Unit 11 Solubility Solutions Ch 17 18 I

Unit 11 - Solubility & Solutions, Ch. 17 & 18

I. Water A. The Molecule 1. O—H bond is highly polar 2. Bond angle 105° making it Bent shaped 3. Water Molecule as a whole is polar 4. Attracted to each other by intermolecular hydrogen bonds Greater electronegativity

I. Water (cont. ) B. Important Properties 1. High surface tension 2. low vapor pressure l hydrogen bonds hold molecules to one another, tendency to escape surface is low 3. high specific heat capacity l 4. 184 J/g×°C 4. high melting and boiling points l 0°C and 100°C

I. Water (cont. ) C. Surface Tension – inward force, or pull, that tends to minimize the surface area of a liquid l Surfactant – wetting agent such as soap or detergent that decreases the surface tension by interfering with hydrogen-bonding Responsible for high surface tension

I. Water (cont. ) D. Atypical Ice 1. As a typical liquid cools, density increases b/c Volume decreases as the mass stays constant 2. As water cools it first behaves like a typical liquid until it reaches 4°C 3. Below 4°C the density of water starts to decrease **Ice is one of only a few solids that float in their own liquid.

Atypical Ice Do not need to write. Density of Liquid Water and Ice Temperature (°C) Density (g/cm 3) 100° (liquid water) 0. 9584 50° 0. 9881 25° 0. 9971 10° 0. 997 4° 1. 000 (most dense) 0° (liquid water) 0° (ice) 0. 9998 0. 9168

Atypical Why does ice behave Ice so differently? Open framework arranged like a honeycomb. Framework collapses, molecules packed closer together, making it more dense

II. The Solution Process A. Solution - homogeneous mixture consisting of the same properties throughout (exists in a single phase) Solute - substance being dissolved - smaller quantity substance Solvent - dissolving medium - larger quantity substance - usually water

II. The Solution Process (cont. ) l Aqueous Solution (aq) – a solution in which the solvent is water HCl(g) + H 2 O(l) → H 3 O(aq) + Cl-(aq) potassium chromate, cobalt(II) nitrate, copper(II) sulfate, K 2 Cr. O 4 (yellow) Co(NO 3)2 (red) Cu. SO 4 (blue) potassium dichromate nickel(II) chloride, permanganate, K 2 Cr 2 O 7 (orange) Ni. Cl 2 (green) KMn. O (purple) 4

II. The Solution Process (cont. ) B. Solvation– the process of dissolving 1 st solute particles (salt) are surrounded by solvent particles (water) 2 nd solute particles (salt) are separated and pulled into solution (salt water)

II. The Solution Process (cont. ) C. “Like Dissolves Like” NONPOLAR Nonpolar solvents dissolve nonpolar compounds POLAR Polar solvents dissolve polar molecules and ionic compounds

“Polar Dissolves Polar”

Polar vs. Nonpolar l Polar Molecule – uneven electron forces acting on the central atom. l l Water Nonpolar Covalent Bond – when 2 atoms are joined by a covalent bond and the bonding electrons are shared equally

III. Water of Hydration Water molecules that are integral part of the crystal structure l Compound that contains water of hydration is called a hydrate l l (ex. Cu. SO 4. 5 H 2 O) Effloresce – losing the water of hydration l Deliquescent – removes water from air to form a solution l

III. Water of Hydration Calculate the percent by mass of water in sodium carbonate decahydrate (Na 2 CO 3 · 10 H 2 O) Percent H 2 O = mass of water x 100% mass of hydrate l Mass of 10 H 2 O = 180 g Mass of Na 2 CO 3 · 10 H 2 O = 286 g Percent H 2 O = 180 g 286 g x 100% = 62. 9% or 63%

IV. Electrolytes A. Electrolytes – compounds that conduct an electric current in solutions l All ionic compounds are electrolytes l Compounds that don’t conduct an electric current are called nonelectrolytes – not composed of ions, includes many molecular compounds (covalent compounds)

