The Nature of Solutions

Chemistry, Canadian Edition Chapter 09: Student Study Guide

Chapter 9: Properties of Solutions

Learning Objectives

Upon completion of this chapter you should be able to:

• work with and interconvert concentration units

• predict the relative solubilities of a solute in various solvents, and explain in terms of intermolecular forces.

• calculate and explain solubilities in water.

• understand the reason for and calculate the magnitudes of colligative properties.

• understand some aspects of colloidal suspensions and surfactant solutions.

Practical Aspects

This chapter completes the basic concepts of the theme “structure determines function.” You have learned how atoms combine to form compounds, and the shapes that these compounds have. In this chapter, you will learn how to assess properties of solutions qualitatively (based on the structures of the solution components) and quantitatively (by performing a wide variety of calculations). Give special attention to the calculations involving unit conversions for solutions and the intermolecular forces explanations behind why solutions gain or lose heat energy when they are prepared as these topics are necessary for every chemistry course you ever take, and will surely help you in other science courses as well.

9.1 the nature of solutions

Skills to Master:

·  Calculate concentrations in mass %, molarity, molality, mole fraction.

·  Converting from one concentration unit to another.

Key Terms:

·  Solvent – the major component of the solution. (Water is considered the “universal solvent.”)

·  Solute – substance present in minor amounts within a solution.

Key Concepts:

·  The words “dissolve” and “melt” are not interchangeable. Melting is a phase change of a single substance from the solid phase to the liquid phase. Dissolving is the even distribution of one substance in another to make a solution.

Helpful Hints

·  See the summary table on the next page for common units used to describe solution concentrations.

·  To convert between concentration units, you may need the density of the solution. (Recall, D=m/V). This will allow you to convert between volume of solution and mass of solution.

·  Concentration is independent of the amount of solution you have, so it is fine to assume you have a given quantity of solution if it will make your conversion calculations easier. If you make the following assumptions, you will always know how many moles of solute you’re starting with:

o  If you are converting from molarity units, then assume you have one litre of solution.

o  If you are converting from molality units, then assume you have one kilogram of solvent.

o  If you are converting from mole fraction units to other units, then assume you have 1 mole total.

Common Units for Solution Concentrations

Units / Expressed in / Notes
Molarity, M / / Volume in denominator must be in litres.
Molality, b / / Useful for solutions that are changing temperature, because it doesn’t depend upon volume.
Mole fraction, X / / The sum of the mole fractions of all components in the solution will equal 1.

Exercise 1: What is the mole fraction of ethanol in an aqueous solution of ethanol that has a molality of 3.88mol/kg?

Strategy: Molality is moles of solute (ethanol) per kilogram of solvent (water, because the problem specifies that it is aqueous). We’re starting with molality, so assume we have one kilogram of solvent. Thus, we have 3.88 moles of ethanol. We now need to find total moles of solution.

Total moles = moles of ethanol + moles of water

Solution: The mole fraction of ethanol in the solution is 0.0653 (3 sig figs from molality data). This number may sound small, but it makes sense when you consider how many moles of water are in 1 kg. Note that we did not need density of the solution for this exercise.

Exercise 2: What is the molality (b) of solute particles for a solution that is 2.165 M Na2SO4 (aq)? (Given: density of solution = 1.145 g/mL, density of water = 1.00 g/mL)

Strategy: We’re given molarity (M) and we’re asked to find molality. Molarity is moles of solute (Na2SO4) per litre of solution (water, because the problem specifies that it is aqueous). We should assume that we are starting with 1L of solution, which will mean we have 2.165 moles of Na2SO4 solute. We now need to find how many kilograms of solvent are present in the one litre of solution, so we can divide that into our moles of solute to get molality.

We have one litre of solution, and we know its density, so we can find the mass of solution.

