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Home , Ideal solution, Lattice energy, Solution

CHAPTER ELEVEN

PROPERTIES OF

For Review

1. Mass percent: the percent by mass of the solute in the .

Mole fraction: the ratio of the number of moles of a given component to the total number of moles of solution.

Molarity: the number of moles of solute per liter of solution.

Molality: the number of moles of solute per kilogram of .

Volume is dependent, whereas mass and the number of moles are not. Only molarity has a volume term so only molarity is temperature dependent.

+ − + − + − 2. KF(s) → K (aq) + F (aq) ∆H = ∆Hsoln; K (g) + Cl (g) → K (aq) + F (aq) ∆H = ∆Hhyd

+ − KF(s) → K (g) + F (g) ∆H1 = !∆HLE + − + − K(g) + F (g) → K (aq) + F (aq) ∆H2 = ∆Hhyd ______+ − KF(s) → K (aq) + F (aq) ∆H = ∆Hsoln = !∆HLE + ∆Hhyd

It is true that ∆H1 and ∆H2 have large magnitudes for their values; however, the signs are opposite (∆H1 is large and positive because it is the reverse of the lattice and ∆H2, the hydration energy, is large and negative). These two ∆H values basically cancel out each other

giving a ∆Hsoln value close to zero.

3. “Like dissolves like” refers to the nature of the intermolecular . Polar solutes and ionic solutes dissolve in polar because the types of intermolecular forces present in solute and solvent are similar. When they dissolve, the strength of the intermolecular forces in solution are about the same as in pure solute and pure solvent. The same is true for nonpolar solutes in nonpolar solvents. The strength of the intermolecular forces (London forces) are about the same in solution as in pure solute and pure solvent. In all cases of like

dissolves like, the magnitude of ∆Hsoln is either a small positive number (endothermic) or a small negative number (exothermic). For polar solutes in nonpolar solvents and vice versa,

∆Hsoln is a very large, unfavorable value (very endothermic). Because the energetics are so unfavorable, polar solutes do not dissolve in nonpolar solvents and vice versa.

4. Structure effects refer to solute and solvent having similar polarities in order for solution formation to occur. Hydrophobic solutes are mostly nonpolar substances that are “- fearing.” Hydrophilic solutes are mostly polar or ionic substances that are “water-loving.”

Pressure has little effect on the of or ; it does significantly affect the of a . Henry’s law states that the amount of a gas dissolved in a solution is directly proportional to the of the gas above the solution (C = kP). The equation for CHAPTER 11 PROPERTIES OF SOLUTIONS 2

Henry’s law works best for dilute solutions of that do not dissociate in or react with the solvent. HCl(g) does not follow Henry’s law because it dissociates into H+(aq) and Cl−(aq) in solution (HCl is a strong ). For O2 and N2, Henry’s law works well since these gases do not react with the water solvent.

An increase in temperature can either increase or decrease the solubility of a solute in water. It is true that a solute dissolves more rapidly with an increase in temperature, but the amount of solid solute that dissolves to form a saturated solution can either decrease or increase with temperature. The temperature effect is difficult to predict for solid solutes. However, the temperature effect for gas solutes is easier to predict as the solubility of a gas typically decreases with increasing temperature.

o 5. Raoult’s law: Psoln = χ solventPsolvent ; When a solute is added to a solvent, the of

a solution is lowered from that of the pure solvent. The quantity χsolvent, the of solvent, is the fraction that the solution vapor pressure is lowered.

For the experiment illustrated in Fig. 11.9, the beaker of water will stop having a net transfer of water molecules out of the beaker when the equilibrium vapor pressure, P o , is reached. H2O This can never happen. The beaker with the solution wants an equilibrium vapor pressure of P , which is less than P o . When the vapor pressure over the solution is above P, a H2O H2O H2O net transfer of water molecules into the solution will occur in order to try to reduce the vapor pressure to P . The two beakers can never obtain the equilibrium vapor pressure they want. H2O The net transfer of water molecules from the beaker of water to the beaker of solution stops after all of the water has evaporated.

If the solute is volatile, then we can get a transfer of both the solute and solvent back and forth between the beakers. A state can be reached in this experiment where both beakers have the same solute and hence the same vapor pressure. When this state is reached, no net transfer of solute or water molecules occurs between the beakers so the levels of solution remain constant.

When both substances in a solution are volatile, then Raoult’s law applies to both. The total vapor pressure above the solution is the equilibrium vapor pressure of the solvent plus the equilibrium vapor pressure of the solute. Mathematically:

o o PTOT = PA + PB = χ APA + χ BPB

where A is either the solute or solvent and B is the other one.

6. An ideal -liquid solution follows Raoult’s law:

o o PTOT = χ APA + χ BPB A nonideal liquid-liquid solution does not follow Raoult’s law, either giving a total pressure greater than predicted by Raoult’s law (positive deviation) or less than predicted (negative deviation).

