What is the value of pMg for 50.0 ml of a 0.0500 M Mg*2 solution buffered at pH 10.00 and titrated with 0.0500 m EDTA when 51.0 mL of EDTA is added (Kf = 6.2x108)(ay4. = 0.30 )? 1. O6.57 2. O2.88 3. O4.94 4. O1.39 5. 07.96
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- The accompanying data (1.00-cm cells) were obtained for the spectrophotometric titration 10.00 mL of Pd(II) with 2.44 10-4 M Nitroso R(O. W Rollins and M. M. Oldham, Anal. chem .,1971, 43, 262, DOI: 10.1021/ac60297a026). Calculate the concentration of the Pd(II) solution, given that the ligand-to-cation ratio in the colored product is 2:1Consider the titration of 25.0 mL of 0.020 0 M MnSO4 with 0.010 0 M EDTA in a solution buffered to pH 6.00. Calculate pMn21 at the following volumes of added EDTA and sketch the titration curve: 0, 20.0, 40.0, 49.0, 49.9, 50.0, 50.1, 55.0, and 60.0 mL.Calculate the PCa vs. EDTA titration curve for 50.0 mL of 0.0060 M Ca2+ being titrated with 0.0150 M EDTA in a solution buffered to a constant pH of 10.00. After the addition of 0.00 mL After the addition of 26.0 mL
- Calculate the PCa vs. EDTA titration curve for 50.0 mL of 0.0060 M Ca2+ being titrated with 0.0150 M EDTA in a solution buffered to a constant pH of 10.00. After the addition of 0.00 mL After the addition of 5.00 mL After the addition of 26.00 mLIn order to titrate EDTA into a water sample of unknown water hardness, the EDTA of known concentration is first diluted. If 25.0 mL of a 0.0632 M EDTA solution is transferred into a 250.0 mL volumetric flask, what is the concentration of the diluted EDTA solution? 0.00632 M 0.006320 M 3.950e-4 M 0.01580 M 0.00158 M 3.95e-4 M 0.6320 MAn 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 α₄ value under these conditions?
- Consider the titration of 25.0mL 0.020M of Co(NO3)2 with .0100M EDTA in a solution buffered to pH 10.00. Calculate pCo^2+ at the following volumes of added EDTA. a. 0.0 mL b. 20.0 mL c. 40.0 mL d. 49.0 mL e. 50.0 mL f. 50.1 mL g. 55.0 mL h. 60.0 mLA 50.00 ml aliquot of a solution containing Ca2+ and Mg2+ was buffered at pH 10 and titrated with 0.0499 M EDTA. The endpoint volume was 40.17 ml. A second aliquot of the same mixture was made strongly basic by the addition of NaOH – this causes the Mg2+ to precipitate as Mg(OH)2. The solution was then titrated with the 0.0499 M EDTA and the endpoint volume was found to be 34.70 ml.What is the equivalence volume when 0.0500 M EDTA is titrated with 100.0 mL of 0.0500 M Mn+ buffered to a pH of 9.00?
- A 50.00 ml aliquot of a solution containing Ca2+ and Mg2+ was buffered at pH 10 and titrated with 0.0418 M EDTA. The endpoint volume was 44.36 ml. A second aliquot of the same mixture was made strongly basic by the addition of NaOH – this causes the Mg2+ to precipitate as Mg(OH)2. The solution was then titrated with the 0.0418 M EDTA and the endpoint volume was found to be 30.02 ml. Calculate the molar concentration of Mg.A 30-mL portion of a solution containing Ca2+ and Mg2+ was titrated with 28.19 mL of 0.213 M EDTA at pH 10. Another 30-mL aliquot of the same Ca-Mg mixture was treated with NaOH to make the solution strongly alkaline and precipitate Mg(OH)2. This solution was then titrated with the same EDTA solution. What would be the required EDTA volume (in mL) to reach the endpoint of the second aliquot if it was found that there was 0.061 M of Mg2+ in the sample?(a) Sketch a titration curve for a 50.0 mL of 0.01M Sr2+ solution, buffered at pH 11 when 0.0,10.0, 24.0, 25.0, 26.0, 30.0, and 40.0 mL of 0.02 M EDTA are added. Given the KSrY2- is 4.3X108 and α4 is 0.85.