Notice that we needed density of solution, not density of the solvent, water. We now know mass of solution and we know that the solution is made up of solvent and solute:

Mass of solution = mass of solvent + mass of solute = mass of water + mass of Na2SO4

Calculate mass of solute, Na2SO4, and then find mass of solvent.

So, mass of water (solvent) = 1145 g – 307.5 g = 838 g

Solution: The molality of the solution is 2.59 mol Na2SO4/kg water (3 sig figs because of subtraction step). Note that density of solution had to be used to answer this exercise.

Try It #1: Calculate the mole fraction of sodium chloride in a solution that is 2.69 M NaCl (aq). Given: the density of the solution is 1.142 g/mL and the density of water is 1.00 g/mL.

9.2 determinants of solubility

Key Terms:

·  Miscible – ability of two substances to dissolve in each other in any proportion.

·  Saturated – describes a solution that contains the maximum amount of solute possible for the amount of solvent present.

·  Alloy – mixture of two or more metals dissolved in each other.

Key Concepts:

·  “Like dissolves like” – two substances that have similar intermolecular forces will typically dissolve in each other. Substances with different types of intermolecular forces typically won’t dissolve in each other.

·  The rule “like dissolves like” is a general rule; keep in mind that it won’t apply to every situation.

·  Molecules with dipole moments contain a permanent partial charge separation and can therefore interact with ionic compounds.

·  The ability of an ionic compound to dissolve in water depends upon the relative strengths of attractive forces between ions within the crystal lattice versus attractive forces between individual ions and the polar water molecules.

·  Covalent network solids (such as diamond) are insoluble in all solvents, because to dissolve them would mean breaking covalent bonds.

·  Metals are insoluble in water because they contain delocalized bonding networks that would have to be broken down (in a reaction) to disperse in a solvent.

Exercise 3: Both KBr and NH3 dissolve in water. Describe why these chemicals dissolve in water, and then draw molecular pictures to illustrate the intermolecular forces in aqueous solutions of these chemicals.

Strategy and Solution:

KBr is an ionic compound. Ionic compounds can typically dissolve in solvents that contain a permanent dipole. In water, the KBr dissociates and yields one K+ for every one Br-; water dipoles are arranged around ions to maximize attractions.
NH3 is a polar covalent compound that is capable of hydrogen bonding. Water also has these structural features. NH3 molecules, because they are covalently bonded, will not dissociate into ions like the KBr does. Instead, the NH3 molecules will align themselves so that they can form hydrogen bonds with water molecules. In the picture here, hydrogen bonds are shown with dotted lines.

Exercise 4: Which pairs of substances should dissolve in each other? a) Sn and Fe (molten);

b) H2S and CCl4; d) CCl4 and oil; e) CH3CH2OH and NH3.

Strategy: “Like dissolves like.” Look to see if the intermolecular forces are similar.

Solution:

a) Both Sn and Fe are metallic solids. When molten, they should dissolve in each other due to their similar structures – both contain delocalized electrons. (Note: in the solid phase, they wouldn’t be able to dissolve in each other.)

b) H2S is a polar covalent molecule, so dipolar forces predominate among H2S molecules. CCl4 is non-polar covalent; although each bond is polar, the net result is a zero dipole moment, because the molecule is symmetrical and the dipoles cancel out. CCl4 molecules are attracted to each other by dispersion forces. H2S and CCl4 should not dissolve in each other because their intermolecular forces are different.

c) CCl4 and oil – these two substances should dissolve in each other. Oil is a carbon-based compound, and we know that oil and water don’t mix, which indicates that oil must be a non-polar molecule. CCl4 is also a non-polar molecule. Both substances have dispersion forces as their dominant intermolecular force, so they should dissolve in each other.

d) CH3CH2OH and NH3 – both of these substances are polar covalent molecules that can hydrogen bond. They should dissolve in each other because they have similar intermolecular forces.

Try It #2: Which pairs of substances should be able to dissolve in each other? a) N2 and O2; b) H2O and CsCl; c) H2O and Fe.

