Water and Aqueous Systems Chapter 17

Objectives Describe the hydrogen bonding that occurs in water Explain the high surface tension and low vapor pressure of water in terms of hydrogen bonding Account for the high heat of vaporization and the high boiling point of water in terms of hydrogen bonding Explain why ice floats on water

The Water Molecule O-H bonds are highly polar covalent bonds. the O atom has a slightly negative charge. H has a slightly positive charge. Hydrogen bonding- occurs when hydrogen is attracted to a very electronegative element (N, O, and F).

Hydrogen bonding in water

The Unique Properties of Water High surface tension Low vapor pressure High specific heat capacity High heat of vaporization High boiling point Low density of ice

Surface properties Surface tension: the inward pull that tends to minimize the surface area of a liquid. Surfactant: surface active agent- surface tension is decreased by detergents (interferes with hydrogen bonding)

Why are water droplets spherical as they fall?

Surface tension

Surface tension

Water on your car….

Vaporization and Vapor Pressure Hydrogen bonding allows few water particles to vaporize, resulting in low vapor pressure above the surface of water

Water’s low vapor pressure

Specific Heat Capacity Water’s high specific heat capacity helps to moderate daily air temperatures around large bodies of water. (Water’s specific heat is more than 4 times that of most metals.) This is due to the hydrogen bonding

Water’s high heat of vaporization Because of hydrogen bonding, water absorbs large amounts of heat as it vaporizes. Water releases large amounts of heat as it condenses.

Water’s high boiling point Hydrogen bonding causes water’s high boiling point (Stronger IMFs to overcome).

Ice Hydrogen bonding causes water molecules to freeze in a certain arrangement Ice has an open framework structure, causing space to be trapped between the molecules The density of ice is lower than that of water. This is why ice floats.

Objectives Explain the significance of the statement “like dissolves like” Distinguish between strong electrolytes, weak electrolytes, and nonelectrolytes, giving examples of each

Solution – homogeneous mixture What is a solution? Solution – homogeneous mixture Solvent – substance present in largest amount (the dissolving medium) Solutes – other substances in the solution Aqueous solution – water is the solvent

Solvation… the process that occurs as a solute dissolves. There are two types of solvation: Ionic compounds dissolve by dissociation. Covalent compounds dissolve by molecular solvation.

Solubility of Ionic Substances: Dissociation The positive and negative ions of a salt come apart (dissociate) as a salt dissolves.

Dissociation vs. Molecular Solvation NaCl(s)  Na +(aq) + Cl -(aq) Note: the ions come apart from each other (electrolytes) Molecular solvation: C6H12O6(s)  C6H12O6(aq) Note: no dissociation occurs (nonelectrolytes)

“Like dissolves like” Polar solutes dissolve in polar solvents (water is polar, so it dissolves polar substances, either ionic or molecular). Nonpolar solutes dissolve in nonpolar solvents (oil dissolves in kerosene).

Solubility of Polar Substances Ethanol is soluble in water because of the polar OH bond.

Why is solid sucrose (C12H22O11), table sugar, soluble in water?

Substances Insoluble in Water Nonpolar oil does not interact with polar water.

Electrolytes Electrolytes- conduct an electric current in solution or in molten state. (ionic compounds) Nonelectrolytes- do not conduct an electric current in solution or in molten state. (molecular compounds)

Electrolytes Strong electrolytes: solute completely breaks apart in solution [includes soluble salts (such as KCl), inorganic acids (such as HNO3), inorganic bases (such as NaOH)]

Electrolytes Weak electrolytes: only a fraction of the solute breaks apart into solution [includes poorly soluble salts (such as PbCl2), organic acids (such as HC2H3O2), and organic bases (such as NH3)]

Electrolytes Nonelectrolyte – does not conduct when in solution- does not break apart into ions. [includes most organic compounds (such as glucose)]

Water of hydration The water in a crystal is called water of hydration. A hydrate is a compound that includes water of hydration.

Water of hydration A hygroscopic substance removes water from the air. These substances are called dessicants.

Heterogeneous Aqueous Systems Suspensions are mixtures from which particles settle out upon standing. (The particles are much larger than those in a solution.) Colloids are heterogeneous mixtures that contain particles that are intermediate in size between suspensions and true solutions

Colloidal Systems Colloids exhibit the Tyndall effect, which is the scattering of light in all directions. Colloids include milk, mayonnaise, marshmallows, egg white, blood, and paint.

Solutions and their Behavior Chapter 18

Identify factors that determine the rate at which a solute dissolves Objectives Identify factors that determine the rate at which a solute dissolves Identify factors that affect the solubility of a solute in solution Calculate the solubility of a gas in a liquid under various pressure conditions

Factors affecting the rate of dissolving How could you speed up the dissolving of sugar in a glass of iced tea? Which dissolves faster, table salt or rock salt?

