I need help completing this page mier moTitration and Equilibria of Diprotic Acids Hbin dgu ais y olsomalor niog nolesi Section: brue (1lo dossName:bacqanPre-Lab Problemseldat steb uovBelow is a student's titration curve of the weak acid H2SO3.Doy solav qg bua q oili mo1.218 ni ben1. Identify the regions on the graph corresponding to equivalence point #1, equivalence point#2, the approximate pKal and pK2 points.Con dul uo Titration of 0.10 M H2SO,brte t 2o151413.Cbr soy biou sdi 1012 -11-10 -ntao muindilhp orls otin W.CTo sulav ordi esob woH.EH SAil bioo gnone a109.Ao lo 2 oz vlolit tai 1210305040iay19ne vileimad to loodbmflmL NaOH2. If the pKal point has a value of 1.85, what is Kal?opoigotuda d3. What is the pH at pKal?

Equivalence point: point in titration at which the amount of titrant added is just enough to…

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Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter31: Introduction To Analytical Separations

Section: Chapter Questions

Problem 31.17QAP

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Two students were asked to determine the Kb of an unknown base. They were given a bottle with a solution in it. The bottle was labeled aqueous solution of a monoprotic strong acid. They were also given a pH meter, a buret, and an appropriate indicator. They reported the following data: volume of acid required for neutralization = 21.0 mL pH after 7.00 mL of strong acid added = 8.95 Use the students' data to determme the Kb of the unknown base.
The indicator dinitrophenol is an acid with a Ka of 1.1104. In a 1.0104 -M solution, it is colorless in acid and yellow in base. Calculate the pH range over which it goes from 10% ionized (colorless) to 90% ionized (yellow).
A solution is made by dissolving 15.0 g sodium hydroxide in approximately 450 mL water. The solution becomes quite warm, but after it is allowed to return to room temperature, water is added to bring the volume to 500.0 mL of solution. (a) Calculate the pH and pOH in the final solution. (b) Why would we wait for it to return to room temperature? (c) If the mass of the water used to initially dissolve the sodium hydroxide were exactly 450 g and the temperature of the water increased by 8.865 C, how much heat was given off by the dissolution of 15.0 g of solute? Assume the specific heat of the solution is 4.184 J/g. K. What is the molar heat change for the dissolution of sodium hydroxide (known as the enthalpy of solution, Hsol)?
The pH of a 0.20-M solution of HP is 1.92. Determine Ka far HF from these data.
A diprotic acid, H2B(MM=126g/moL), is determined to be a hydrate, H2B xH2O. A 10.00-g sample of this hydrate is dissolved in enough water to make 150.0 mL of solution. Twenty-five milliliters of this solution requires 48.5 mL of 0.425 M NaOH to reach the equivalence point. What is x?
Consider the following acids and bases: HCO2H Ka = 1.8 104 HOBr Ka = 2.0 109 (C2H5)2NH Kb = 1.3 103 HONH2 Kb = 1.1 108 Choose substances from the following list that would be the best choice to prepare a pH = 9.0 buffer solution. a. HCO2H b. HOBr c. KHCO2 d. HONH3NO3 e. (C2H5)2NH f. (C2H5)2NH2Cl g. HONH2 h. NaOBr
In dilute aqueous solution HF acts as a weak acid. However, pure liquid HF (boiling point = 19.5 C) is a strong acid. In liquid HF, HNO3 acts like a base and accepts protons. The acidity of liquid HF can be increased by adding one of several inorganic fluorides that ale Lewis acids and accept F- ion (for example, BF3 or SbF5]. Write balanced chemical equations for the reaction of pure HNO3 with pure HF and of pure HF with BF3.
Amino acids are an important group of compounds. At low pH, both the carboxylic acid group (CO2H) and the amine group (NHR) are protonated. However, as the pH of the solution increases (say, by adding base), the carboxylic acid proton is removed, usually at a pH between 2 and 3. In a middle range of pHs, therefore, the amine group is protonated, but the carboxylic acid group has lost the proton. (This is called a zwitterion.) At more basic pH values, the amine proton is dissociated. What is the pH of a 0.20 M solution of alanine hydrochloride, [NH3CHCH3CO2H]Cl?

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