Chapter 12 Solutions

3.8: Solutions Solute: The substance that dissolves (the minor component of a solution). KMnO4

3.8: Solutions Solvent: The substance in which the solute dissolves (the major component of a solution).

Solution: A homogeneous mixture of the solute and solvent. KMnO4 solution

3.8: Solutions Dilute Concentrated

Solubility The solubility of a solute is the quantity that will dissolve in a given amount of solvent to produce a saturated solution. Expressed as grams of solute in 100 grams of solvent, usually water. g of solute 100 g water

Dissolving of ionic compounds

Solution Conductivity

Types of solutes Strong Electrolyte - 100% dissociation, high conductivity Strong Electrolyte - 100% dissociation, all ions in solution Na+ Cl-

Strong Electrolytes When an ionic compound, a strong electrolyte dissolves it dissociates completely into ions. Strong electrolytes can also be molecular. These substances also dissociate completely into ions when they dissolve, no intact molecules remain. This solution conducts electricity well.  Strong electrolytes are either strong acids, strong bases or soluble salts.

Strong Electrolytes → Here, the ions in the solid are being broken apart, being pulled away from the other ions in the solid, and are being surrounded by water molecules. An ion-dipole attraction occurs between the polar water molecules and the dissociated ions.

Types of solutes Weak Electrolyte - partial dissociation, slight conductivity Weak Electrolyte - partial dissociation, molecules and ions in solution CH3COOH CH3COO- H+

Weak Electrolytes A weak electrolyte is a compound that when dissolved in water only partially ionizes or dissociates into ions. That is, the compound exists in water as a mixture of individual ions and intact molecules. This solution conducts electricity weakly. Weak electrolytes include weak acids and weak bases.

Weak Electrolytes → Molecules which dissolve in water are polar molecules, so there will be dipole-dipole, or possibly even hydrogen bonding, between the solute and the solvent. For the molecules which do dissociate into ions, you will have ion-dipole attractions.

Types of solutes Non-electrolyte - No dissociation, no conductivity Non-electrolyte - No dissociation, all molecules in solution sugar

Nonelectrolytes A nonelectrolyte is a compound that when dissolved in water does not ionize or dissociate into ions at all. In water, this compound exists entirely as intact molecules. The solution does not conduct electricity at all. Nonelectrolytes would be molecular compounds other than the acids or bases. Examples: alcohol, sugar.

Nonelectrolytes → The molecules of solute mix with the water but stay together as molecules.

Whether a compound is an electrolyte (strong or weak), or a nonelectrolyte does not mean soluble or insoluble.

Percent Solution by mass Represents the number of grams of solute per 100 grams of solution. mass of solute % solution = X 100 mass of solution

A solution contains 130g of HCl and 750g of water A solution contains 130g of HCl and 750g of water. What is the percent by mass this HCl solution? 15 %

A concentrated HCl solution has a density of 1. 19 g/mL and is 37 A concentrated HCl solution has a density of 1.19 g/mL and is 37.2% HCl by mass. What mass of HCl is needed to make exactly 1L of this solution? 443 g HCl

What volume of a concentrated HCl solution that is 37 What volume of a concentrated HCl solution that is 37.2% HCl by mass and has a density of 1.19 g/mL is needed to obtain 125g of HCl?

Molarity (M) Moles of solute per liter of solution. moles of solute liters of solution

What is the molarity of a solution prepared by dissolving 10 What is the molarity of a solution prepared by dissolving 10.0 g of NaCl in 250.0 mL of solution? 0.684M

Concentrated sulfuric acid has a density of 1. 84 g/ml and is 98 Concentrated sulfuric acid has a density of 1.84 g/ml and is 98.3% H2SO4 by mass. What is the molarity of this solution? 18.5M

How would you prepare 500. ml of a 6.0M NaOH solution?

