Chem 104 1st Edition Lecture 12 Outline of Last Lecture I. Continuation on solutions, their solubility, and their propertiesA. Heat of solution and their energetics of a solution formation and solution processesII. Heat of hydration and its equationIII. Solution equilibrium and solubility limitIV. SolidsA. The temperature dependency of solids solubility in waterB. A solid’s purification by crystallizationOutline of Current Lecture I. Practice with solving for lattice energy and hydrationII. Henry’s Law and useIII. Molarity and dissociation’s role in ionic compounds and how to determine.IV. Solution concentration, parts solute, parts solution, and percent concentration:V. Molarity (m), parts per solution, and ppmVI. Equations for solving concentrationCurrent LectureI. In order to make a solution the bonds must be broken between solutes and solvent. Solute-solute and solvent–solvent are endothermic. A solute-solvent mix is exothermic. The equation that represents this is:∆ Hsolution=∆ Hsolute+∆ Hsolvent+HmixII. Remember that solute and solvent are both positive while the mix is negative. The combination of solvent and mix is the change in heat of hydration. The solute by itself represents the lattice energy.A. Example:What is the lattice energy of KI if Hsol’n= + 21.5 kJ/mol and the Hhydration= -583 kJ/mol?∆ Hsolution=∆ Hhydration−∆ Hlattice−∆ Hlattice=−583kJmol−(+21.5kJmol)¿−604.5=−605kJmolIII. Pressure dependence of solubility is dependent of the partial pressure of a gas in contact with a liquid. The larger the partial pressure, the more soluble the gas is in the liquid. Henry’s Law is the measure of solubility of a gas as it is directly proportional to its partial pressure. This is represented by:Sgas=kHPgasA. Example 1:What is the pressure of CO2 is required to keep [CO2] = 0.12 M in soda at 25°C?Sgas=kHPgasPgas=SgaskHPgas=0.12 M3.4∗10−2Matm=3.5 atmB. Example 2:How many grams of NH3 will dissolve in 0.10 L of solution when its partial pressure is 7.6 torr? (kH= 58 M/atm)First convert torr to atm.7.6torr∗1 atm760torr=.01 atmThen solve for Sgas.Sgas=kHPgasSgas=(58matm).01 atmSgas=.58 MThen convert M to moles and then to grams.MN H3=SN H3=¿moles N H3V(l).58 M=¿moles N H30.10 L.058 mols=¿ moles N H3.058moles∗17 g1mol=.989 gIV. Concentrations have variable composition. A concentration is the amount of solute in a given solution. To describe a concentration, uses dilute and concentrated. Dilute is making asolution weaker or thinner by adding water or another solvent. Concentrated is having a reducing agent or water removed. This is represented by the equation:molarity , M =moles of soluteLof solutionV. Molarity and dissociation revolve around the molarity of the ionic compound. The molarity allows you to determine the molarity of dissolved ions.A. Example 1:CaCl2(aq) = Ca2+(aq) + Cl-(aq)1.0 M CaCl2 = 1.0 mol CaCl2 per liter[Ca2+] = 1.0 M[Cl-] = 2 * 1.0 M = 2.0 MB. Example 2:Al2Cl3(aq) = 0.3 M[Al3+] = 2 * 0.3 M = 0.6 M[Cl-] = 3 * 0.3 M = 0.9 MVI. Solution Concentration is based on molarity, m. This is the number of moles per 1 kilogram of solvent not solution. It also does not vary with temperature. This is represented as:molarity , m=moles of solutekg of solventParts solute in parts solution is measured by mass or volume. Parts are generally in the sameunits. A percentage is the parts solute in every 100 parts solution In parts per million (ppm) it is the parts of solute for every 1 million parts solution.A. Example:solution is 0.9% by mass = 0.9 grams solute in every 100 grams solutionB. Example:solution is 36 ppm by volume = 36 ml solute in 1 million ml solutionPercentage concentration is measured in many equations. Below are a few mentioned in class today:percent=part (solute)whole(solution)∗100 %pe rcentmassvolume=mass of solute , gvolumeof solution ,ml∗100 %mass of solute +volume of solvent ≠ volume of solutionvolume percent =v olume of solutevolume of solution∗100