9.3 characteristics of aqueous solutions

Skill to Master:

·  Using Henry’s law to calculate gas solubility.

Key Terms:

·  Heat of Solution (DHsoln) – the energy released or required to dissolve a specific quantity of a given substance in water. (Molar heat of solution refers to 1 mole of substance.)

·  Heat of Dilution – energy released or required to dilute an existing solution.

Key Concepts:

·  Water is an excellent solvent. The fact that it has a strong dipole moment and can hydrogen bond allows it to interact with many chemicals.

·  Solubility of solids in water usually increases as temperature increases, but some exceptions do exist.

·  Solubility of gases in water decreases as temperature increases. This can be quantitatively assessed using Henry’s Law.

·  The solute molecules in a saturated solution are constantly moving between the aqueous and pure solute phases. The molecules are said to be in dynamic equilibrium.

·  Energy changes occur when a solute is dissolved or diluted in a solvent because a net overall change in intermolecular forces occurs. For example, when salt is dissolved in water, ion-ion attractions are broken (energy is required) and new ion-dipole attractions are formed (energy is released). Figure 9-7 in the text details the various factors that influence the overall change.

Useful Relationship:

·  (Pgas)eqKH = [gas(aq)]eq / This is Henry’s Law. It shows the relationship between the amount of gas that can be dissolved in water and the pressure of the gas at equilibrium. KH is Henry’s Law constant, and it depends upon the identity of the gas. Table 9-2 in the text provides constants for common gases. To use this equation, gases must have pressure units of “atm” and aqueous solutions must have concentration units of “mol/L”.

Exercise 5: The molar heat of solution of calcium chloride is –81.8 kJ/mol. a) How many grams of CaCl2 must be added to 65.0 mL of water at 19.0°C to get the water’s temperature to change by 5.0°C? b) Explain the value of the molar heat of solution for calcium chloride.

Strategy: a) We’re asked to find the mass of calcium chloride. When the solid calcium chloride comes in contact with the water, it will dissolve, releasing heat energy in the process (The heat of solution of calcium chloride is a negative value, indicating the heat is released when CaCl2 dissolves). As we learned in Chapter 3, the heat released by the CaCl2 will be absorbed by the water. This means that the water temperature will rise. Here is what we know:

·  qsoln = -qwater

·  DTwater = +5.0°C, mwater = 65.0 g (recall that density of water is 1.00 g/mL), Cwater = 75.291 J/molK

·  DHsoln = -81.8 kJ/mol

Working backwards from the goal, we see that we need moles of CaCl2 to find mass of CaCl2. We can find moles of CaCl2 from comparing the qsoln to the heat released by one mole of CaCl2.

qsoln = -qwater = -[nwater Cwater DTwater]

Set up a ratio to determine moles of CaCl2:

(2 sig figs based on temperature data)

Part b) strategy: Determine the types of intermolecular forces being broken and formed in the process. Consider the fact that the negative value for calcium chloride’s DHsoln indicates that the net change in these forces involves an overall release in energy. (Keep in mind that positive value for DHsoln would indicate the opposite: it would require more energy to break apart attractions than the amount of energy released forming new attractions.)

Solution: a) 1.8 g of CaCl2 must be added to the water. b) The net attractive forces that are formed (ion-dipole between calcium ions/water and chloride ions/water) must be stronger than the net attractive forces being broken (ion-ion between calcium ions and chloride ions, hydrogen bonding between some water molecules) to make the value for molar heat of solution of calcium chloride to be a negative value.

Exercise 6: a) Draw a molecular picture to illustrate a saturated solution of oxygen dissolved in water. b) The earth’s atmosphere is about 21% oxygen. The concentration of dissolved oxygen in water changes with temperature. Different types of fish are biologically-adapted with varying capabilities to absorb dissolved oxygen in the ocean, thus different types of fish are found in different water climates. Calculate the concentration of oxygen that can dissolve in water at 0°C and at 25 °C.