How would you prepare 500. ml of a 6.0M NaOH solution? 120 g of NaOH dissolved in water and diluted to a volume of 500. ml

This is a dilution problem I need to make 250 ml of a 0.50M NaOH solution using the 6.0M NaOH prepared in the previous example. How do I prepare this solution? This is a dilution problem

Dilution Problems Mi x Vi = Mf x Vf

21 ml of the 6.0M NaOH diluted to a total volume of 250 ml. I need to make 250 ml of a 0.50M NaOH solution using the 6.0M NaOH prepared in the previous example. How do I prepare this solution? 21 ml of the 6.0M NaOH diluted to a total volume of 250 ml.

850 ml of a 5. 00M Cu(NO3)2 solution is diluted to a total volume of 1 0.850 ml of a 5.00M Cu(NO3)2 solution is diluted to a total volume of 1.80L. What is the molarity of the resulting diluted solution? 0.00236M

How many milliliters of water are needed to dilute 12ml of a 0 How many milliliters of water are needed to dilute 12ml of a 0.44M acetic acid solution to 0.12M? 32 ml of water added

molality (m) Moles of solute per kilogram of solvent. moles of solute kg of solvent

What is the molality of a solution made from dissolving 20 What is the molality of a solution made from dissolving 20.4 g KBr in 195 g of water? 0.879m

Mole Fraction (X) moles of component per total moles of solution. total moles in solution

A gas mixture contains 45. 6g CO and 899g CO2 A gas mixture contains 45.6g CO and 899g CO2. What is the mole fraction of CO? 0.0741

What is the mole fraction of solute in a 3.0m NaCl aqueous solution? 0.051

Colligative properties Properties determined by the number of solute particles in solution rather than the type of particles. Vapor pressure lowering Freezing point depression Boiling point elevation Osmotic pressure

How Vapor Pressure Lowering Occurs Solute particles take up space in a solution. Solute particles on surface decrease number of solvent particles on the surface. Less solvent particles can evaporate which lowers the vapor pressure of a liquid.

Consider two beakers in a bell jar. Beaker “A” contains pure water. Beaker “B” contains saltwater. What change in volume would you observe in beaker “A” as time passes? What change in molarity would you observe in beaker “B” as time passes? Explain your answers. A B

The volume in “A” decreases? The molarity in “B” decreases. The rate of evaporation (vapor pressure) is greater in “A” than in “B” however the rates of condensation are equal. A B B

Raoult’s Law Vapor Pressure Lowering = (mole fractionsolute)(Vapor Pressuresolvent)

Example: The vapor pressure of water at 20oC is 17. 5 torr Example: The vapor pressure of water at 20oC is 17.5 torr. Calculate the vapor pressure of an aqueous solution prepared by adding 36.0 g of glucose (C6H12O6) to 14.4 g of water. Answer: 14.0 torr

Boiling Point Elevation Boiling point elevation is a colligative property related to vapor pressure lowering. The boiling point is defined as the temperature at which the vapor pressure of a liquid equals the atmospheric pressure. Due to the lowering of the vapor pressure of the solvent when a solution forms, a solution will require a higher temperature than the pure solvent to reach its boiling point.

Freezing Point Depression Every liquid has a freezing point - the temperature at which a liquid undergoes a phase change from liquid to solid. When solutes are added to a liquid, forming a solution, the solute molecules disrupt the formation of crystals of the solvent. That disruption in the freezing process results in a depression of the freezing point for the solution relative to the pure solvent.

Freezing-Point Depression & Boiling-Point Elevation

Boiling Point Elevation ∆Tb = Kbm

Freezing Point Depression ∆Tf = Kfm

Kb and Kf

Calculate the freezing point of a solution that contains 5 Calculate the freezing point of a solution that contains 5.15 g of benzene (C6H6) dissolved in 50.0 g of CCl4.

Calculate the freezing point of a solution that contains 5 Calculate the freezing point of a solution that contains 5.15 g of benzene (C6H6) dissolved in 50.0 g of CCl4. -63°C

Find the boiling point of a solution containing 92 Find the boiling point of a solution containing 92.1g of iodine, I2, in 800.0g of chloroform.