IV. Electrolytes (cont. ) - - + salt Strong Electrolyte solute exists as ions only + - water Weak Electrolyte solute exists as ions and molecules + sugar Non. Electrolyte solute exists as molecules only

Some Examples of Strong Electrolytes, Weak Electrolytes and Nonelectrolytes Strong Electrolyte Weak Electrolyte Nonelectrolyte Acids (HCl, HBr, HI, HNO 3, HCl. O 4) Heavy metal halides Hg. Cl 2, Pb. Cl 2 Most organic Compounds Bases (Na. OH, KOH) Water (very weak) Glucose Ions, Ionic compounds (Na. Cl, KCl, Ca. Cl 2, KCl. O 3, Mg. SO 4) Also called salts Organic (acids & bases) Acetic acid, aniline Glycerol

V. Heterogeneous Mixtures A. Suspensions – mixtures from which particles settle out upon standing and the average particle size is greater than 100 nm in diameter. l l Clearly identified as two substances Gravity or filtration will separate the particles B. Colloids – heterogeneous mixtures containing particles that are between 1 nm and 100 nm in diameter l l Appear to be homogeneous but particles are dispersed through medium Ex: paint, aerosol spray, smoke, marshmallow, whipped cream

V. Heterogeneous Mixtures C. Tyndall Effect - phenomenon observed when beam of light passes through a colloid or suspension l Colloids exhibit the Tyndall effect Colloid Solution

VI. Solubility • defined as the maximum grams of solute that will dissolve in 100 g of solvent at a given temperature • based on a saturated solution

Solubility SATURATED SUPERSATURATED UNSATURATED SOLUTION over max amount capable of dissolving no more solute becomes unstable, more solute crystals form dissolves Concentration Increasing l l Supersaturated solutions are not in equilibrium with the solid substance and can quickly release the dissolved solids. Saturated solution is one that is in equilibrium with respect to the dissolved substance. These conditions can quickly change with temperature.

VI. Solubility (cont. ) B. Factors Affecting Solubility 1. Stirring (agitation) l Increases solubility b/c fresh solvent is brought in contact with the surface of the solute l Surface phenomenon

VI. Solubility (cont. ) 2. Temperature l Increases solubility by increasing kinetic energy, which increases the collisions b/w molecules of solvent and the surface of the solute l Increases amount and rate of solute dissolved

VI. Solubility (cont. ) 3. Surface Area l A smaller particle size dissolves more rapidly than larger particles size l Surface phenomenon l More surface area exposed, faster rate of dissolving

VI. Solubility (cont. ) A. Solubility Curve l l shows the dependence of solubility on temperature Note: the solubility of the gases are greater in cold water than in hot water Out of the solids, which has the lowest solubility at 40°C? KCl. O 3 Which has the highest solubility at 20°C? KI

VI. Solubility (cont. ) l l l How many grams of KNO 3 can be dissolved in 100 g of water at 60°C? 110 g What is the solubility of NH 4 Cl at 50°C? 50 g/L Which salt shows the least change in solubility from 0°C to 100°C? Na. Cl

Output side 1. Which of the following compounds dissolved the highest at 20°C? 2. The lowest at 20°C? 3. Overall which compound dissolved the fastest from 0°C to 100°C? 4. Name a compound in the graph that is a gas? How do you know it’s a gas?