Factors affecting the rate of dissolving Temperature – increasing the temperature speeds up the rate of dissolving Agitation – stirring speeds up the rate of dissolving Particle size – smaller particles dissolve faster than large particles (surface area)

Solubility The solubility- the amount that dissolves in a given quantity of solvent at a given temperature.

Solubility A saturated solution contains the maximum amount of solute at a constant temperature.

Solubility An unsaturated solution contains less solute than a saturated solution. A supersaturated solution contains more solute than a saturated solution. (This occurs when a solution is saturated and then allowed to cool but all of the solid remains dissolved. It is an unstable solution, adding a crystal causes precipitation.)

What would happen… …if you added more sugar to a saturated sugar solution and stirred? …if you added more sugar to an unsaturated sugar solution and stirred?

Factors affecting the solubility of a substance How could you increase the amount of sugar that would eventually dissolve in a glass of tea?

Factors affecting the solubility of a substance Only two factors affect the amount of solute that can dissolve. Temperature affects solubility of both solids and gases in liquid solvents. Pressure affects solubility of gases in liquid solvents.

Factors affecting the solubility of a substance The solubility of most solid substances increases as the temperature of the solvent increases. For a few substances, the reverse occurs.

The Effect of Temperature on the Solubility of Solids

The Effect of Temperature on Gas Solubility Increasing the temperature of a dissolved gas solution decreases the concentration of the gas. Have you ever tried a hot Dr. Pepper? Heat it in a pan on the stove, pour a cup, and it has no bubbles! Thermal pollution occurs when hot water is added to a lake, the dissolved oxygen levels fall in the water and it kills the fish.

The Effect of Temperature on Gas Solubility

The Effect of Pressure on Gas Solubility Increasing the pressure of a gas over the surface of a solvent increases the solubility of the gas in the solvent. In a bottled soda, the pressure of CO2 over the liquid is high and when the cap is opened, the pressure is reduced and bubbles begin to come out of the solution.

The Effect of Pressure on Gas Solubility Henry’s Law: at a given temperature the solubility (S) of a gas in a liquid is directly proportional to the pressure (P) of the gas above the liquid. S1/P1 = S2/P2

Question: If the solubility of a gas in water is 0.77 g/L at 3.5 atm pressure, what is the solubility (in g/L) at 1.0 atm? (The temperature is held constant at 25oC.) Answer: S1/P1 = S2/P2 0.77 g/L / 3.5 atm = S2 / 1.0 atm S2 = 0.22 g/L

Solve problems involving the molarity of a solution Objectives Solve problems involving the molarity of a solution Describe how to prepare dilute solutions from concentrated solutions of known molarity.

Solution Composition: Molarity Concentration of a solution is the amount of solute in a given volume of solution. A concentrated solution has a high concentration of solute. A diluted solution has only a low concentration of solute.

A unit of solution concentration A common unit of concentration: Molarity (M) (Notice that the ratio includes liters of total solution, not liters of solvent.)

Solution Composition: Molarity Consider both the moles of solute and the volume of solution to find concentration.

Solution Composition: Molarity To find the moles of solute in a given volume of solution of known molarity use the definition of molarity.

Solution Preparation: Molarity To make a molar solution: Weigh out a sample of solute that contains the necessary moles of solute. Transfer to a volumetric flask. Add only enough solvent to mark on flask to get total solution volume.

Problem: Calculate the molarity when 3.0 mol of solute is dissolved in enough water to make 2.0 L of solution. Answer: mol solute/L solution = molarity 3.0 mol / 2.0 L = 1.5M

Problem A solution contains 17.0 grams of sodium chloride in 200. mL of solution. What is the molarity of the solution? Answer: (convert grams to moles) 17.0 grams . (1 mol/58.5 g) = 0.2906 mol mol solute/L solution = molarity 0.2906 mol / 0.200 L = 0.145 molar NaCl

Problem How many grams of sodium chloride are needed to prepare 275 mL of a 0.200M solution of NaCl? Answer: molarity . L solution = mol solute 0.200M . 0.275 L = 0.0550 moles of NaCl 0.0550 mol . (58.5 g/1 mol) =3.22 grams NaCl

Another unit of solution concentration Another common unit of concentration: Molality (m) (Notice that the ratio includes liters of solvent, not liters of total solution.)

