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 Zn(NH3)n 2+: pKfn, = 2.21, 2.29, 2.36, 2.03 ZnY2- : Kf = 3.0 x 1016 EDTA: Ka1 = 1.02 x 10-2 , Ka2 = 2.14 x 10-3 , Ka3 = 6.92 x 10-7 , Ka4 = 5.50 x 10-11

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 Zn(NH3)n 2+: pKfn, = 2.21, 2.29, 2.36, 2.03 ZnY2- : Kf = 3.0 x 1016 EDTA: Ka1 = 1.02 x 10-2 , Ka2 = 2.14 x 10-3 , Ka3 = 6.92 x 10-7 , Ka4 = 5.50 x 10-11

Appl Of Ms Excel In Analytical Chemistry

2nd Edition

ISBN:9781285686691

Author:Crouch

Publisher:Crouch

Chapter10: Potentiometry And Redox Titrations

Section: Chapter Questions

Problem 7P

See similar textbooks

Similar questions

0.0322 M EDTA solution is used as the titrant in the titration of a 50.00 mL solution that is 0.0129 M in Cu2+. The solution is held at pH = 10.00 by a buffer that is 0.050 in ammonia (NH3). At what volume (in mL) of EDTA added will the equivalence point be reached?
Consider the titration of 25.00 mL of 0.03555 M Co2+ by 0.02784 M EDTA at pH 10.00. Kf is 1045. Calculate the pCo2+ when 35.00 mL of EDTA are added.
50 mL of a solution of 0.0200 M Zn2+ will be titrated with 0.0100 M EDTA in 0.0100 M NH3 at pH 6.0. Ethylenediaminetetraacetic acid (EDTA) can be considered as a tetraprotic acid (H4Y). The stepwise acid dissociation constants are: K1 = 1.02 x 10-2, K2= 2.14 x 10-3, K3 = 6.92 x 10-7 and K4 = 5.50 x 10-11. The alplia value of the un-deprotonated species in a solution buffered to a certain pH is given by the following equation: a0 = [H+]4/([H+]4 + K1[H+]3 + K1K2[H+]2 + K1 K2K3[H+] + K1 K2K3K4) Calculate the alpha value of the fully deprotonated species (Y4- ) in a solution buffered to a pH of 6.0.
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 A 50.00 mL aliquot is withdrawn, the pH adjusted to 1.5, and the bismuth titrated with 30.26 mL of 0.01024 M EDTA.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.
Titration of 25.0 mL of a 0.0500 M Zn2+ solution with 0.0550 M EDTA in a solution buffered at pH 8. Assume that the temperature is 25 oC and that the formation constant for Zn2+ is 3.13 x 1016 at this temperature. What is the pZn of the solution after 30 mL of titrant have been added?
Calamine, which is used for relief of skin irritations, is a mixture of zinc and iron oxides. A 1.022-g sample of dried calamine was dissolved in acid and diluted to 250.0 mL. A 50.00 mL aliquot was suitably buffered and titrated with 2.40 mL of 0.002727 M ZnY-2 (Zn-EDTA) solution to allow the following reaction: Fe+3 + ZnY-2 → FeY- + Zn+2. Molecular mass: Fe2O3 = 165.74 a. The weight of sample in the aliquot portion is _________g ? b. the percentage composition of Fe2O3 in the sample is ______ % ?
A 100.0 mL100.0 mL solution of 0.0200 M Fe3+0.0200 M Fe3+ in 1 M HClO41 M HClO4 is titrated with 0.100 M Cu+0.100 M Cu+, resulting in the formation of Fe2+Fe2+ and Cu2+Cu2+. A PtPt indicator electrode and a saturated Ag∣∣AgClAg|AgCl electrode are used to monitor the titration. Write the balanced titration reaction. titration reaction: Fe3++Cu+⟶Fe2++Cu2+Fe3++Cu+⟶Fe2++Cu2+ Complete the two half‑reactions that occur at the PtPt indicator electrode. Write the half‑reactions as reductions. half‑reaction: ?∘=0.161 V half‑reaction: ?∘=0.767 V Select the two equations that can be used to determine the cell voltage at different points in the titration. ?E of the Ag∣∣AgClAg|AgClelectrode is 0.197 V.0.197 V. ?=0.767 V−0.05916×log([Cu2+][Cu+])−0.197 V E=0.767 V−0.05916×log([Cu2+][Cu+])−0.197 V ?=0.767 V−0.05916×log([Fe3+][Fe2+])−0.197 V E=0.767 V−0.05916×log([Fe3+][Fe2+])−0.197 V ?=0.767 V−0.05916×log([Cu+][Cu2+])−0.197 V E=0.767…
A 50.0ml sample of water containing both Ca+2 and Mg+2 is titrated with 16.54ml of 0.01104M EDTA in an ammonia buffer of pH 10. Another 50.0ml sample is treated with NaOH to precipitate the Mg(OH)2 and then titrated at pH 13 with 9.26ml of the same EDTA solution. Calculate the ppm each of CaCO3 and MgCO3 in the sample.
Average [EDTA] = C: Determination of the hardness of an unknown water sample: Using an unknown water sample, three 25 mL samples in 3 separate 250 mL Erlenmeyer flasks are prepared. To each flask, 5 mL of pH pH10 buffer, 3 to 5 drops of Eriochrome Black T indicator, and 15 drops of 0.03 M MgCl2 solution are added. Each flask is titrated to the end point using EDTA.
To adjust the EDTA solution, a 0.01 M 10 mL Ca+2 sample was taken and 10 mL of pH 10 buffer was added to it, and the volume was completed to 100 mL. Then, two drops of EBT indicators were added and titrated with EDTA. Since the consumption is 10.4 mL, which of the following is the concentration of the EDTA solution in terms of molarity? A. 0.0076B.0.0101C. 0.0096D.0.0067
A 0.4071-g sample of pure CaCO3 was transferred to a 500-mL volumetric flask, dissolved using a minimum of 6 M HCl, and diluted to volume. After transferring a 50.00-mL portion of this solution to a 250-mL Erlenmeyer flask, the pH was adjusted by adding 5 mL of a pH 10 buffer. After adding an indicator, the solution was titrated with the EDTA, requiring 42.63 mL to reach the end point. What is the concentration of EDTA? Choices: 0.0191 M, 0.0955 M, 0.00955 M, 0.477 M
A 100.0 mL100.0 mL solution of 0.0200 M Fe3+0.0200 M Fe3+ in 1 M HClO41 M HClO4 is titrated with 0.100 M Cu+0.100 M Cu+, resulting in the formation of Fe2+Fe2+ and Cu2+Cu2+. A PtPt indicator electrode and a saturated Ag∣∣AgClAg|AgCl electrode are used to monitor the titration. Calculate the values of ? for the cell after each of the given volumes of the Cu+Cu+ titrant have been added.