2. From the structure of EDTA (and knowing that it is ethylene diamine tetraacetic acid) explain why the EDTA is much more soluble after the addition of the basic ammonia buffer. Is there a chemical reaction that is responsible for this solubility change? If so explain what the reaction is and how it would be responsible for the increased solubility of EDTA in water.
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- From the structure of EDTA (and knowing that it is ethylene diamine tetraacetic acid) explain why the EDTA ismuch more soluble after the addition of the basic ammonia buffer. Is there a chemical reaction that isresponsible for this solubility change? If so explain what the reaction is and how it would be responsiblefor the increased solubility of EDTA in water.17. Which of the following statements is not true? (a) EDTA is a hexaprotic acid containing four carboxyl protons and two ammonium protons. (b) EDTA forms 1:1 complexes with most metal ions. (c) The concentration of a free metal ion can be determined through titration with EDTA. (d) If the pH of a metal-EDTA complex solution is fixed by a buffer, the conditional formation constant (K,) can be determined. (e) The formation constant, K, for an EDTA complex is typically a small value.Explain why in Mohr's method neutral medium should be used in titration of AgNO3?
- Give information about the structure of EDTA and the titrations with EDTA (indicators used, determination of the end point, reactions during the titration).Calculate pCd²+ at each of the given points in the titration of 60.00 mL of 0.0060 M Cd²+ with 0.0060 M EDTA in the presence of the auxiliary complexing agent NH,. The solution is buffered at a pH of 11.00 and the NH, concentration is fixed at 0.100 M. The fraction of EDTA in the form Y as a function of pH can be found in this table. The formation constant for the Cd²+ -EDTA complex can be found in this table. The cumulative formation constants for the Cd(NH,)2+ complexes can be found in this table. 0 mL. pCd²+ = 17.00 ml pCd²+ 55.00 mL 60.00 ml. pCd²+ = 65.00 ml. pCd²+ = pCd²+ = = [1Sometimes it is not possible to indicate the end point of a titration.a) How can one proceed then and what is the name of the type of titration that can be performed? Briefly describe. An example in which this method can be used is in the determination of mercury, which forms strong complexes with EDTA, but for which there is no suitable indicator that can indicate the end point. b) You are given the task of determining the Hg2 + concentration in a sample solution? After adding an excess of EDTA, the sample solution is titrated with a magnesium solution. 20.00 ml of a 0.0452 M EDTA solution was added to 30.00 ml of sample solutionThe excess EDTA was determined by adding 0.0500 M Mg 2+ solution, consuming 4.37 ml to the end point.
- The solubility constant Ksp of Ag2CrO4 is related to its solubility in water (s) as Kspthe total solubility of amoxicillin hydrate a weekly acid acidic drug with a PK of 7.11 in SO 00.0 1M in acquiesce buffered solution of a pH equaling 7.2 is(ii) Consider the titration of 25.00 mL of 0.02000 M CaSO4 with 0.01000 M EDTA at pH 10.00. Write the chemical equation for this titration and calculate the conditional formation constant for this reaction. (iii) From (ii) calculate the concentration of Ca2+ and pCa2+ at the volume of 20.0 mL of EDTA added.
- Chromel is an alloy composed of nickel, iron and chromium.A 0.6472 g sample was dissolved and diluted to 250 mL. When a50 mL aliquot of 0.05182 M EDTA was mixed with an equal volumeof the diluted sample and all the three ions were chelated, a 5.11 mLback titration with 0.06241 M copper (II) was required.The chromium in a second 50 mL aliquot was masked through theaddition of hexamethylenetetramine, titration of the Fe and Nirequired 36.28 mL of 0.05182 M EDTA.Iron and chromium were masked with pyrophosphate in a third50 mL aliquot and the nickel was titrated with 25.91 mL of theEDTA solution.Calculate the percentage of nickel, chromium and iron in thealloy.Arsenic(III) oxide (As2O3) is available in pure form and is a useful (but carcinogenic) primary standard for oxidizing agents such as MnO4-. The As2O3 is dissolved in base and then titrated with MnO4- in acidic solution. A small amount of iodide (I - ) or iodate (IO3-) is used to catalyze the reaction between H3AsO3 and MnO4-. As2O3 + 4 OH - 2 HAsO32- + H2OHAsO32- + 2 H+ H3AsO35 H3AsO3 + 2 MnO4- + 6 H+ 5 H3AsO4 + 2 Mn2+ + 3 H2O (a) A 3.209 g aliquot of KMnO4 (FM 158.034) was dissolved in 1.000 L of water, heated to cause any reactions with impurities to occur, cooled and filtered. What is the theoretical molarity of this solution if no MnO4- was consumed by impurities? M(b) What mass of As2O3 (FM 197.84) would be just sufficient to react with 25.00 mL of the KMnO4 solution in part (a)? g(c) It was found that 0.146 8 g of As2O3 required 29.98 mL of KMnO4 solution for the faint color of unreacted MnO4- to appear. In a blank titration, 0.03 mL of MnO4- was required to produce enough…A cyanide solution with a volume of 12.99 ml was treated with 30.00 ml. of Ni" solution (containing excess Nit) to convert the cyanide into tetracyanonickelate(): 4 CN + Ni?+ Ni(CN) The excess Nit was then titrated with 11.96 ml. of 0.01357 M ethylenediaminetetraacetic acid (EDTA): Nit + EDTA Ni(EDTA)- Ni(CN) does not react with EDTA. If 39.40 ml. of EDTA were required to react with 30.65 ml. of the original Nit solution, calculate the molarity of CN in the 12.99 mL cyanide sample. 0.12055 (CN = Incorvect