MCAT General Chemistry Review - Alexander Stone Macnow, MD 2019-2020

Solutions
Answers to Concept Checks

· 9.1

1. Solvation refers to the breaking of intermolecular forces between solute particles and between solvent particles, with formation of intermolecular forces between solute and solvent particles. In an aqueous solution, water is the solvent.

2. Solubility is the amount of solute contained in a solvent. Saturation refers to the maximum solubility of a compound at a given temperature; one cannot dissolve any more of the solute just by adding more at this temperature.

3. Solubility of solids can be increased by increasing temperature. Solubility of gases can be increased by decreasing temperature or increasing the partial pressure of the gas above the solvent (Henry’s law).

4. Group I metals, ammonium, nitrate, and acetate salts are always soluble.

· 9.2

1.

Compound

Molality

Molarity

Normality

Glucose Image

Image

Cannot be calculated directly because final volume is unknown. In dilute solutions, M ≈ m.

At higher concentrations,

M < m because

solute particles increase

overall volume of solution.

1 N (glucose does not dissociate)

Carbonate Image

Image

Approximately 24 N (twice the molality and approximate molarity)

2. Image

Thus, start with 25 mL of the stock solution and add 75 mL pure water to get 100 mL of solution with 25 ppm Cl2.

3. Image

· 9.3

1. First, write out the balanced equation: Image

Next, identify that the molar solubility x represents the amount of Ni(OH)2 that dissociates, creating x of Ni2+ and 2x of OH—. Write out the Ksp equation, and plug in the values of x to solve for Ksp:

Image

2. Start with the balanced reaction and calculation for Ksp, which can also be used to calculate Q. Keep in mind that for every x of Ba(OH)2 that dissolves, x of Ba2+ will be produced, and 2x of OH—.

Image

[Ba2+]

Ion Product

Behavior of Solution

0.5 M

(0.5 M)(1 M)2 = 0.5

0.5 > 5.0 × 10−3 → precipitation

0.1 M

(0.1 M)(0.2 M)2 = 4.0 × 10−3

4.0 × 10−3 < 5.0 × 10−3 → dissolution

0.05 M

(0.05 M)(0.1 M)2 = 5.0 × 10−4

5.0 × 10−4 < 5.0 × 10−3 → dissolution

Note that the concentration of hydroxide is double that of barium. While there will be a very small contribution of hydroxide from the autoionization of water, this amount is negligible compared to the values given in the question.

3. Start by writing the balanced reaction for the least soluble salt in the problem: Zn(OH)2 --> Zn2+ + 2OH— . Next, write out the Ksp equation and enter the variables for the concentrations. The Zn2+ concentration will equal x, the molar solubility under these conditions, but the OH- concentration will come from two contributors: the dissociated Zn(OH)2, and the 0.1 M NaOH solution. This results in the following Ksp expression: Ksp = [Zn2+][OH−]2 = (x)(0.1 + 2x)2. Since the Ksp for Zn(OH)2 is 4.1 x 10—17, x will be negligible compared to the 0.1 M from NaOH. The Ksp expression simplifies to: Ksp = (x)(0.1)2. Thus, 4.1 × 10−17 = 0.01x. x = molar solubility of Zn(OH)2= 4.1 × 10−15.

· 9.4

1. Colligative properties are those that depend on the amount of solute present, but not the actual identity of the solute particles. Examples include vapor pressure depression, boiling point elevation, freezing point depression, and osmotic pressure.

2. Molarity (M) and molality (m) are nearly equal at room temperature. This is only because 1 L solution is approximately equal to 1 kg solvent for dilute solutions (the denominators of the molarity and molality equations, respectively). For other solvents, molarity and molality differ significantly because their densities are not Image like water.

3. Image

4. Image

The new boiling point will be 373 + 2 = 375 K.

5. Image