The formation constant (Kf) for calcium reacting with EDTA is 6.2 x108. A calcium-EDTA titration was done at a pH = 10.00, where the fraction of species of Y is approximately 0.39, what would be the value for the conditional formation constant (Kf') for the reaction? 1.0 x 1010 O 6.2 x 108 O 4.5 x 1011 O 2.4 x 108
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- An antacid tablet, weighing 1.25 g, was dissolved in a 1.0 L volumetric flask to allow for determination of calcium carbonate and magnesium carbonate contents. After preparing the solution, an aliquot of 10.0 ml was transferred to an Erlenmeyer flask containing a buffer solution with pH = 10. This aliquot was then titrated with EDTA 0.0040 mol/L and the average volume spent was 15.01 mL. To measure calcium after precipitation After the magnesium was fractionated, a second 10.00 mL aliquot of the stock solution was transferred to an Erlenmeyer flask. and the pH of the present solution was adjusted to a value of approximately 13. In EDTA titration, the volume consumed for observation of the end point was equal to 13.03 mL. Based on this information, (a) Outline the two steps involved, representing the related reactions. (b) Calculate the concentrations of CaCO3 and MgCO3 present in the initial solution. (c) Calculate the masses of CaCO3 and MgCO3 present in the pellet. (d) Calculate the…. An antacid tablet, weighing 1.25 g, was dissolved in a 1.0 L volumetric flask to allow for determination of calcium carbonate and magnesium carbonate contents. After preparing the solution, an aliquot of 10.0 ml was transferred to an Erlenmeyer flask containing a buffer solution with pH = 10. This aliquot was then titrated with EDTA 0.0040 mol/L and the average volume spent was 15.01 mL. To measure calcium after precipitation. After the magnesium was fractionated, a second 10.00 mL aliquot of the stock solution was transferred to an Erlenmeyer flask, and the pH of the present solution was adjusted to a value of approximately 13. In EDTA titration, the volume consumed for observation of the end point was equal to 13.03 mL. Based on this information, a) Outline the two steps involved, representing the related reactions./5 b) Calculate the concentrations of CaCO3 and MgCO3 present in the initial solution./6 c) Calculate the masses of CaCO3 and MgCO3 present in the pellet./6 d) Calculate…A 0.5745 g sample of an alloy containing principally bismuth and lead is dissolved in nitric acid and diluted to 250.0 mL in a volumetric flask. A 50.00 mL aliquot is withdrawn, the pH adjusted to 1.5, and the bismuth titrated with 30.26 mL of 0.01024 M EDTA. The pH of the solution is then increased to 5.0 and the lead titrated with 20.42 mL of the same EDTA solution. Calculate the percentages of lead and bismuth in the alloy.
- A 0.9352g sample of ore containing Fe³+, Al³+ and Sr²+ was dissolved and made up to 500.00 mL. The analysis of metals was performed using complexation volumetry. Initially, an aliquot of 50.00 mL had its pH adjusted to 1.0 and titrated with a standard 0.03145 mol/L EDTA solution, requiring 6.95 mL to reach the end point. Subsequently, another 25.00 mL aliquot was buffered at pH=5 and titrated with the same EDTA solution, requiring 6.24 mL to reach the end point. Finally, a third aliquot of 25.00 mL was titrated at pH=11, requiring 11.10 mL of the same EDTA solution to complete the titration. Given the molar masses: Fe=55.845 g/mol; Al-26.982 g/mol and Sr-87.620 g/mol. a) Determine the percentage of each of the metals in the sample. b) Explain why the change in pH allows the determination of the three ions in this sample.An EDTA solution was allowed to react with Pb²⁺ to produce 0.25 M PbY²⁻, 2.67×10⁻⁸ M Pb²⁺ and an excess of 0.10 M at equilibrium (K = 1.1×10⁸). What will be the hydronium ion concentration of the resulting solution?The formation constant (Kf) for calcium reacting with EDTA is 6.2 x108. A calcium-EDTA titration was done at a pH = 10.00, where the fraction of species of Y4- is approximately 0.39, what would be the value for the conditional formation constant (Kf') for the reaction? Group of answer choices 6.2 x 108 2.4 x 108 4.5 x 1011 1.0 x 1010
- Titration of a 25.00 mL sample of mineral water (containing both Ca2+ and Mg2+) at pH 10 required 19.18 mL of 0.01125M EDTA solution. Another 25.00 mL aliquot of the same mineral water was rendered strongly alkaline to precipitate the magnesium. Titration with a calcium-specific indicator required 14.92 mL of the EDTA solution. Calculate the mass of MgCO3 (FW= 84.314 g/mol) in the mineral water in mg (keep 2 decimals).25.00 mL 0.01000 M Ni2+ is titrated with 0.01000 M EDTA in a solution buffered to pH 5.0. Given that the formation constant for the Ni-EDTA (NiY2–) chelate is 4.2 x 1018 and the 4 value at pH 5.0 is 3.54 x 10–7. What is EDTA?A 0.7352g sample of ore containing Fe3+, Al3+ and Sr2+ was dissolved and made up to 500.00 mL. The analysis of metals was performed by a chemistry using complexation volumetry. Initially, an aliquot of 50.00 mL had its pH adjusted to 1.0 and titrated with a standard solution of EDTA 0.02145 mol/L, requiring 6.95 mL to reach the end point. Subsequently, another 25.00 mL aliquot was buffered at pH=5 and titrated with the same EDTA solution, requiring 6.24 mL to reach the end point. Finally, a third aliquot of 25.00 mL was titrated at pH=11, requiring 11.10 mL of the same EDTA solution to complete the titration. Determine the percentage of each of the metals in the sample Given the molar masses: Fe=55.845 g/mol; Al=26.982 g/mol and Sr=87.620 g/mol.
- Explain the analogies between the titration of a metal with EDTA and the titration of a strong acid (H1) with a weak base (A2). Make comparisons in all three regions of the titration curve.As part of a geological team that studied a local cave, you brought with you a bunch of 1.00 g rock samples to be studied. Each rock was prepared and titrated against 0.050 M EDTA.a. Calculate the percent calcite (CaCO3) content of rock A if it was titrated with 48.0 mL EDTAb. Calculate the percent brucite (Mg(OH)2) content of rock B if it was titrated with 76.5 mL EDTASometimes 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.