CHEM 111 1st Edition Lecture 7 Past Lecture Review:Limiting Reactants, How to calculate Theoretical yield, Actual Yield, Percentage Yield.Current Lecture: 2/11/2014When Aq is denoted in parenthesis, it designates water solutions. Na+ (aq)When S is denoted in parenthesis, it means it is a solid.When L is denoted in parenthesis, it means it is a pure liquid.For chemical equations, you can sometimes cancel out the elements that are repeated and don’t change.Ex: H+ (aq) + Cl- (aq) + K+ (aq) + OH-(aq) ---------> H2O (aq) + K+ (aq) + Cl- (aq)The Potassium and Chlorine would not have to be present in the final equation as no changes were made. Instead, it would be written as the following…H+ (aq) + OH- (aq) ---------> H2O (aq)Molarity – The number of moles of solute in one liter of solution.How to calculate: moles of solute / liters of solutionEx: What is the molar concentration of NaF in a solution prepared by dissolving 2.51 g of NaF in enough water to form 200 mL of solution?Rule: Always convert mL into L2.51 g / 41.97 g/mol = 0.06 mol (41.97 comes from the molar mass of NaF)0.06 mol / 0.2 L = 0.3 M Dilution: Solutions of lower concentration can be prepared by dilution of more concentrated solutions of known molarity.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.Equation to calculate Dilution (only used for dilution problems, not problems involving equations)Molarity (conc) x volume (conc) = Molarity (dil) x Volume (dil)Titration: The concentration and volume of a solution of known concentration is used to determine the concentration of an unknown solution.Equivalence Point: The point in a titration where stoichiometrically equivalent amounts of the two reactants have been added.Ex: Calculate the molarity of a hydrochloric acid solution if 26.4 mL of the solution neutralizes 30.0 mL of 0.120 M Barium Hydroxide.1. Balance the Equation: 2 HCl + Ba(OH)2 -----------------> 2H2O + BaCl22. Multiply the liters of solution by the moles per liter. 0.30 L x 0.120 mol/L = 0.0036 moles.Since there is a 2 in front of the HCl, there is a mole ratio of 2:1, so you must multiply 0.0036 by 2 to get .0072 moles.3. Divide .0072 mol / .0264 L = 0.273 MCurrent Lecture Review:How to calculate Titration/Dilution, How to calculate Molarity, How to denote aqueous/solid/pure liquid solutions, how to write shorthand chemical