Calculate the pZn2+ for solutions prepared by adding 0.00, 5.00, 10.00, 15.00, 20.00, 25.00 and 30.00 mL of 0.0100 M EDTA to 25.00 mL of 0.00250 M Zn2+. Assume that both the Zn2+ and EDTA are 0.0100 M in NH3 to provide a constant pH of 9.0
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Calculate the pZn2+ for solutions prepared by adding 0.00, 5.00, 10.00, 15.00, 20.00, 25.00 and 30.00 mL of 0.0100 M EDTA to 25.00 mL of 0.00250 M Zn2+. Assume that both the Zn2+ and EDTA are 0.0100 M in NH3 to provide a constant pH of 9.0
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- Calculate the pZn2+ for solutions prepared by adding 0.00, 5.00, 10.00, 15.00, 20.00, 25.00and 30.00 mL of 0.0100 M EDTA to 25.00 mL of 0.00250 M Zn2+. Assume that both the Zn2+and EDTA are 0.0100 M in NH3 to provide a constant pH of 9.0 Zn(NH3)n2+: pKfn, = 2.21, 2.29, 2.36, 2.03ZnY2-: Kf = 3.0 x 1016EDTA: Ka1 = 1.02 x 10-2 , Ka2 = 2.14 x 10-3 , Ka3 = 6.92 x 10-7 , Ka4 = 5.50 x 10-11Construct a titration curve that is expected to be obtained by titration 50.00mL of 0.1000M Sr2+ with 0.1000M EDTA at pH 11. Kf = 4.3 x 10^8. Use the following volumes to construct the curve: 0.00, 10.00, 30.00, 40.00, 45.00, 50.0, 55.00, 60.00 and 70.00mL.State in your own words what the abbreviation α Y4- means with regards to EDTA speciation in water. Calculate α Y4- for EDTA at (a) pH 3.50 and (b) 10.50.
- In 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 MA 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?What is the rationale of maintaining the analyte solution at high pH (~10) in the titration process with EDTA? At this pH, EDTA is at its fully deprotonated form, Y-4. At this pH, EDTA is partially deprotonated with 1 ionizable H+. At this pH, EDTA is at its fully protonated form, H4Y. At this pH, EDTA is partially deprotonated with 2 ionizable H+.
- A standard CaCO3 solution is prepared by dissolving 0.4193g in enough dilute HCl to effect solution and then diluted to 500ml solution. A 25.00ml aliquot requires 23.62ml EDTA solution for titration. A 25.00ml water sample determined for total hardness, requires 8.45ml of the EDTA solution using Eirochrome Black T indicator. Calculate the total hardness in the water sample expressed in ppm CaCO3.A 5.00 mL tap water sample was measured out with a volumetric pipette, and added to a 25 mL Erlenmeyer flask. It was then titrated with a 0.0100 M Na2EDTA.2H2O solution, and found to take 0.635 mL of the EDTA solution to reach the blue endpoint.You are asked to titrate a Mn3+ solution with EDTA at pH 9.00. The overall ionic strength of the solution is 0.10 M. Mn3+ +EDTA4- ⇌ MnEDTA- log K = 25.2 a. Calculate the conditional formation constant for MnEDTA- at pH 9.00 b. Calculate the equilibrium [Mn3+] at pH 9.00 for total Mn3+ = 2.0 mM i) total EDTA = 0.50 mM ii) total EDTA = 5.00 mM
- Is this for Average Molarity of EDTA Standard Solution? Then, would it be always the same with average Molarity of EDTA? How about the Ca Titer (mg Ca/mL of EDTA Solution? Show step by step solution.Please answer this NEATLY, COMPLETELY, and CORRECTLY for an UPVOTE. A 50.00-mL sample of water containing Ca2+ and Mg2+ is titrated with 10.28 mL of 0.01001 M EDTA in an ammonia buffer at pH 10.00. Another 50.00-mL sample is titrated with NaOH to precipitate Mg(OH)2 and then titrated at pH 13.00 with 6.75 mL of the same EDTA solution. Calculate the concentration of CaCO3 and MgCO3 in the sample in ppm.The Cr plating on a surface that measured 3.00 cm x 4.00 cm was dissolved in HCl. The pH was suitably adjusted, and 10.19 mL of 0.01441 M EDTA was then introduced. The excess reagent required a 2.91-mL back-titration with 0.007171 M Cu2+. Calculate the average weight (in mg) of Cr (51.9961 g/mol) on each square centimeter of surface.