Chapter 15 SI Activity: Acids & Bases CHM 1220Problems1. Phenol (shown below) is an organic alcohol that can act as a weak acid. The pKa for phenol is 9.90 at 298 K.Ⓐ Calculate the Ka for phenol. Ⓑ Calculate the Kb and pKb of phenolate.Ⓒ Calculate the pH of 50.00 mL a 0.150 M aqueous solution of phenol.Ⓓ Would you expect a solution of sodium phenolate to be acidic, basic, or neutral? Why?Ⓔ 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. OHH3OO+phenolphenolateH2O+ANSWERSQuestions? Comments? Suggestions? Email me at [email protected] 10a1.310 10 10KK−−−×= = =( )514wb10a5bb7.7 104.111.0 101.3 10p log log 7.7 10KKKKK−−−−×= = =×=− =− ×=×[ ][ ]( )( )( )( )( )2310a10 211 10 266H O phenolate1.3 10phenol 0.150 0.1501.3 10 0.1501.9 10 1.3 10 04.35 10 or 4.36 10xxxKxxxxxxxx+−−−−−−= = = = ×−−× −=× −× −== × =−×( )( )63pH log H O log log 4.35 10 5.36x+−=− =− =− ×=As is common with many of the acid-base problems, the ICE table used to generate the K expression was omitted.basicThe 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.[ ][ ][ ]( )( )( )initial2b1 mol sodium phenolate 1 mol phenolate58.5 g sodium phenolate116.1 g sodium phenolate 1 mol sodium phenolatephenolate 0.252 M2.000 L solutionphenol OHphenolate 0.252 0.25xxxKx−      = == = =−( )55 52337.7 1021.9 10 7.7 10 04.3 10 or 4.4 10xxxxx−−−−−= ×−× − × −== × =−×( )( )311.6pH 14 pOH 14 log OH 14 log 14 log 4.3 10 4x−−=−=+ =+ =+ ×=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?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 principal species at pH = 1.000? At pH = 4.140?Rev. 1 – Fall 2017 – J. Stevens3. A 0.100 M solution of bromoacetic acid (FM 138.948 g/mol) is 13.2% ionized. Calculate the Ka for bromoacetic acid. 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.Acids / Bases ___ HClO4___ HCl ___ HClO2___ HClO ____ HClO3Acids / Bases ___ F–___ Cl–___ Br–___ H–___ I–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? Are they strong or weak? Explain. (Hint: What is the Kb?) MgBrethylmagnesium bromideEstimate:Use the formula for % Ionization to calculate the bromoacetate concentration at equilibrium:Plug values into our K expression:83pH log H O log 1.275 10 7.8945+−  =− =− ×=  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.[ ][ ][ ]( )[ ]( )( )equilibriuminitialequilibrium initialbromoacetate% Ionizationbromoacetic acidbromoacetate % Ionization bromoacetic acid 0.132 0.100 M 0.0132 M=ℵSince 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.More O atoms withdraw the electron density from the acidic proton, stabilizing the anion, and make the acid stronger.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.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!  This can be done conceptually (as I have shown) or can be done with the Henderson-Hasselbalch Equation[ ][ ]( )( )( )3a3H O bromoacetate0.0132 0.0132bromoacetic acid 0.100 0.01322.01 10K−+×= = =−Acids5 1 3 2 4BasesbasesSince 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.strong54 3 2 122 3223H Ox H O HOx H OHOx H O Ox H O−+− −+++++a1 a2p 1.25; p 4.14KK= =@ 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 =