Molality Calculate the molality of a solution prepared by dissolving 45.0 grams of KCl in 500. grams of water. Answer: 45.0 grams KCl . (74.55 g/1 mol) =0.604 mol m = mol solute/kg solvent 0.604 mol KCl/ .500 kg water = 1.21 molal

Molality How many grams of sodium chloride must be dissolved in 250.0 grams of water to prepare a 0.75 molal NaCl solution? Answer: mol solute = m . kg solvent mol = 0.75m . 0.250 kgwater = 0.19 mol NaCl 0.19 mol NaCl . (58.5 g/1 mol) = 11 grams NaCl

Dilution When diluting a solution to a lower concentration, only water is added in the dilution – the amount of solute remains the same. (remember: mol solute = molarity . L solution) (mol solute)conc = (mol solute)diluted Mconc . Lconc = Mdiluted . Ldiluted Mconc . Vconc = Mdiluted . Vdiluted M1. V1 = M2. V2

Dilution Diluting a solution Transfer a measured amount of original solution to a flask containing some water. Add water to the flask to the mark (with swirling) and mix by inverting the flask. M1 . V1 = M2 . V2

Problem How many mL of a concentrated stock solution of 5.0M HCl is needed to prepare 250. mL of 1.75M HCl? Answer: M1 . V1 = M2 . V2 5.0M . V1 = 1.75M . 250 mL V1 = 87.5 mL (notice: you do not have to convert to liters in dilution problems)

Calculate the molality and mole fraction of a solution Objectives Explain on a particle basis why a solution has a lower vapor pressure than the pure solvent of that solution Explain on a particle basis why a solution has an elevated boiling point and a depressed freezing point compared to the pure solvent Calculate the molality and mole fraction of a solution Calculate the molar mass of a molecular compound from the freezing point depression or boiling point elevation of a solution of the compound

Colligative Properties Colligative property – a solution property that depends on the number, but not the type, of solute particles present There are three colligative properties of solutions: Vapor pressure lowering Boiling point elevation Freezing point depression

Vapor Pressure Lowering When solute particles are added to a solvent, the solute forms associations with the solvent. Fewer solvent particles are free to vaporize, so the vapor pressure is lower over a solution than over the pure solvent.

Colligative Properties Adding solute particles causes the liquid range to become wider. The lowering of vapor pressure causes the boiling point to rise and the freezing point to fall. ( interferes with crystal formation) Boiling point increases Freezing point decreases

Colligative Properties The higher the number of particles (ions, molecules, etc.) the greater the change in the colligative property. Which would have the greatest elevation in boiling point? 1.0M C6H12O6 1.0M NaCl 1.0M CaCl2 1.0M AlCl3

Answer: All 4 solutions have the same concentration, but the number of particles that form is not the same: C6H12O6(s)  C6H12O6(aq) (1 particle) NaCl(s)  Na+(aq) + Cl-(aq) (2 particles) CaCl2(s)  Ca2+(aq) + 2Cl-(aq) (3 particles) AlCl3(s)  Al3+ (aq)+ 3Cl-(aq) (4 particles, greatest effect) (Remember: ionic compounds dissociate, breaking down into multiple particles, molecular compounds do not.)

Calculating Solution Boiling Point The boiling point is raised by the addition of a nonvolatile solute. To calculate the change in boiling point, you must consider the concentration of the particles in solution. Remember: water normally boils at 100oC. DT = Kb i m change in boiling point = (boiling point constant) (number if ions) (molality)

Problem What is the boiling point of 2.75m NaCl(aq) solution? (Kb for water is 0.512oC/m) Answer: DT = Kb i m DT = (0.512oC/m) (2) (2.75m) DT = 2.82oC (this is the change in boiling point) Boiling point = 100oC + 2.82oC = 102.82oC

Calculating Solution Freezing Point You calculate the freezing point of a solution in a similar way. Remember: water normally freezes at 0.0oC. DT = Kf i m change in freezing point = (freezing point constant) (number if ions) (molality)

Problem What is the freezing point of a solution that contains 2.50 mol of CaCl2 in 1250 g of water? (Kf for water is 1.86oC/m) Answer: DT = Kf i m DT = (1.86oC/m) (3) (2.50 mol/1.250 kg) = 11.2oC (this is the freezing point change) Freezing point = 0.0oC - 11.2oC = -11.2oC

Determining Molar Mass You can use the change in boiling point or freezing point to calculate the molar mass of an unknown solute. If you are given a mass of an unknown solute, you can use the boiling point elevation to determine the molality of the solution, then use the volume of solvent to determine the moles of solute. Use the given mass and the moles of solute to determine the molar mass of the solute.

Problem: The boiling point of water is raised to 102.5oC when 65.50 grams of a nonvolatile molecular solute is dissolved in 500. g of water. Calculate the molar mass of the solute. Answer: DT = Kf i m 2.5oC = (.512oC/m) (1) (mol solute/.500 kg) mol solute = 2.44 mol 65.50 grams/2.44 mol = 26.8 g/mol