Wright CHM 1220 - CHM 1220 SI_Acids and Bases_Rev 1_Answers (2 pages)

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CHM 1220 SI_Acids and Bases_Rev 1_Answers



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CHM 1220 SI_Acids and Bases_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 ANSWERS Chapter 15 SI Activity Acids Bases CHM 1220 Problems 1 Phenol shown below is an organic alcohol that can act as a weak acid The pKa for phenol is 9 90 at 298 K OH O H2O H3O phenol phenolate Calculate the Ka for phenol pK a 9 90 K a 10 10 1 3 10 10 Calculate the Kb and pKb of phenolate K w 1 0 10 14 Kb 7 7 10 5 10 K a 1 3 10 pK b 4 11 log K b log 7 7 10 5 Calculate the pH of 50 00 mL a 0 150 M aqueous solution of phenol H 3O phenolate Ka phenol x x 0 150 x x2 1 3 10 10 0 150 x x2 1 3 10 10 0 150 x 1 9 10 11 1 3 10 10 x x 2 0 x 4 35 10 6 or x 4 36 10 6 As is common with many of the acid base problems the ICE table used to generate the K expression was omitted pH log H 3O log x log 4 35 10 6 5 36 Would you expect a solution of sodium phenolate to be acidic basic basic or neutral Why The salt is expected to be basic because the phenolate anion is the conjugate base of the weak acid phenol The sodium ion does not influence pH because it is the cation of the strong base NaOH Verify your answer to part by calculating the pH of a solution made of 58 5 grams of sodium phenolate FM 116 1 g mol diluted to 2 000 L 1 mol sodium phenolate 1 mol phenolate 58 5 g sodium phenolate 116 1 g sodium phenolate 1 mol sodium phenolate 0 252 M phenolate initial 2 000 L solution Kb phenol OH x x x2 phenolate 0 252 x 0 252 x 7 7 10 5 1 9 10 5 7 7 10 5 x x 2 0 x 4 3 10 3 or x 4 4 10 3 pH 14 pOH 14 log OH 14 log x 14 log 4 3 10 3 11 64 2 Estimate the pH of a 1 275 10 8 M solution of HBr Does the answer make sense The true pH calculated by a more rigorous method is 6 9723 Why might your answer be different log H 3O log 1 275 10 8 7 8945 Estimate pH This answer does not make sense because it indicates that adding an acid to the water creates a slightly basic pH 7 solution This estimate is different than the true pH because the autoprotolysis of water must be considered when the acid concentration is as dilute as it is 3 A 0 100 M solution of bromoacetic acid FM 138 948 g mol is 13 2 ionized Calculate the Ka for bromoacetic acid Use the formula for Ionization to calculate the bromoacetate concentration at equilibrium bromoacetate equilibrium bromoacetic acid initial bromoacetate equilibrium Ionization bromoacetic acid initial 0 132 0 100 M Ionization 0 0132 M Since the bromoacetate and H3O are produced in a 1 1 stoichiometric ratio in the reaction this concentration is the same for both bromoacetate and H3O This is x from an ICE table It then follows that our equilibrium concentration of bromoacetic acid has to be 0 100 M 0 0132 M 0 0868 M Plug values into our K expression H 3O bromoacetate Ka bromoacetic acid 0 0132 0 0132 0 100 0 0132 2 01 10 3 4 Identify the species below as acids or bases Then rank them in increasing order of strength with 1 being the weakest and 5 being the strongest More O atoms withdraw the electron density from the acidic proton stabilizing the anion and make the acid stronger Acids Bases 5 HClO4 1 HCl 3 HClO2 2 HClO 4 HClO3 Acids Bases 4 F 3 Cl 2 Br 5 H 1 I The basicity of the halogens increases with decreasing atomic size and H is a stronger base than F b c H2 is a weaker acid than HF 5 Organic chemists often use organomagnesium halide compounds called Grignard reagents in the formation of carbon carbon bonds An example is shown below Given that Grignards have pKa values around 60 would you expect them to behave as acids or bases bases Are they strong or weak Explain Hint What is the Kb Since the conjugate acids of Grignards have such a high pKa value they are EXTREMELY weak essentially nonacidic and their Ka values must be 1 Thus their Kb values are 1 making them very strong bases Note The pKa I have associated with the Grignard reagent is actually that of its conjugate acid which is the hydrocarbon chain from which the Grignard is made I could not specify that in the problem statement because that would immediately tell you the answer to the first question MgBr ethylmagnesium bromide 6 Oxalic acid is the simplest diprotic organic acid Representing oxalic acid as H2Ox write the stepwise deprotonation reactions for oxalic acid Given that Ka1 5 623 10 2 and Ka2 7 244 10 5 what would be the pK a1 1 25 pK a2 4 14 principal species at pH 1 000 At pH 4 140 H 2 Ox HOx H 2 O HOx H 2 O Ox 2 Rev 1 Fall 2017 J Stevens H 3O H 3O pH 1 000 H2Ox is the principal species b c pH pKa1 pH 4 14 HOx and Ox2 are present in equal proportions b c pH pKa2 This can be done conceptually as I have shown or can be done with the Henderson Hasselbalch Equation


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