pH means that fewer H⁺ ions are present in solution. Lowering the concentration of H⁺ will shift the equilibrium in equation 2 to the right, and fewer HOCl molecules will be in solution. Because the HOCl molecules can oxidize harmful agents, the oxidizing power has been reduced as a result of the pH increase. An increase in pH will not break the HOCl compound (A), though over time it will break down. Although salts do act as buffers (B), it is still possible to change the pH of the pool. As for (D), an increase in pH will decrease [H⁺] and thus result in fewer H⁺ ions available to associate with OCl^-.

7. D

To make a supersaturated solution you must first heat the solution, which allows additional salt to dissolve. (While not all salts have a higher solubility at higher temperatures, this statement holds true for the majority of salts; an increase in temperature will generally increase the solubility of a collection of salts.) This occurs because at higher temperatures, the K_sp generally increases. The solution can then be cooled and the salt will remain dissolved, creating a supersaturated solution.

8. B

Using the top equation we can write the following:

We can manipulate the equation to solve for [H⁺], where

Using the bottom equation we can write the following: K_sp = [Ca²⁺][CO₃^2-]

We can rearrange this equation to solve for [CO₃^2-], which can then be substituted into the top equation:

Substituting into the top equation gives:

The K_a equation includes the reactant in the denominator because the reactant is aqueous (in contrast to the K_sp equation which doesn’t include the reactant in the denominator). The K_sp reactant is a salt in its solid form; solids are not included in equilibria equations.

9. B

The color change will occur slightly before the pK_a of phenol red is reached. This is because the basic form, which is prevalent above the pK_a, has a higher absorptivity than the acidic form. This means that the basic form will absorb light more strongly than the acidic form. At the pK_a the basic and acidic forms are equal (definition of pK_a), but because of the higher absorptivity of the basic form, the color will begin to change when there still is more acidic than basic molecules of phenol red. (C) and (D) are above the pK_a value, after the color change has occurred. (A) is under the pK_a but is too low. At pH 4, acidic molecules heavily dominate the solution so the absorptivity difference between acidic and basic forms does not come into play.

PASSAGE II

10. D

Hydrogen is diatomic in its elemental state, i.e., H₂ (g). The passage refers to the fact that each gaseous hydrogen atom has a single electron, for two total electrons in H₂, but the proper electron configuration of a diatomic substance requires using the bonding-antibonding model from molecular orbital theory. None of the choices are that specific, so they are not correct. (A) is a distortion; substances in their elemental state are presumed to keep their electrons in the ground state. (B) is a distortion as well; hydrogen is diatomic in its elemental state, meaning two electrons.

11. A

(B) is opposite; light is absorbed when the electron moves away from the nucleus. While electrons make the energy transitions, the energy transitions result in light radiation, not the emission of the electron itself, so (C) and (D) are incorrect.

12. B

The Lyman series transitions occur in the UV range (? = 200-400 nm). The Balmer series corresponds to four visible wavelengths (? = 400-700 nm), though the Balmer constant itself is in the UV range. Both the infrared and X-ray ranges of the light spectrum [(C) and (D)] fall far outside the transitions which characterize these spectra. Infrared rays are low-energy and have higher wavelengths, while X-rays are high-energy and have very short wavelengths. It is not necessary to memorize the specific wavelengths in different kinds of radiation as long as you have a general sense of differences in magnitude.

13. A