Find the boiling point of a solution containing 92 Find the boiling point of a solution containing 92.1g of iodine, I2, in 800.0g of chloroform. 62.9°C

Semi-permeable Membrane a partition that allows solvent particles to pass through but not solute particles

Osmosis Osmosis: the net movement of a solvent through a semipermeable membrane toward the solution with greater solute concentration.

Osmosis Examples

Other everyday examples of osmosis: A cucumber placed in brine solution loses water and becomes a pickle. A wilting flower placed in water will perk up because water enters by osmosis. Eating large quantities of salty food can cause edema (swelling in the body caused by the accumulation of abnormally large amounts of fluid in the spaces between the body's cells or in the circulatory system).

Osmotic Pressure Osmotic Pressure (π) – the pressure required to prevent osmosis from occurring.

Osmotic Pressure (π) π = MRT

R values a

The average osmotic pressure of blood is 7. 7 atm at 25 C The average osmotic pressure of blood is 7.7 atm at 25 C. What concentration of glucose will be isotonic with blood? 0.31M

Molar Mass Calculations

A sample of a human hormone weighing 0. 546g was dissolved in 15 A sample of a human hormone weighing 0.546g was dissolved in 15.0g benzene. The freezing point of the benzene was lowered by 0.240 C. What’s the molecular weight of the hormone? Kf(benzene) = 5.12 °C.kg/mol 776 g/mol

A solution containing 4.5 g of glycerol, a nonvolatile nonelectrolyte, in 100.0 g of ethanol has a boiling point of 79.0oC. If the normal BP of ethanol is 78.4oC, calculate the molar mass of glycerol. (Kb of ethanol = 1.22 °C/m) 92 g/mol

One liter of an aqueous solution containing 20 One liter of an aqueous solution containing 20.0g of hemoglobin has an osmotic pressure of 5.9 torr at 22°C. What is the molecular mass of hemoglobin? 62,500 g/mol

Colligative properties Properties determined by the of solute particles in solution rather than the type of particles. Vapor pressure lowering Freezing point depression Boiling point elevation Osmotic pressure number

Ionic Solute vs. Molecular Solute Compare the properties of a 1.0 M aqueous sugar solution to a 0.5 M solution of table salt (NaCl) in water. Despite the concentrations, both solutions have precisely the same number of dissolved particles because each sodium chloride unit creates two particles upon dissolution - a sodium ion, Na+, and a chloride ion, Cl-. Therefore their colligative properties, which depend on the number of dissolved particles, would be identical. Both solutions have the same freezing point, boiling point, vapor pressure, and osmotic pressure because those colligative properties of a solution only depend on the number of dissolved particles.

Formula Adjustment for Ionic Solutes The formulas for colligative properties must take into account a multiplier to account for the number of dissolved particles for ionic compounds. Thus NaCl and NaNO3 would have multipliers of two, while Na2SO4 and CaCl2 would have multipliers of three.

Formula Adjustment for Ionic Solutes (i) = # of ions in the ionic substance Vapor Pressure Lowering = (mole fractionsolute)(Vapor Pressuresolvent)(i) ∆Tb = Kbm(i) ∆Tf = Kfm(i) π = MRT(i)

Calculate the boiling point of a 0.40m aqueous solution of Ca(NO3)2.

Calculate the boiling point of a 0.40m aqueous solution of Ca(NO3)2.

Colligative Properties and Ionic Solutes Colligative effects in ionic solutions are almost always smaller than the effect predicted on the basis of complete dissociation due to the fact that ions in solution to not behave completely independently of one another, especially at higher concentrations. This is known as the Van’t Hoff Factor.

Ion Pairs Ion Pair

Van’t Hoff factor In dilute solutions, ionic compounds have simple multiple effects, but as the solution concentration increases the multiplier effect diminishes. As concentration increases, some of the ions floating in solution find one another and form ion pairs, in which two oppositely charged ions briefly stick together and act as a single particle. Obviously, the higher the concentration, the more likely it is that two ions will find one another.

For example: While we might expect NaCl to lower the freezing point of water twice as much as an equal amount of a nonelectrolyte, we may actually get about 1.8 times the effect. Ion Pair