Analytical Chem istry - DePauw University

Chapter 6 Equilibrium Chemistry269ous answer is acceptable, then a ladder diagram may help you estimate theequilibrium system’s composition.Solutions containing relatively similar amounts of a weak acid and itsconjugate base experience only a small change in pH upon adding a smallamount of a strong acid or a strong base. We call these solutions buffers. Abuffer can also be formed using a metal and its metal–ligand complex, oran oxidizing agent and its conjugate reducing agent. Both the systematicapproach to solving equilibrium problems and ladder diagrams are usefultools for characterizing buffers.A quantitative solution to an equilibrium problem may give an answerthat does not agree with experimental results if we do not consider the effectof ionic strength. The true, thermodynamic equilibrium constant is afunction of activities, a, not concentrations. A species’ activity is related toits molar concentration by an activity coefficient, γ. Activity coefficients canbe calculated using the extended Debye-Hückel equation, making possiblea more rigorous treatment of equilibria.6NProblems1. Write equilibrium constant expressions for the following reactions.What is the value for each reaction’s equilibrium constant?+ −a. NH ( aq) + HCl( aq) NH ( aq) + Cl ( aq)3 42−−b. PbI () s + S ( aq) PbS() s + 2I( aq)22− − −c Cd(EDTA) ( aq) + 4CN ( aq) Cd(CN) ( aq)+ EDTA− ( aq)+ −d. AgCl() s + NH ( aq) + Ag(NH ) ( aq) + Cl ( aq )23 3 22 44+ 2+e. BaCO () s + HO ( aq) Ba ( aq) + HCO ( aq) + H O()l33 2 32Answers, but not worked solutions, tomost end-of-chapter problems are availablehere.Most of the problems that follow require one ormore equilibrium constants or standard state potentials.For your convenience, here are hyperlinksto the appendices containing these constantsAppendix 10: Solubility ProductsAppendix 11: Acid Dissociation ConstantsAppendix 12: Metal-Ligand Formation ConstantsAppendix 13: Standard State Reduction Potentials2. Using a ladder diagram, explain why the first reaction is favorable andthe second reaction is unfavorable.−−HPO ( aq) + F ( aq) HF( aq) + HPO ( aq)3 4 2 4−2−HPO ( aq) + 2F ( aq) 2HF( aq) + HPO ( aq)3 4 4Determine the equilibrium constant for these reactions and verify thatthey are consistent with your ladder diagram.3. Calculate the potential for the following redox reaction for a solutionin which [Fe 3+ ] = 0.050 M, [Fe 2+ ] = 0.030 M, [Sn 2+ ] = 0.015 M and[Sn 4+ ] = 0.020 M.3+ 2+ 4+ 2+2Fe ( aq) + Sn ( aq) Sn ( aq) + 2Fe( aq)