Wright CHM 1220 - CHM 1220_Exam3Review_InClass_Fall2017_answers (2 pages)

CHM 1220 In Class Activity Aqueous Equilibria Complex Ions and Electrochemistry Fall 2017 ANSWERS 1 A 25 0 mL solution of quinine was titrated with 1 00 M hydrochloric acid HCl It was found that the solution initially contained 0 125 moles of quinine What was the pH of the solution after 75 00 mL of the HCl were added Quinine is monobasic with pKb 5 10 quinine H aq OH aq QH 1 00 mol HCl Need to determine where in the titration we are 75 00 mL HCl soln 0 075 mol HCl added by comparing moles of analyte and titrant 1000 mL soln HCl Initially quinine aq H2O l Q mol analyte does not equal mol titrant 0 125 mol quinine 0 05 mol HCl added NOT at equivalence point or start can use Henderson Hasselbalch equation pH pK a log Q QH Need concentrations or mol since volume identical for Q and QH mol HCl added mol QH formed 0 075 mol QH mol QH formed mol Q consumed 0 075 mol mol Q remaining 0 125 mol Q initially 0 075 mol Q consumed 0 05 mol Q Plug in values solve for pH pH pK a log 2 Q QH 0 05 mol 14 5 10 log 8 90 0 176 8 72 0 075 mol Calculate Ecell for an electrochemical cell based on the two half reactions given below when the reaction occurs in an acidic aqueous solution in which the pH is buffered at 7 000 and BrO 3 0 880 M Cl 0 144 M and 0 260 atm Use the smallest wholenumber coefficients possible when balancing the overall reaction BrO3 aq 6 H aq 5 e Br2 3 H2O E 1 52 V reduction at cathode Cl2 g 2 e 2 Cl aq E 1 36 V oxidation at anode Net redox reaction 2 BrO3 aq 12 H aq 10 Cl aq Br2 6 H2O 5 Cl2 g Number of electrons transferred n 10 mol e Non standard conditions need Nernst equation to calculate cell potential Standard cell potential o o o Ecell Ecathode Eanode E o cell 1 52 V 1 36 V o Ecell 0 16 V Ecell E o cell Plug in values 0 260 0 0592 0 16 V log 10 0 880 2 1 10 7 12 0 144 10 5 Ecell Cl PCl2 0 0592 0 0592 o log Q Ecell log 2 n n BrO3 H Ecell 0 16 0 530 0 37 V 5 12 10 CHM 1220 3 In Class Activity Aqueous Equilibria Complex Ions and Electrochemistry What mass of Ca NO3 2 must be added to 1 0 L of a 1 0 M HF solution to begin precipitation of CaF2 s For CaF2 Ksp 4 0 10 11 and Ka for 7 2 10 4 Assume no volume change on addition of Ca NO3 2 s Dissolution of Ca NO3 2 Ca NO3 2 s Ca2 aq 2 NO3 aq Should be complete all nitrates are soluble Dissolution of CaF2 CaF2 s Ca2 aq 2 F aq Solubility low depends on Ca and F Acid dissociation of HF HF aq H2O l F aq H3O aq First figure out how much F is in solution Acid dissociation equilibrium expression F H O 7 2 10 Ka Set up an ICE table or jump directly here 3 HF x x 7 2 10 4 Will provide common ion effects CaF2 solubility Second figure out how the F effect the solubility of CaF2 CaF2 s 4 solubility Equilibrium of solubility for CaF2 1 0 x y since Ka is small x 1 0 M 0 0268 2y K sp Ca 2 F 4 0 10 11 y 0 0268 2 y 2 4 0 10 11 since Ksp is small y 0 0268 M y 0 0268 2 4 0 10 11 x 7 2 10 4 1 0 x F 0 0268 M y Ca 2 5 56 10 8 M Finally figure out how much Ca NO3 2 is needed to provide the Ca Ca2 aq 2 F aq y 2 5 56 10 8 mol Ca 2 1 mol Ca NO3 2 164 088 g 1 0 L soln 9 1 10 6 g Ca NO3 2 2 L 1 mol Ca 1 mol Ca NO3 2 anything greater than 4 Fall 2017 9 1 g Ca NO3 2 EDTA is used as a complexing agent in chemical analysis Solutions of EDTA usually containing the disodium salt Na 2H2EDTA are used to treat heavy metal poisoning The equilibrium constant for the reaction below is 1 0 1023 Calculate Cr3 at equilibrium in a solution originally 0 0010 M in Cr3 and 0 050 M in H2EDTA2 and buffered at pH 6 00 Cr3 aq H2EDTA2 aq CrEDTA aq 2H aq This is going to be an ICE table type problem We need to make some necessary assumptions 1 Kf is large so initially all the Cr3 will be complexed 2 the pH does not change as it is buffered Cr3 H2EDTA2 CrEDTA H Initial Concentration M 0 0 049 0 001 1 10 6 Change in Concentration M x x x Equilibrium Concentration M x 0 049 x 0 001 x 1 10 6 Notice a small amount of the EDTA is consumed to complex the Cr 3 complex ion formation equilibrium expression Substitute from the table CrEDTA H 1 10 Cr H EDTA 0 001 x 1 10 1 10 2 Kf 3 23 2 2 6 2 x 0 049 x 23 since Kf is huge x any of these concentrations 0 001 1 10 6 2 x 0 049 1 10 23 2 04 10 x Cr 3 2 04 10 37 M 37 essentially no Cr3 at equilibrium 3 6 02 1023 ions 13 Cr ions M Cr 1 23 10 1 mol liter 3