2018年第一学期UCL大学本科化学课件.pptx

AP ChemistryExam Review;Big Idea #6;What is chemical equilibrium? ;Manipulating Q and K;Kinetics and Equilibrium;Q vs. K;Calculating K;Calculating K;0.497 atm Cl2 and 0.00752 atm Cl;Magnitude of K;Le Chatelier’s Principle;Experimentally Examining Le Chatelier’s Principle;Changes to Q and K for a System at Equilibrium;Acid/Base Particulates;pH of Weak or Strong Acid;Titrations;Kw and Temperature;Acid/Base Mixtures and its pH;pH and Acid/Base Equilibria;Acid/Base reaction species;How to Build a Buffer:;A 50.0 mL sample of 0.50 M HC2H3O2 is titrated to the half equivalence point with 25.0 mL of 0.50 M NaOH. Which of the following options shows the correct ranking of the molarities of the species in solution? (pKa for HC2H3O2 is 4.7) a. [HC2H3O2] [C2H3O2 1-] [ H+ ] [ OH- ] b. [HC2H3O2] = [C2H3O2 1-] [ H+ ] [ OH- ] c. [HC2H3O2] [C2H3O2 1-] = [ H+ ] [ OH- ] d. [C2H3O2 1-] [HC2H3O2] [ OH- ] [ H+ ] Option B is correct. At the half equivalence point, the concentrations of the weak acid and its conjugate base are equal because the OH- ion has reacted with half of the original acetic acid. The pH of the buffer is equal to the pKa at that point, so the [H+] is 10-4.7 which is far lower than the concentrations of the conjugates but far higher than the [OH-] which has a value of 10-9.3.;The Buffer Mechanism;Ksp and Solubility Calculations;Find a Ksp from solubility data;Common Ion Effect;Salt dissolution: ?H and ?S;K, ?G° and thermodynamic favorability;Science Practices;Gravimetric Analysis;How It’s Done/Analysis: Volatile liquid heated until completely vaporized PV=nRT to solve for n: Temperature of water bath=T Small hole in stopper means Pressure = room pressure Volume=volume of flask Divide Mass of recondensed unknown by moles, compare to known molar mass to ID unknown Assume vapor completely recondenses ?If lose vapor, then mass would be too low, and moles would be low;Virtual Lab;Filter and collect alum crystals Possible Error Sources: crystal