CHEM 1212 1nd Edition Lecture 4 Outline of Last Lecture I Structures and Formulas of Ionic Solids II Bonding in Metals and Semiconductors III Semiconductors IV Bonding in Ionic Compounds Lattice Energy V The Solid State Other types of solid materials VI Phase Changes Involving Solids VII Phase Diagrams Outline of Current Lecture I Units of Concentration II The Solution Process III Factors Affecting Solubility Pressure and Temperature Current Lecture I Units of Concentration a Molarity M is the number of moles of solute per liter of solution but this does not tell us the exact amount of solvent used to make the solution b Three concentration units that reflect the number of molecules of solute per solvent molecule molality mole fraction and weight percent i Molality m is the amount of solute mol per kilogram of solvent 1 Molality concentration c amount of solute mol mass of solvent kg These notes represent a detailed interpretation of the professor s lecture GradeBuddy is best used as a supplement to your own notes not as a substitute II 2 The molarity and molality of a given solution cannot be the same although when the solution is dilute the difference can be very small ii Mole Fraction X is the amount of the component nA divided by the total amount of all components of the mixture 1 Mole fraction of liquid A XA nA nA nB NC 2 Example consider a solution that contains 1 00 mol of ethanol in 9 00 mol of water a Xethanol 1 00 mol ethanol 1 00 mol ethanol 9 00 mol water b Xwater 9 00 mol water 1 00 mol ethanol 9 00 mol water 3 The sum of the mole fractions of the components in the solution must equal 1 000 iii Weight Percent is the mass of one component divided by the total mass of the mixture multiplied by 100 1 Weight A mass of A mass of A mass of B mass of C 2 Example an alcohol water mixture has 46 1 g of ethanol and 162 g of water the total mass is 208g and the weight percent for alcohol is a 46 1g ethanol 46 1g ethanol 162g water x 100 22 2 3 A lot of consumer products use weight percent on their packaging to list amounts of ingredients in product 4 Because naturally occurring solutions are often very dilute scientist often use parts per million ppm to express concentrations a 1 ppm 1 0g of a substance in a same with a total mass of 1 million g The Solution Process a Saturated when a solution is saturated no more of the solid placed in the solute will dissolve and the concentration of the ions in the solute will not increase i Ex If you put CuCl2 in a beaker of water it will begin to dissolve and the concentration of Cu2 aq and Cl aq in the solution will increase Eventually it will appear that no more of the CuCl2 will dissolve into the water and any more of the compound added to the water after this point will remain a solid at the bottom of the beaker ii Even though it appears that there is no more dissolving of the compound taking place the solution has actually just reached the point of dynamic equilibrium The amount of CuCl2 that is dissolving into the liquid is equal to the amount of CuCl2 being formed by the ions in the solution b c d e 1 Such substances can be linked in their equations by a set of double arrows iii Saturation gives us a way to determine the precise solubility of a solid in a liquid Solubility the concentration of solute in equilibrium with un dissolved solute in a saturated solution i Ex the solubility of CuCl2 is 70 6g in 100 mL of water at 0 degrees Celsius so if you add 100g of CuCl2 to a solution in this state you could expect 70 6g to dissolve and 29 4g of the solid would remain Liquids dissolving in liquids i Miscible when two liquids mix to an appreciable extent to form a solution ii Immiscible when liquids do not mix to form a solution iii These rules led to like dissolves like 1 Miscible ethanol water octane tetrachloride 2 Immiscible C8H18 CCl4 iv The reason two nonpolar molecules or two polar molecules can dissolve together is because the bond strength in each molecule is similar so there is no great energy change v When a polar and nonpolar molecule are mixed they generally separate into two distinct layers Solids Dissolving in Water i Like dissolves like also applies to a solid molecule mixing with a liquid ii Ex the O H bonds in sugar allow it to interact easily with water molecules through hydrogen bonding iii Ionic solids sometimes dissolve in water but not always iv Predicting ionic compound solubility 1 Enthalpy and entropy together determine the extent of solubility of an ionic compound a A more negative enthalpy usually leads to solubility However and unfavorable enthalpy does not guarantee that a substance will not dissolve 2 Network solids i e diamonds graphite and quartz do not dissolve in water covalent bonds are too strong to be broken so the lattice remains in tact Enthalpy of Solution illustrated by Potassium Fluoride KF dissolving in water i KF has an ionic crystal lattice held together by forces of attraction between opposite charges ii In water ions are separated and hydrated surrounded by water molecules iii Ion dipole forces bind water molecules to each ion III iv The energy change that occurs going from the reactants KF to the products K and F is the sum of the energies of two individual steps 1 Energy must be supplied to separate the ions in the lattice against their forces of attractions Highly endothermic process 2 Energy is evolved when the individual ions dissolve in water This is called hydration when water is the solvent Highly exothermic process a The enthalpy of the overall reaction is called the enthalpy of solution sum of two enthalpies v To be soluble and ionic compound will usually have an enthalpy of solution that is exothermic or only slightly endothermic vi The two energy quantities are affected by ion size and charge 1 A salt composed of smaller ions is expected to have a greater more negative lattice enthalpy because the ions can be closer together and have higher attractive forces this will also allow a greater solvation energy Factors Affecting Solubility Pressure and Temperature a Both affect solubility of a gas in liquid but only temperature is a factor in the solubility of solids in liquids b Dissolving Gases in Liquids Henry s Law i Henry s Laws the solubility of gas in a liquid is directly proportional to the gas pressure ii Sg kHPg 1 Sg is the gas solubility mol kg 2 Pg is the partial pressure of the gaseous solute 3 KH is Henry s law constant iii At equilibrium the rate at which gas escapes