Solution (Homogeneous Mixture) Evenly mixed, uniform particles Solute - gets dissolved Solvent - does the dissolving Aqueous (water) Tincture (alcohol) Amalgam (mercury) Organic (carbon based) Polar Non-polar Dental filling Solutions are always homogeneous – evenly mixed.

Types of Solutions Solute Solvent Solution Gaseous Solutions liquid air (nitrogen, oxygen, argon gases) humid air (water vapor in air) Liquid Solutions solid carbonated drinks (CO2 in water) vinegar (CH3COOH in water) salt water (NaCl in water) Solid Solutions (Alloys) dental amalgam (Hg in Ag) sterling silver (Cu in Ag) Charles H.Corwin, Introductory Chemistry 2005, page 369

Pure water does not conduct an electric current Source of electric power Pure water Pure water doesn’t conduct electricity because it contains no ions. Ions (cations have (+) charges) carry electrons in solution. The flow of electrons is called electricity. Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 215

Ionic Solutions conduct a Current Source of electric power Free ions present in water To make a solution that conducts electricity (an electrolyte) – an ionic salt is added. An alternative method is to add acid. In truth, most acids are oxysalts dissolved in water. Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 215

Electrolytes Electrolytes - solutions that carry an electric current strong electrolyte weak electrolyte nonelectrolyte NaCl(aq) Na+ + Cl- HF(aq) H+ + F- Timberlake, Chemistry 7th Edition, page 290

Solubility Solubility maximum grams of solute that will dissolve in 100 g of solvent at a given temperature varies with temp based on a saturated solution

Dissolving of NaCl Timberlake, Chemistry 7th Edition, page 287

Oil and Water Don’t Mix Oil is nonpolar Water is polar “Like dissolves like” Lycopodium Powder demonstration. Lycopodium powder is a nonpolar hydrocarbon that doesn’t mix with water. It is dry to the touch and not oily feeling. Sprinkle lycopodium powder on top of a beaker of water. Slowly, push your finger through the surface of the lycopodium powder into the water. Slowly withdraw your finger, it remains dry the entire time. Why? Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 470

of solubility on temperature Solubility vs. Temperature 140 KI 130 120 gases solids NaNO3 110 Solubility Curves 100 KNO3 90 80 HCl NH4Cl shows the dependence of solubility on temperature 70 Solubility (grams of solute/100 g H2O) 60 NH3 KCl 50 “Solubility Curves for Selected Solutes”   Description: This slide is a graph of solubility curves for 10 solutes. It shows the number of grams of solute that will dissolve in 100 grams of water over a temperature range of 0cC to 10 cC. Basic Concepts The maximum amount of solute that will dissolve at a given temperature in 100 grams of water is given by the solubility curve for that substance. When the temperature of a saturated solution decreases, a precipitate forms. Most solids become more soluble in water as temperature increases, whereas gases become less soluble as temperature increases. Teaching Suggestions Use this slide to teach students how to use solubility curves to determine the solubilities of various substances at different temperatures. Direct their attention to the dashed lines; these can be used to find the solubility of KClO3 at 50 cC (about 21 g per 100 g of H2O). Make sure students understand that a point on a solubility curve represents the maximum quantity of a particular solute that can be dissolved in a specified quantity of solvent or solution at a particular temperature. Point out that the solubility curve for a particular solute does not depend on whether other solutes also are present in the solution (unless there is a common-ion effect; this subject usually is covered at a later stage in a chemistry course). Questions Determine the solubilities (in water) of the following substance at the indicated temperatures: NH3 at 50 oC; KCl at 90 oC; and NaNO3 at 0 oC. Which of the substances shown on the graph is most soluble in water at 20 oC? Which is lease soluble at that temperature? For which substance is the solubility lease affected by changes in temperature? Why do you think solubilities are only shown between 0 oC and 100 oC? In a flask, you heat a mixture of 120 grams of KClO3 and 300 grams of water until all of the KClO3 has just been dissolved. At what temperature does this occur? You then allow the flask to cool. When you examine it later, the temperature is 64 oC and you notice a white powder in the solution. What has happened? What is the mass of the white powder? Compare the solubility curves for the gases HCl, NH3, and SO2) with the solubility curves for the solid solutes. What generalizations(s) can you make about the relationship between solubility and temperature? According to an article in an engineering journal, there is a salt whose solubility in water increases as the water temperature increases from 0 oC to 65 oC. The salt’s solubility then decreases at temperatures above 65 oC, the article states. In your opinion, is such a salt likely to exist? Explain your answer. What could you do to verify the claims of the article? 40 30 NaCl KClO3 20 10 SO2 0 10 20 30 40 50 60 70 80 90 100 temp (0C) LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 517

