Chapter 16 SI Activity: Lewis Model, Complexation, and Review CHM 1220Problems1. 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?Ⓑ 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 Ⓐ?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.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 BrØnsted-Lowry acid or BrØnsted-Lowry base.SOOOHp-toluenesulfonic acidOHHSOOOp-toluenesulfonate++OHH HLewis Model BrØnsted-Lowryp-toluenesulfonic acid Acid / Base Acid / Basewater Acid / Base Acid / BaseANSWERSQuestions? Comments? Suggestions? Email me at [email protected] you mix an acid with its conjugate base, you get what you mix!When you mix an acid with a strong base, it reacts.Since TfOH is a strong acid, it dissociates completely. Therefore H+ is a common ion!Note that we can use mmol here instead of concentrations!Use Henderson-Hasselbalch Eqn:Before reaction:Need ICE Table for Common Ion Effect:Henderson-Hasselbalch Again:NaOH Converts Acid into Base:[ ][ ]( )( )abase12.50 mL 0.375 MpH p log 0.23 logacid 15.00 m0.318L 0.255 MK×=+=+ =×3315.00 mL 0.255 M = 3.825 mmol CF COOH12.50 mL 0.375 M = 4.688 mmol CF COO03.75 mL 0.600 M = 2.250 mmol NaOH−×××33 3.825 mmol 2.250 mmol = 1.575 mmol CF COOH 4.688 mmol 2.250 mmol = 6.938 mmol CF COO−−+pH pbaseacida=+[][]=+ =K log.log...0236 9381 5750 874HNO2H++ NO2–I 1.25 M 1.0 M 0C – x + x + xE 1.25 – x 1.0 + x xKxxxKKxxxKKKapaaaaa==+()−()=−=+−+(−−101125101251251342....))−== ×−xxx24049710. MRev. 1 – Fall 2017 – J. Stevens4. 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. 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 × 108.6. 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.The reactions have different stoichiometries, so it is not immediately clear which species will precipitate first. We shall consider both possibilities.Since the concentration of chromate required to precipitate 99.9% Ag is greater than what is required for Ba, it is not possible to pre-cipitate Ag without precipitating Ba.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 ini-tially complexes and try to do this calculation, x is actually very large, and there is no way to approximate the higher-order polynomial.Note that Ni2+ exists in water as Ni(H2O)62+. We therefore only need to find [Ni2+], which is x.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+.Since there is a 1:1 mole ratio of H+ to B, their concentra-tions are equal.Relevant Equilibria:Need to find [Chromate] that can ppt. 99.9% Ba:Need an ICE Table:Need [BH+]:Find [H+] and [B] from pH:Write Ka Expression:Find Kb from Ka:Use K expression to find x:Need to find [Chromate] that can ppt. 99.9% Ag:Will this much chromate cause Ag to ppt.? Need Q:BaCrOBaCrOAg CrO Ag CrOsp42421024 4221 102sKsK()+=×()++− −+−.ssp=×−12 1012.KKspspBa CrOCrOBa==×==+− −−+2421042221 1021..×××=×−−=−1010 1021 10105015...% M M of 0.010 MQspAg CrO 2.1Since ==()×()=×+−−−2422590 010 10 21 102...11101210912×=>=×−−QKsp sp Ag would ppt..,KKspspAg CrOCrOAg==×==+− −−+2421242212 1012..×××=−−=1010 100 012125201...% M M of 0.010 M Not Possible!Ni2++ 3 PhenNi(phen)32+I 0 1.4989 M 1.1 × 10–3C + x + x – xE + x 1.4989 + x 1.1 × 10–3 – xKxxxfNi phenNi phen=()[]=×−()()+(++−3223311 101 4989..))=×→×()()()=×→−383387951011 101 498979510....xxis smallxx =×−41 1013. MBH mmol mL M+=+()=4 162537 250 067137..pH log.pH .=−== =×=[]+−− −HHB+10 10 7 7625 104 1100 5KaBHBH=[]=×()×()()=++−−7 765 10 7 765 100 0671378955....775 108×−KKKbwa==××= ×−−−100108 975 101 114