In an , the strength of the intermolecular forces in solution are equal to the

strength of the intermolecular forces in pure solute and pure solvent. When this is true, ∆Hsoln CHAPTER 11 PROPERTIES OF SOLUTIONS 3

= 0 and ∆Tsoln = 0. For positive deviations from Raoult’s law, the solution has weaker intermolecular forces in solution than in pure solute and pure solvent. Positive deviations

have ∆Hsoln > 0 (are endothermic) and ∆Tsoln < 0. For negative deviations, the solution has stronger intermolecular forces in solution than in pure solute or pure solvent. Negative

deviations have ∆Hsoln < 0 (are exothermic) and ∆Tsoln > 0. Examples of each type of solution are:

ideal: benzene-toluene positive deviations: - negative deviations: acetone-water

7. are properties of a solution that depend only on the number, not the identity, of the solute particles. A solution of some concentration of glucose (C6H12O6) has the same colligative properties as a solution of (C12H22O11) having the same con- centration.

A substance freezes when the vapor pressure of the liquid and solid are identical to each other. Adding a solute to a substance lowers the vapor pressure of the liquid. A lower temperature is needed to reach the point where the vapor of the solution and solid are identical. Hence, the freezing point is depressed when a solution forms.

A substance boils when the vapor pressure of the liquid equals the external pressure. Because a solute lowers the vapor pressure of the liquid, a higher temperature is needed to reach the point where the vapor pressure of the liquid equals the external pressure. Hence, the is elevated when a solution forms.

The equation to calculate the freezing point depression or boiling point elevation is: ∆T = Km where K is the freezing point or boiling point constant for the solvent and m is the of the solute. Table 11.5 lists the K values for several solvents. The solvent which shows the largest change in freezing point for a certain concentration of solute is camphor; it has the largest Kf value. Water, with the smallest Kb value, will show the smallest increase in boiling point for a certain concentration of solute.

To calculate , you need to know the mass of the unknown solute, the mass and identity of solvent used, and the change in temperature (∆T) of the freezing or boiling point of the solution. Since the mass of unknown solute is known, one manipulates the freezing point data to determine the number of moles of solute present. Once the mass and moles of solute are known, one can determine the molar mass of the solute. To determine the moles of solute present, one determines the molality of the solution from the freezing point data and multiplies this by the kilograms of solvent present; this equals the moles of solute present.

8. : the pressure that must be applied to a solution to stop ; osmosis is the flow of solvent into the solution through a . The equation to calculate osmotic pressure, π, is:

π = MRT

CHAPTER 11 PROPERTIES OF SOLUTIONS 4

where M is the molarity of the solution, R is the gas constant, and T is the Kelvin temperature. The molarity of a solution approximately equals the molality of the solution when 1 kg solvent ≈ 1 L solution. This occurs for dilute solutions of water since d = 1.00 H2O g/cm3.

With addition of or , the osmotic pressure inside the fruit cells (and bacteria) is less than outside the cell. Water will leave the cells which will dehydrate bacteria present, causing them to die.

Dialysis allows the transfer of solvent and small solute molecules through a membrane. To purify blood, the blood is passed through a cellophane tube (the semipermeable membrane); this cellophane tube is immersed in a dialyzing solution which contains the same concen- trations of and small molecules as in blood, but has none of the waste products normally removed by the kidney. As blood is passed through the machine, the unwanted waste products pass through the cellophane membrane, cleansing the blood.

Desalination is the removal of dissolved from an . Here, a solution is subjected to a pressure greater than the osmotic pressure and occurs, i.e., water passes from the solution through the semipermeable membrane back into pure water. Desalination plants can turn sea-water with its high salt content into drinkable water.

9. A strong completely dissociates into ions in solution, a weak electrolyte only partially dissociates into ions in solution, and a nonelectrolyte does not dissociate into ions when dissolved in solution. Colligative properties depend on the total number of solute particles in solution. By measuring a property such as freezing point depression, boiling point elevation, or osmotic pressure, we can determine the number of solute particles present from the solute and thus characterize the solute as a strong, weak, or nonelectrolyte.

The van’t Hoff factor, i, is the number of moles of particles (ions) produced for every mol of + − solute dissolved. For NaCl, i = 2 since Na and Cl are produced in water; for Al(NO3)3, i = 4 3+ − since Al and 3 NO3 ions are produced when Al(NO3)3 dissolves in water. In real life, the van’t Hoff factor is rarely the value predicted by the number of ions a salt dissolves into; i is generally something less than the predicted number of ions. This is due to a phenomenon called pairing where at any instant a small percentage of oppositely charged ions pair up and act like a single solute particle. Ion pairing occurs most when the concentration of ions is large. Therefore, dilute solutions behave most ideally; here i is close to that determined by the number of ions in a salt.

10. A colloidal dispersion is a of particles in a dispersing medium. See Table 11.7 for some examples of different types of .

Both solutions and colloids have suspended particles in some medium. The major difference between the two is the size of the particles. A is a suspension of relatively large particles as compared to a solution. Because of this, colloids will scatter light while solutions will not. The of light by a colloidal suspension is called the Tyndall effect.

Coagulation is the destruction of a colloid by the aggregation of many suspended particles to form a large particle that settles out of solution.