VII. Concentrations of Solutions l l Concentration of a solution is a measure of the amount of solute that is dissolved in a given quantity of solution. Dilute solution – contains a low concentration of solute. Concentrated solution – contains a high concentration of solute. Molarity (M) – number of moles of a solute dissolved per liter of solution l a. k. a. molar concentration

A. Molarity Calculates the number of moles in 1 L of the solution Molarity (M) = moles of solute M = n liters of solution L l Ex 1 Calculate the molarity when 2 mol of glucose is dissolved in 5 L of solution. M=n L 2 mol glucose 5 L solution = 0. 4 mol/L = 0. 4 M

Ex 2 How many moles of solute are present in 1. 5 L of 0. 24 M Na 2 SO 4? M = n of solute L of solution 1. 5 × 0. 24 M = n 1. 5 L 0. 36 mol = n “Triangle Trick” × 1. 5 n Divide 0. 24 M 1. 5 L Multiply

Ex 3 A saline solution contains 0. 90 g Na. Cl in exactly 100. 0 m. L of solution. What is the molarity of the solution? M = n of solute L of solution What are you solving for? Molarity (M) 1. CONVERT GRAMS TO MOLES!! 2. CONVERT m. L to L. 0. 90 g Na. Cl 1 mol Na. Cl = 0. 016 mol 58 g Na. Cl 100. 0 m. L 1 L 1000 m. L Kh. DMdcm L = 0. 1000 L M = 0. 016 mol 0. 1000 L Molarity = 0. 16 M

Ex 4 How many grams of solute are in 2. 40 L of 0. 650 M HCl. O 2? M=n 2. 40 L× 0. 650 M = n × 2. 40 L L 2. 40 L 1. 56 mol = n 1. 56 mol HCl. O 2 68 g HCl. O 2 1 mol HCl. O 2 = 106. 08 g HCl. O 2

Title: Molarity Problems You do not have to write the problem. You MUST show your work. BOX in your answer! 1. A solution has a volume of 2. 0 L and contains 36. 0 g of glucose. If the molar mass of glucose is 180 g, what is the molarity of the solution? 2. A solution has a volume of 250 m. L and contains 0. 70 mol Na. Cl. What is its molarity? 3. How many moles of ammonium nitrate are in 335 m. L of 0. 425 M NH 4 NO 3? 4. How many moles of solute are in 250 m. L of 2. 0 M Ca. Cl 2? How many grams of Ca. Cl 2 is this? (molar mass = 110 g)

VII. Concentrations of Solutions (cont. ) B. Making Dilutions l l You can make a solution less concentrated by diluting it with a solvent. A dilution reduces the moles of solute per unit volume, however, the total moles of solute in solution does not change M 1 × V 1 = M 2 × V 2 l Volumes can be in L or m. L, as long as the same units are used for both V 1 & V 2

Ex 1 You need 150 m. L of 0. 40 M Na. Cl and you have a 1. 0 M of Na. Cl solution. Calculate the volume of the Na. Cl solution. V 1 = 150 m. L of 0. 40 Na. Cl M 1 = 0. 400 M Na. Cl M 2 = 1. 0 M Na. Cl Unknown = V 2 M 1 × V 1 = M 2 × V 2 0. 400 M × 150 m. L = 1. 0 M × V 2 60. m. L = V 2

Ex 2 What volume of 5. 00 M sulfuric acid is required to prepare 25. 00 L of 0. 400 M sulfuric acid? M 1 = 5. 00 M V 2 = 25. 00 L M 2 = 0. 400 M M 1 × V 1 = M 2 × V 2 5. 0 M × V 1 = 0. 40 M × 25 L V 1 = 2. 0 L

Ex 3 How many milliliters of a stock solution of 2. 00 M Mg. SO 4 would you need to prepare 100. 0 m. L of 0. 400 M Mg. SO 4? M 1 = 2. 00 M Mg. SO 4 M 2 = 0. 400 M Mg. SO 4 V 2 = 100 m. L of 0. 400 M Mg. SO 4 Unknown = V 1 M 1 × V 1 = M 2 × V 2 2. 00 M × V 1 = 0. 400 M × 100 m. L V 1 = 20 m. L

LEFT SIDE Title: Dilution Problems You do not have to write the problem. You MUST show your work. BOX in your answer. 5. In making a dilution how many m. L of 4. 0 M HCl are needed to make 250 m. L of 0. 30 M HCl? 6. In making a dilution how many m. L of a 3. 00 M Na. Br solution are needed to make 175 m. L of 0. 400 M Na. Br? 7. How many milliliters of a stock solution of 2. 00 M Mg. SO 4 would you need to prepare 100 m. L of 0. 400 M Mg. SO 4? 8. How many milliliters of a stock solution of 4. 00 M KI would you need to prepare 250. 0 m. L of 0. 760 M KI?

Adding to the mole road map!! Solutions ol m ? L 1 mol ? Molarity: 5. 0 M = 5. 0 mol 1 L Molecule Atoms (ions) a ul rm o F it un

VIII. Using Molarity in Stoichiometry Zn + 2 HCl → Zn. Cl 2 + H 2 How many milliliters of 3. 00 M HCl are required to react with 17. 35 g of zinc? 17. 35 g Zn 1 mol Zn 2 mol HCl 1 L HCl 1000 m. L HCl 65 g Zn 1 mol Zn 3. 00 mol HCl 1 L HCl = 177. 95 m. L HCl

Using Molarity in Stoichiometry Cu + 2 Ag. NO 3 → 2 Ag + Cu(NO 3)2 How many grams of copper are required to react with 40. 0 m. L of 9. 0 M Ag. NO 3? 1 L Ag. NO 3 40. 0 m. L Ag. NO 3 1000 m. L Ag. NO 3 9. 0 mol 1 mol Cu Ag. NO 64 g Cu 3 1 L Ag. NO 3 2 mol Ag. NO 3 1 mol Cu = 11. 52 g Cu

IX. Percent Solutions If both solute and solvent are liquids the concentration of the solute is expressed as a percent. Percent by volume (%(v/v)) = volume of solute solution volume X 100 For solutions of solids dissolved in liquids is percent (mass/volume). Percent (mass/volume) (%(m/v)) = mass of solute (g) X 100 solution volume (m. L)

IX. Percent Solutions Ex 1 What is the percent by volume of ethanol (C 2 H 6 O), or ethyl alcohol, in the final solution when 85 m. L of ethanol is diluted to a volume of 250 m. L with water? % (v/v) = 85 m. L ethanol X 100 250 m. L solution l = 34% ethanol (v/v)

IX. Percent Solutions Ex 2 A solution of glucose (C 6 H 12 O 6) contains 2. 8 g in 20. 0 m. L of solution. What is the percent (m/v) of the solution? % (m/v) = 2. 8 g X 100 20. 0 m. L l = 14% glucose (m/v)

Why do we get to make ice cream? X. Freezing-Point Depression l the difference in temperature between the freezing point of a solution and that of the pure solvent. l The salt depresses the freezing point of water because it disrupts the crystal formation of the water

Why do we get to make ice cream? When you add salt to the ice, it lowers the freezing point of the ice, so even more energy has to be absorbed from the environment in order for the ice to melt. This makes the ice colder than it was before, which is how your ice cream freezes Compounds that break into ions upon dissolving, like Na. Cl breaks into Na+ and Cl-, are better at lowering the freezing point than substances that don't separate into particles because the added particles disrupt the ability of the water to form crystalline ice. The more particles there are, the greater the disruption and the greater the impact on particle-dependent properties (colligative properties) like freezing point depression.

Ice Cream Lab 1. Put all ingredients on my desk. 2. While I am mixing the ingredients. You need to prepare your coffee can and freezer bag. 3. To prepare your coffee can, put a 5 cm layer of ice then sprinkle 1 cm of rock salt, then add another layer of ice then rock salt. (3/4 th full, leave room to put in the ice cream bag) 4. Then you will get your freezer zip lock bag and duct tape the 3 side edges. 5. I will then put ice cream in your bag and you will seal the opening with duct tape. 6. Place the taped bag of ice cream inside the middle of the coffee can and duct tape the lid onto the coffee can. 7. Roll your coffee can outside for 20 minutes! 8. When complete, take bag out of can and rinse off bag. 9. Then enjoy ice cream!