SUPERSATURATED SOLUTION Solubility UNSATURATED SOLUTION more solute dissolves SATURATED SOLUTION no more solute dissolves SUPERSATURATED SOLUTION becomes unstable, crystals form increasing concentration

Classify as unsaturated, saturated, or supersaturated. 100 g H2O 80 g NaNO3 @ 30oC 40 g KCl @ 60oC 50 g NH3 @ 10oC 70 g NH4Cl @ 70oC Per 500 g H2O, 120 g KNO3 @ 40oC saturation point @ 40oC for 100 g H2O = 55 g KNO3 So sat. pt. @ 40oC for 500 g H2O = 5 x 55 g = 303 g 120 g < 303 g

Classify as unsaturated, saturated, or supersaturated. 100 g H2O 80 g NaNO3 @ 30oC unsaturated 40 g KCl @ 60oC saturated 50 g NH3 @ 10oC unsaturated 70 g NH4Cl @ 70oC supersaturated Per 500 g H2O, 120 g KNO3 @ 40oC saturation point @ 40oC for 100 g H2O = 55 g KNO3 So sat. pt. @ 40oC for 500 g H2O = 5 x 55 g = 303 g 120 g < 303 g unsaturated

Factors Affecting Solubility As To , rate (except gases) 1. temperature As size , rate 2. particle size More mixing, rate 3. mixing 4. nature of solvent or solute Polar vs. non-polar vs. ionic

Gas Solubility Higher Temperature …Gas is LESS Soluble CO2 O2 2.0 O2 Higher Temperature …Gas is LESS Soluble CO Solubility (mM) 1.0 The general rule of thumb is that the solubility of gases decreases when temperature increases. He 10 20 30 40 50 Temperature (oC)

Concentration…a measure of solute-to-solvent ratio concentrated vs. dilute “lots of solute” “not much solute” “watery” Add water to dilute a solution; boil water off to concentrate it.

Concentration “The amount of solute in a solution” A. mass % = mass of solute X 100% (% solution) mass of solution B. parts per million (ppm)  also, ppb, ppt, pph – commonly used for minerals or ppm = mass of solute X 1,000,000 contaminants in water supplies mass of solvent C. molarity (M) = moles of solute L of solution – used most often in this class mol L M D. molality (m) = moles of solute kg of solvent E. mole ratio (or mole fraction) = moles (n) solute (or solvent) moles (n) solution

One molar solution.

Molarity Find the molarity of a solution containing 75 g of MgCl2 in 250 ml of solution. 75 g MgCl2 1 mol MgCl2 95.21 g MgCl2 0.25 L solution = 3.2M MgCl2

Molality Find the molality of a solution containing 75 g of MgCl2 in 250 mL of water. 75 g MgCl2 1 mol MgCl2 95.21 g MgCl2 0.25 kg water = 3.2m MgCl2

Molality How many grams of NaCl are req’d to make a 1.54m solution using 0.500 kg of water? 0.500 kg water 1.54 mol NaCl 1 kg water 58.44 g NaCl 1 mol NaCl = 45.0 g NaCl

Preparing Solutions 500 mL of 1.54M NaCl molality molarity 1.54m NaCl in 0.500 kg of water 500 mL of 1.54M NaCl mass 45.0 g of NaCl add 0.500 kg of water mass 45.0 g of NaCl add water until total volume is 500 mL 500 mL water 500 mL volumetric flask 45.0 g NaCl 500 mL mark

Making a Dilute Solution remove sample moles of solute initial solution same number of moles of solute in a larger volume mix Making a Dilute Solution diluted solution Timberlake, Chemistry 7th Edition, page 344

Dilution Preparation of a desired solution by adding water to a concentrate (i.e stock solution). Moles of solute remain the same.

