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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- 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?Solve the following problem: Calamine, which is used to relieve skin irritations, is a mixture of zinc and iron oxides. A 1.056 g sample of dry calamine was dissolved in acid and diluted to 250.0 mL. Potassium fluoride was added to a 10.00 mL aliquot of the diluted solution to mask the iron. After adjusting the pH, the Zn2+ consumed 38.37 mL of EDTA 0.01133 M. A second aliquot of 50.00 mL was buffered and titrated with 2.30 mL of a solution of ZnY2- 0.002647 M: Fe3+ + ZnY2- → FeY- + Zn2+ Calculate the percentages of ZnO and Fe2O3 in the sample.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 1.000 mL aliquot of a solution containing Cu2t and Ni2+ is treated with 25.00 mL of a 0.05220 M EDTA solution. The solution is then back titrated with 0.02082 M Zn+ solution at a pH of 5. A volume of 20.42 mL of the Zn2t solution was needed to reach the xylenol orange end point. A 2.000 mL aliquot of the Cu2+ and Ni2+ solution is fed through an ion-exchange column that retains Ni²+. The Cu2+ that passed through the column is treated with 25.00 mL of 0.05220 M EDTA. This solution required 15.92 mL of 0.02082 M Zn+ for back titration. The Ni2+ extracted from the column was treated witn 25.00 mL of 0.05220 M EDTA. How many milliliters of 0.02082 M Zn2+ is required for the back titration of the Ni2+ solution? volume: mLA 1.000 mL aliquot of a solution containing Cu²+ and Ni2+ is treated with 25.00 mL of a 0.03992 M EDTA solution. The solution is then back titrated with 0.02019 M Zn²+ solution at a pH of 5. A volume of 21.77 mL of the Zn²+ solution was needed to reach the xylenol orange end point. A 2.000 mL aliquot of the Cu²+ and Ni²+ solution is fed through an ion-exchange column that retains Ni2+. The Cu²+ that passed through the column is treated with 25.00 mL of 0.03992 M EDTA. This solution required 19.13 mL of 0.02019 M Zn²+ for back titration. The Ni2+ extracted from the column was treated witn 25.00 mL of 0.03992 M EDTA. 2+ How many milliliters of 0.02019 M Zn2+ is required for the back titration of the Ni2+ solution? volume: mL
- A 25.00 ml of Ni2* solution was diluted in HCI and treated with 25.00 ml of 0.05283 M NazEDTA. The solution was neutralized with NaOH followed by addition of acetate buffer until the pH 5.5. The solution turns yellow after addition of few drops of xylenol orange indicator. Back titration using standard 0.022 MZn2* at pH 5.5 requires 17.61 ml until end point, on which the solution will turn red. Calculate for the molarity of the unknown. Ans. 0.03664 MA solution is prepared by mixed 50.0 mL of 0.0400 M Ni2+ with 50.00 mL of 0.0400 M EDTA.Calculate the concentration of Ni 2+ after mixing. The mixture is buffered to a pH of 3.0. KNiY =4.2x1018, alpha4 =2.5x10 -11.A 1.000 mL aliquot of a solution containing Cu²+ and Ni2+ is treated with 25.00 mL of a 0.03835 M EDTA solution. The solution is then back titrated with 0.02240 M Zn2+ solution at a pH of 5. A volume of 15.56 mL of the Zn2+ solution was needed to reach the xylenol orange end point. A 2.000 mL aliquot of the Cu2+ and Ni2+ solution is fed through an ion- exchange column that retains Ni²+. The Cu²+ that passed through the column is treated with 25.00 mL of 0.03835 M EDTA. This solution required 19.69 mL of 0.02240M Zn2+ for back titration. The Ni2+ extracted from the column was treated witn 25.00 mL of 0.03835 M EDTA. How many milliliters of 0.02240 M Zn2+ is required for the back titration of the Ni2+ solution? volume: mL
- 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?(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.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.