Wright CHM 1220 - CHM 1220 SI_Lewis, Complexes, & Review_Rev 1_ANSWERS (2 pages)

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CHM 1220 SI_Lewis, Complexes, & Review_Rev 1_ANSWERS



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CHM 1220 SI_Lewis, Complexes, & Review_Rev 1_ANSWERS

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Wright State University
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Chm 1220 - General Chemistry II
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Questions Comments Suggestions Email me at stevens 161 wright edu Chapter 16 SI Activity Lewis Model Complexation and Review CHM 1220 ANSWERS Problems 1 Consider 15 00 mL of a 0 255 M solution of trifluoroacetic acid TFA pKa 0 23 What is the pH of the solution resulting from the addition of 12 50 mL of a 0 375 M trifluoroacetate solution When you mix an acid with its conjugate base you get what you mix Use Henderson Hasselbalch Eqn 12 50 mL 0 375 M base pH pK a log 0 23 log 0 318 15 00 mL 0 255 M acid Note that we can use mmol here instead of concentrations What is the pH of the solution resulting from the addition of 3 75 mL of a 0 600 M NaOH solution to the solution from part When you mix an acid with a strong base it reacts Before reaction NaOH Converts Acid into Base 3 825 mmol 2 250 mmol 1 575 mmol CF3COOH 15 00 mL 0 255 M 3 825 mmol CF3COOH 12 50 mL 0 375 M 4 688 mmol CF3COO 4 688 mmol 2 250 mmol 6 938 mmol CF3COO 03 75 mL 0 600 M 2 250 mmol NaOH Henderson Hasselbalch Again pH pK a log base 0 23 log 6 938 1 575 acid 0 874 2 Consider a solution that is 1 25 M in nitrous acid pKa 3 4 and 1 0 M in the monoprotic superacid trifluoromethanesulfonic acid pKa 14 7 Determine the concentration of nitrite in the solution Since TfOH is a strong acid it dissociates completely Therefore H is a common ion Need ICE Table for Common Ion Effect I C E HNO2 1 25 M x 1 25 x H 1 0 M x 1 0 x K a 10 pKa NO2 0 x x 1 x x 10 3 4 1 25 x 1 25 K a K a x x x 2 1 25 K a 1 K a x x 2 0 x 4 97 10 4 M 3 In the reaction below identify each reactant species as either a Lewis acid or a Lewis base Then identify each reactant species as either a Br nsted Lowry acid or Br nsted Lowry base O O OH H S O O H S O O H p toluenesulfonic acid p toluenesulfonic acid water p toluenesulfonate Lewis Model Acid Base Acid Base Br nsted Lowry Acid Base Acid Base H O H 4 An aqueous solution contains 0 010 M Ba2 and 0 010 M Ag Can 99 90 of one metal ion be precipitated by chromate CrO42 without precipitating the other metal ion If so which one precipitates The Ksp for BaCrO4 is 2 1 10 10 and for Ag2CrO4 is 1 2 10 12 The reactions have different stoichiometries so it is not immediately clear which species will precipitate first We shall consider both possibilities 2 2 10 BaCrO 4 s Ba CrO 4 K sp 2 1 10 2 Ag 2 CrO 4 s 2 Ag CrO 4 K sp 1 2 10 12 Need to find Chromate that can ppt 99 9 Ba Will this much chromate cause Ag to ppt Need Q 2 2 Qsp Ag CrO24 0 010 2 1 10 5 2 1 10 9 K sp Ba 2 CrO 24 2 1 10 10 K sp Since 2 11 10 9 Qsp K sp 1 2 10 12 Ag would ppt 2 1 10 10 5 CrO 24 2 1 10 M Need to find Chromate that can ppt 99 9 Ag Ba 2 1 0 10 5 M 2 CrO 24 1 2 10 12 K Ag sp 0 1 of 0 010 M Relevant Equilibria Since the concentration of chromate required to precipitate 99 9 Ag is greater than what is required for Ba it is not possible to precipitate Ag without precipitating Ba CrO 24 K sp Ag 2 1 2 10 12 2 1 0 10 5 M 0 012 M Not Possible 0 1 of 0 010 M 5 Consider a solution that is 1 1 10 3 in Ni NO3 2 and 0 150 M in the bidentate ligand ortho phenanthroline phen Determine the concentration of Ni H2O 62 after the solution equilibrates ignoring any ion pairing effects The Kf for Ni phen 32 is 7 95 108 Since the formation constant is large we assume that all of the Ni is initially complexed This will simplify the calculation drastically because we can make the x is small approximation to avoid solving a higher order polynomial If we do not assume that the Ni all initially complexes and try to do this calculation x is actually very large and there is no way to approximate the higher order polynomial Use K expression to find x Need an ICE Table I C E Ni2 0 x x 3 Phen Ni phen 32 1 4989 M 1 1 10 3 x x 1 4989 x 1 1 10 3 x Ni phen 2 1 1 10 3 x 3 Kf 7 95 108 3 3 2 Ni phen x 1 4989 x 1 1 10 3 x is small x 1 4989 3 7 95 108 x 4 1 10 13 M Note that Ni2 exists in water as Ni H2O 62 We therefore only need to find Ni2 which is x 6 Consider the titration of 37 00 mL of a 0 1125 M weak base solution which required 25 00 mL of a 0 1665 M strong acid solution to reach the equivalence pH of 4 1100 Determine the Kb of the weak base from these data At the equivalence point all weak base has been converted into its conjugate acid BH which hydrolyzes There were initially 37 00 mL 0 1125 M 4 1625 mmol of B and there are now 4 1625 mmol of BH 4 1625 mmol 0 067137 M Need BH BH 37 25 mL Find H and B from pH Write Ka Expression B H 7 765 10 5 7 765 10 5 Ka 8 975 10 8 BH 0 067137 Find Kb from Ka Rev 1 Fall 2017 J Stevens Kb Since there is a 1 1 mole ratio of H to B their concentrations are equal pH log H H 10 pH 10 4 1100 7 7625 10 5 B K w 1 00 10 14 1 114 10 7 K a 8 975 10 8


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