Chapter 14: Solutions Chemistry 1020: Interpretive chemistry Andy Aspaas, Instructor.

9
Chapter 14: Solutions Chemistry 1020: Interpretive chemistry Andy Aspaas, Instructor

Transcript of Chapter 14: Solutions Chemistry 1020: Interpretive chemistry Andy Aspaas, Instructor.

Page 1: Chapter 14: Solutions Chemistry 1020: Interpretive chemistry Andy Aspaas, Instructor.

Chapter 14: Solutions

Chemistry 1020: Interpretive chemistry

Andy Aspaas, Instructor

Page 2: Chapter 14: Solutions Chemistry 1020: Interpretive chemistry Andy Aspaas, Instructor.

Solutions

• Solution = homogeneous mixture– Components of mixture are equally dispersed

throughout the mixture• Solutions can be solid, liquid, or gas

– Air = solutions of gases– Brass = solution of copper and zinc

• Solvent: component of solution that is present in the largest amount

• Solutes: other components of a solution• Aqueous solutions: water is solvent

Page 3: Chapter 14: Solutions Chemistry 1020: Interpretive chemistry Andy Aspaas, Instructor.

Solubility

• Both sugar and salt readily dissolve in water

– In these cases, a solid solute (salt or sugar) dissolves in the liquid solvent (water)

• Ionic solids like NaCl consist of closely packed ions which are held together by strong ionic bonds

– But, when dissolved in water, the polar water molecules are attracted to the ions, and can overcome ionic bonds

– The ions are separated and surrounded by water molecules (- of H2O attracted to anions, and vice-versa)

Page 4: Chapter 14: Solutions Chemistry 1020: Interpretive chemistry Andy Aspaas, Instructor.

Solubility and molecular substances

• Most water-soluble molecules have some type of hydrogen-bonding source and/or polar bonds– Ethanol contains a polar OH group which can

hydrogen-bond with water– Sucrose (sugar) contains many OH groups + on H of OH groups attracted to — on O of

H2O• Nonpolar molecules like petroleum have no polar

OH groups and are not able to form hydrogen bonds with water; they are insoluble– Petroleum forms a layer on top of water

Page 5: Chapter 14: Solutions Chemistry 1020: Interpretive chemistry Andy Aspaas, Instructor.

Like dissolves like

• Like dissolves like: a solvent will usually dissolve solutes with polarities similar to its own

• Polar solvents like water will dissolve polar molecules

• Nonpolar solvents like hydrocarbons will dissolve nonpolar solutes

– Drycleaners use nonpolar solvents to remove grease stains (greases are composed of very nonpolar molecules)

Page 6: Chapter 14: Solutions Chemistry 1020: Interpretive chemistry Andy Aspaas, Instructor.

Solution composition

• There is a limit to how much solute a particular solvent can dissolve

• A saturated solution contains as much solute as will dissolve at that particular temperature

• An unsaturated solution has not yet reached that limit

• A solution with a relatively high amount of solute is called concentrated

• A solution with a relatively low amount of solute is called dilute

Page 7: Chapter 14: Solutions Chemistry 1020: Interpretive chemistry Andy Aspaas, Instructor.

Mass percent in solutions

• Mass percent (sometimes called weight percent)

[ (grams solute) / (grams solution) ] x 100%

• Often given in nonscientific applications

• Vinegar is approximately 5% by mass acetic acid in water

– That means 100 g of vinegar would contain 5 g acetic acid

• When calculating mass %, don’t forget to include solute mass in the total solution mass

Solution mass = solvent mass + solute mass

Page 8: Chapter 14: Solutions Chemistry 1020: Interpretive chemistry Andy Aspaas, Instructor.

Molarity

• Concentration: amount of solute in a given volume of solution (often more convenient than dealing with masses of solutions)

• Molarity, M, is most common excpression of concentration

M = (moles solute) / (liters solution) = mol / L

• A 1.0 molar solution (written 1.0 M) contains 1.0 mol solute per liter of solution

Page 9: Chapter 14: Solutions Chemistry 1020: Interpretive chemistry Andy Aspaas, Instructor.

Dilution

• In a dilution, water is added to a concentrated solution in order to make it more dilute

• The number of moles of solute is equal before and after the dilution since only water is added

• But the concentration has changed

• M1 x V1 = M2 x V2

– Where M1 and V1 are the molarity and volume of the initial concentrated solution (before dilution)

– And M2 and V2 are the molarity and volume of the final dilute solution (after dilution)