Dilution **Safety Tip: When diluting, add acid or base to water.** Acids (and sometimes bases) are purchased in concentrated form (“concentrate”) and are easily diluted to any desired concentration. Dilutions of Solutions  **Safety Tip: When diluting, add acid or base to water.** C = concentrate D = dilute Dilution Equation: Concentrated H3PO4 is 14.8 M. What volume of concentrate is required to make 25.00 L of 0.500 M H3PO4? VC = 0.845 L = 845 mL

Molarity and Stoichiometry V P mol M = mol L mol = M x L M x L M x L __ 1 Pb(NO3)2(aq) + KI (aq)  PbI2(s) + KNO3(aq) 2 1 2 What volume of 4.0 M KI solution is required to yield 89 g PbI2 with excess lead nitrate?

1 Pb(NO3)2(aq) + 2 KI (aq)  1 PbI2(s) + 2 KNO3(aq) ? L 4.0 M 89 g What volume of 4.0 M KI solution is required to yield 89 g PbI2 with excess lead nitrate? Strategy: (1) Find mol KI needed to yield 89 g PbI2. (2) Based on (1), find volume of 4.0 M KI solution. 1 mol PbI2 2 mol KI X mol KI = 89 g PbI2 = 0.39 mol KI 461 g PbI2 1 mol PbI2 M = mol L L = mol M = 0.39 mol KI 4.0 M KI = 0.098 L of 4.0 M KI

CuSO4(aq) + Al (s)  Cu(s) + Al2(SO4)3(aq) 2 3 1 x mol 11 g How many mL of a 0.500 M CuSO4 solution will react with excess Al to produce 11.0 g Cu? Al3+ SO42– __CuSO4(aq) + __Al (s)  __Cu(s) + __Al2(SO4)3(aq) 3 CuSO4(aq) + Al (s)  Cu(s) + Al2(SO4)3(aq) 2 3 1 x mol 11 g 1 mol Cu 3 mol CuSO4 X mol CuSO4 = 11 g Cu = 0.173 mol CuSO4 63.5 g Cu 3 mol Cu M = mol L L = mol M 0.173 mol CuSO4 0.500 M CuSO4 = 0.346 L 1000 mL 0.346 L = 346 mL 1 L

Cu + 2AgNO3  2Ag + Cu(NO3)2 1.5 L .10 mol AgNO3 1 L 1 mol Cu 2 mol How many grams of Cu are required to react with 1.5 L of 0.10M AgNO3? Cu + 2AgNO3  2Ag + Cu(NO3)2 ? g 1.5L 0.10M 1.5 L .10 mol AgNO3 1 L 1 mol Cu 2 mol AgNO3 63.55 g Cu 1 mol Cu = 4.8 g Cu Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g ? L 0.90 L 2.5M 79.1 g of zinc react with 0.90 L of 2.5M HCl. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP? Zn + 2HCl  ZnCl2 + H2 79.1 g 0.90 L 2.5M ? L Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g ? L 0.90 L 2.5M 79.1 g Zn 1 mol Zn 65.39 g Zn 1 mol H2 Zn 22.4 L H2 1 mol = 27.1 L H2 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g 0.90 L 2.5M ? L 0.90 2.5 mol HCl 1 L 1 mol H2 2 mol HCl 22.4 L H2 1 mol H2 = 25 L H2 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Limiting Reactants Zn: 27.1 L H2 HCl: 25 L H2 Limiting reactant: HCl Excess reactant: Zn Product Formed: 25 L H2 left over zinc Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem