Solutions Edward Wen

Outline Composition of solution: solvent vs. solute Factors affecting Solubility Concentration of solution: Definition & Calculation Mass% Molarity* Solution Stoichiometry

Solution Homogeneous mixtures composition may vary from one sample to another appears to be one substance, though contains multiple materials most Homogeneous materials are actually Solutions Gas state: common air Liquid: Gasoline (dozens of compounds), Soda water (sugar or asparatame, CO2, citric acid, fructose) Solid: Alloy such as brass

Solutions: Solute + Solvent Solute: the dissolved substance. Example: Sugar and CO2 gas in Soda seems to “disappear” “takes on the state” of the solvent Solvent: the substance solute dissolves in. Example: Water in Soda does not appear to change state Aqueous solutions: solutions in which the solvent is water.

Common Types of Solution Solution Phase Solute Phase Solvent Phase Example Gaseous solutions gas air (mostly N2 & O2) Liquid solutions liquid solid soda (CO2 in H2O) vodka (C2H5OH in H2O) seawater (NaCl in H2O) Solid solutions brass (Zn in Cu) Alloys: solutions that contain Metal solutes and a Metal solvent, such as Nickel (5 cents of 1$), Brass, Stainless steel

How Soluble? Solubility Soluble: when one substance (solute) dissolves in another (solvent)  Homogeneous Salt and Sugar are soluble in water: Saline and Soda Acetic acid (HC2H3O2) in water: Vinegar Oxygen gas in Nitrogen gas: Air Insoluble: when one substance does not dissolve in another  Heterogeneous Oil is insoluble in water: Italian salad dressing

Will It Dissolve? Like Dissolves Like Chemist’s Rule of Thumb – a chemical will dissolve in a solvent if it has a similar structure to the solvent when the solvent and solute structures are similar, the solvent molecules will attract the solute particles at least as well as the solute particles to each other

Classifying Solvents: Polar vs. Nonpolar Structural Feature Water, H2O polar O-H Ethyl Alcohol, C2H5OH Acetone, C3H6O C=O Toluene, C7H8 nonpolar C-C & C-H Hexane, C6H14

Solubility in Water, A Polar Solvent? Ionic compound (Yes): Ions are attracted to polar Water. Salt NaCl dissolve in water Polar molecules (Yes): attracted to polar solvents table Sugar, Alcohol, glucose Nonpolar molecules are NOT attracted to Water b-carotene, (C40H56), is not water soluble; it dissolves in fatty (nonpolar) tissues Those molecules with both polar and nonpolar structures: depends on structural features in the molecule

Salt Dissolved in Water

Solubility Definition: the maximum amount of solute that can be dissolved in a given amount of solvent Example, at room temperature, 100. g water can dissolve 200 g sugar. Solubility = 200 g sugar/100g water. Usually a limit to the solubility of one substance in another Exceptions: gases are always soluble in each other two liquids that are mutually soluble are said to be miscible alcohol and water are miscible

Descriptions of Solubility Saturated solution has the maximum amount of solute that will dissolve in that solvent at that temperature. If more solute is added, it _______ dissolve. Unsaturated solution is holding ______ solute than it is capable of. It can dissolve ______ solute. Supersaturated solution is holding ____ solute than it is capable of at that temperature Unstable If more solute is added, ___________________. Will not; less; more; more; more solute will separate from the solution

Adding Solute to various Solutions ___saturated ___saturated _____saturated

Supersaturated Solution is unstable A supersaturated solution has more dissolved solute than the solvent can hold, not stable. When disturbed, all the solute above the saturation level will separate. Online demo: A grain of sodium acetate crystal is placed in the middle of supersatured sodium acetate solution.

Solubility of Solid Increases as Temperature increase Application: To make pure crystals, make saturated solution at high temp cooling leads to supersaturated solution, extra crystals forms at lower temp.

Solubility of Gases Decreases at higher Temperature Observation 1: Warm soda pop fizzes more than cold soda pop Cause: Solubility of CO2 in water is ________ (higher/lower) at high temperature. Observation 2: When water is heated up, gas bubbles appear even before boiling occurs. Cause: Solubility of air in water decreases as temperature increases.

Solubility of Gas depends on Pressure higher pressure = higher solubility CO2 is dissolved under Pressure into bottled/canned soda

Under which conditions does oxygen gas have the best solubility? high temperature and high partial pressure (PO2) high temperature and low partial pressure low temperature and low partial pressure low temperature and high partial pressure

Solution Concentration Descriptions Diluted solutions have low solute concentrations. Soda drink from the soda fountain. Concentrated solutions have high solute concentrations. Syrup in the storage tank in the soda fountain.

Concentrations – Quantitative Descriptions of Solutions Solutions have variable composition. Salt vs. Water in Seawater To describe a solution accurately, you need to describe the components and their relative amounts Concentration = amount of solute in a given amount of solution Seawater: Salt concentration 3.4% Dead Sea: Salt concentration 30% Vinegar: Acetic acid concentration 5%

Mass Percent (%) mass of solute (gram) in every 100 gram of solution if a solution is 0.9% by mass, then there are 0.9 grams of solute in every 100 grams of solution Mass of solution = Mass of solute + Mass of solvent Mass% = _________________ x _____

Information Given: 27.5 g C2H6O; 175 mL H2O Find: % by mass Example: Calculate the mass percent of a solution containing 27.5 g of ethanol (C2H6O) and 175 mL of H2O (assume the density of H2O is 1.00 g/mL) Information Given: 27.5 g C2H6O; 175 mL H2O Find: % by mass Mass of Solvent = 175 g Mass of Solution = 202.5 g Mass of Solute = 27.5 g Mass% = 13.6

Using Concentrations as Conversion Factors concentrations show the relationship between the amount of solute and the amount of solvent 12% by mass sugar(aq) means 12 g sugar  100 g solution The concentration can then be used to convert the amount of solute into the amount of solution, or visa versa

Example: A soft drink contains 11. 5% sucrose (C12H22O11) by mass Example: A soft drink contains 11.5% sucrose (C12H22O11) by mass. What volume of soft drink in milliliters contains 85.2 g of sucrose? (assume the density is 1.00 g/mL) Information Given: 85.2 g C12H22O11 Find: mL sol’n V = 741 mL

Preparing a Solution What we need to know: Amount of solution & Concentration of solution Calculate the mass of solute needed start with amount of solution use concentration as a conversion factor 5% by mass solute Þ 5 g solute  100 g solution

Preparing a Solution by Mass% Example - How would you prepare 250.0 g of 5.00% by mass glucose solution (normal glucose)? 12.5 g of glucose 237.5 g of water

Solution Concentration: Molarity Definition: Moles of solute per 1 liter of solution Purpose: describing how many molecules of solute in each liter of solution Unit: mole/L, abbreviated as “M”. If a sugar solution concentration is 2.0 M , 1 liter of solution contains 2.0 moles of sugar. molarity = moles of solute liters of solution

Why Molarity? Many reagents used in chemistry, even many biology labs, are in the form of solution. Molarity concentration of solution is particularly important and useful because: Easy to use: To obtain given amount (mole) of reagent, just calculate the volume of solution to be used: Easy to prepare a solution to a given molarity 28

Calculations involving Molarity Molarity = mole  Volume (L) Solve for mole: Mole = ___________________ Solve for volume of solution in liters: Volume (L) = ______________ 29

Example – How to prepare 500 mL of 0.020 M NaCl solution? Part A. Calculate the mass NaCl needed: 0.010 mol NaCl 0.58 g NaCl 30

Example: How to prepare 500 mL of 0. 020 M NaCl solution. Part B Example: How to prepare 500 mL of 0.020 M NaCl solution? Part B. Preparation Weigh out 0.58 g NaCl and Add it to a 500mL Volumetric flask. Step 1 Step 2 Add water to dissolve the NaCl, then add water to the mark. Step 3 Put the lid on, Invert the flask to homogenize the solution

Example: Calculate the molarity of a solution made by putting 15 Example: Calculate the molarity of a solution made by putting 15.5 g of NaCl into a beaker and adding water to make 1.50 L of NaCl solution. Information Given: 15.5 g NaCl; 1.50 L sol’n Find: molarity, M CF: 58.44 g = 1 mol NaCl; 0.265 mol NaCl 0.177 M NaCl

Example: How many liters of a 0. 114 M NaOH solution contains 1 Example: How many liters of a 0.114 M NaOH solution contains 1.24 mol of NaOH? Information Given: 1.24 mol NaOH Find: L solution CF: 0.114 mol = 1 L SM: mol → L 10.9 L

Molarity of Ions: Dissociation of Ionic Compound When strong electrolytes dissolve, virtually all the solute particles dissociate into ions From the formula of the compound and the molarity of the solution  Determine the Molarity of the dissociated Ions by simply multiplying the salt Concentration by the Number of Ions

Molarity & Dissociation CaCl2(aq) = Ca2+(aq) + ____ Cl-(aq) 1 mole “molecules” = ___ mole ions + ___ mole ions 1 M CaCl2 =____M Ca2+ ions + ____ M Cl- ions 0.25 M CaCl2 = _____ M Ca2+ + ____ M Cl-

Making a Solution by Dilution Example: A student added 1.00 L water to 2.00 L 1.00 M HCl. The final volume became 3.00 L. mole HCl before mixing = M1 x V1 = 2.00 mole = mole HCl after mixing more water. When mixing more solvent into a solution, the volume of final solution is greater than the original solution The mole of solute remains the same before and after mixing more solvent For dilution, mole solute = M1 x V1 = M2 x V2

Example—What Volume of 12. 0 M KCl Is Needed to Make 5. 00 L of 1 Example—What Volume of 12.0 M KCl Is Needed to Make 5.00 L of 1.50 M KCl Solution? Given: M1 = _____; V2 = ____ L, M2 = ____ M Find: V1 Equation: M1 x V1 = M2 x V2 0.625 L

Example—What is the final concentration (molarity) if 10. 0 mL 12 Example—What is the final concentration (molarity) if 10.0 mL 12.0 M HCl is diluted to a final volume of 5.00 L? Given: M1 = _____; V1 = ____ L, V2 = ____ Find: V1 Equation: M1 x V1 = M2 x V2 0.0240 M

Solution Stoichiometry Stoichiometry: Balanced chemical equation shows the mole ratio among reactants and products in a reaction. 2 H2(g) + O2(g) → 2 H2O(l) Solution molarity  moles of solute and liters of solution Measure the moles of a material in a reaction in solution by knowing its molarity and volume.

Example: How much 0.115 M KI solution, in liters, is required to completely precipitate all the Pb2+ in 0.104 L of 0.225 M Pb(NO3)2? 2 KI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2 KNO3(aq) 0.115 mol KI  1 L solution 0.225 mol Pb(NO3)2  1 L solution 2 mol KI  1 mol Pb(NO3)2 Solution Map: L Pb(NO3)2 → mol Pb(NO3)2 → mol KI → L KI = 0.407 L

Full solution for Examples

Example: Calculate the mass percent of a solution containing 27 Example: Calculate the mass percent of a solution containing 27.5 g of ethanol (C2H6O) and 175 mL of H2O (assume the density of H2O is 1.00 g/mL) Information Given: 27.5 g C2H6O; 175 mL H2O Find: % by mass Eq’n: CF: 1.00 g H2O = 1 mL H2O Design a Solution Map: Mass Solute & Vol Solvent Mass Percent density Mass Solution Mass Solvent Solution = solute + solvent

Apply the Solution Maps Example: Calculate the mass percent of a solution containing 27.5 g of ethanol (C2H6O) and 175 mL of H2O (assume the density of H2O is 1.00 g/mL) Information Given: 27.5 g C2H6O; 175 mL H2O Find: % by mass Eq’n: CF: 1.00 g H2O = 1 mL H2O SM: mass sol & vol solv → mass solv → mass sol’n → mass percent Apply the Solution Maps Mass of Solution = Mass C2H6O + Mass H2O = 27.5 g C2H6O + 175 g H2O = 202.5 g

Apply the Solution Maps - Equation Example: Calculate the mass percent of a solution containing 27.5 g of ethanol (C2H6O) and 175 mL of H2O (assume the density of H2O is 1.00 g/mL) Information Given: 27.5 g C2H6O; 175 mL H2O Find: % by mass Eq’n: CF: 1.00 g H2O = 1 mL H2O SM: mass sol & vol solv → mass solv → mass sol’n → mass percent Apply the Solution Maps - Equation = 13.5802% = 13.6%

Example: A soft drink contains 11. 5% sucrose (C12H22O11) by mass Example: A soft drink contains 11.5% sucrose (C12H22O11) by mass. What volume of soft drink in milliliters contains 85.2 g of sucrose? (assume the density is 1.00 g/mL) Information Given: 85.2 g C12H22O11 Find: mL sol’n CF: 11.5 g C12H22O11  100 g sol’n 1.00 g sol’n = 1 mL sol’n Design a Solution Map: Mass Solute Volume Solution mass percent density Mass Solution

Apply the Solution Map = 740.87 mL = 741 mL Example: A soft drink contains 11.5% sucrose (C12H22O11) by mass. What volume of soft drink in milliliters contains 85.2 g of sucrose? (assume the density is 1.00 g/mL) Information Given: 85.2 g C12H22O11 Find: mL sol’n CF: 11.5 g C12H22O11  100 g sol’n 1.00 g sol’n = 1 mL sol’n SM: g sucrose → g sol’n → mL sol’n Apply the Solution Map = 740.87 mL = 741 mL

Preparing a Solution by Mass% Example - How would you prepare 250.0 g of 5.00% by mass glucose solution (normal glucose)? Dissolve 12.5 g of glucose in 237.5 g of water to give the solution

Example: Calculate the molarity of a solution made by putting 15 Example: Calculate the molarity of a solution made by putting 15.5 g of NaCl into a beaker and adding water to make 1.50 L of NaCl solution. Information Given: 15.5 g NaCl; 1.50 L sol’n Find: molarity, M CF: 58.44 g = 1 mol NaCl; Design a Solution Map: Mass Solute Mole Solute Molarity Volume Solution L Solution already liters

Example: Calculate the molarity of a solution made by putting 15 Example: Calculate the molarity of a solution made by putting 15.5 g of NaCl into a beaker and adding water to make 1.50 L of NaCl solution. Information Given: 15.5 g NaCl; 1.50 L sol’n Find: molarity, M CF: 58.44 g = 1 mol NaCl; Apply the Solution Map = 0.177 M NaCl

Example: How many liters of a 0. 114 M NaOH solution contains 1 Example: How many liters of a 0.114 M NaOH solution contains 1.24 mol of NaOH? Information Given: 1.24 mol NaOH Find: L solution CF: 0.114 mol = 1 L SM: mol → L Apply the Solution Map

Find: volume of KI solution, L Example: How much 0.115 M KI solution, in liters, is required to completely precipitate all the Pb2+ in 0.104 L of 0.225 M Pb(NO3)2? 2 KI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2 KNO3(aq) Write down the given quantity and its units. Given: 0.104 L Pb(NO3)2 Write down the quantity to find and/or its units. Find: volume of KI solution, L

Collect needed conversion factors: Example: How much 0.115 M KI solution, in liters, is required to completely precipitate all the Pb2+ in 0.104 L of 0.225 M Pb(NO3)2? 2 KI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2 KNO3(aq) Information: Given: 0.104 L Pb(NO3)2 Find: L KI Collect needed conversion factors: 0.115 M KI  0.115 mol KI  1 L solution. 0.225 M Pb(NO3)2  0.225 mol Pb(NO3)2  1 L solution. Chemical equation  2 mol KI  1 mol Pb(NO3)2.

Example: How much 0.115 M KI solution, in liters, is required to completely precipitate all the Pb2+ in 0.104 L of 0.225 M Pb(NO3)2? 2 KI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2 KNO3(aq) Information: Given: 0.104 L Pb(NO3)2 Find: L KI Conversion Factors: 0.115 mol KI  1 L solution 0.225 mol Pb(NO3)2  1 L solution 2 mol KI  1 mol Pb(NO3)2 Design a solution map: L Pb(NO3)2 mol Pb(NO3)2 mol KI L KI

Apply the solution map: Example: How much 0.115 M KI solution, in liters, is required to completely precipitate all the Pb2+ in 0.104 L of 0.225 M Pb(NO3)2? 2 KI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2 KNO3(aq) Information: Given: 0.104 L Pb(NO3)2 Find: L KI Conversion Factors: 0.115 mol KI  1 L solution 0.225 mol Pb(NO3)2  1 L solution 2 mol KI  1 mol Pb(NO3)2 Solution Map: L Pb(NO3)2 → mol Pb(NO3)2 → mol KI → L KI Apply the solution map: = 0.40696 L = 0.407 L