A 1.000 ml aliquot of a solution containing Cu* and Ni* is treated with 25.00 mL of a 0.03468 M EDTA solution. The solution is then back titrated with 0.02308 M Zn* solution at a pH of 5. A volume of 20.28 ml. of the Zn solution was needed to reach the xylenol orange end point. A 2.000 ml aliquot of the Cu* and Ni* solution is fed through an ion- exchange column that retains Ni. The Cu+ that passed through the column is treated with 25.00 mL of 0.03468 M EDTA. This solution required 23.18 mL of 0.02308 M Zn* for back titration. The Ni?+ extracted from the column was treated witn 25.00 mL of 0.03468 M EDTA. How many milliliters of 0.02308 M Zn* is required for the back titration of the Ni+ solution?

A solid containing tris was dissolved in water and brought to a total volume of 50.00 ml. A 10.00 ml aliquot of the solution was titrated with 0.09978 M HCl to a bromcresol green endpoint. The aliquot consumed 38.93 ml of titrant. Calculate the weight of tris in the original sample.

A 1.000 mL aliquot of a solution containing Cu²+ and Ni²+ is treated with 25.00 mL of a 0.04333 M EDTA solution. The solution is then back titrated with 0.02071 M Zn²+ solution at a pH of 5. A volume of 20.01 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 0.04333 M EDTA. This solution required 18.20 mL of 0.02071 M Zn²+ for back titration. The Ni²+ extracted from the column was treated witn 25.00 mL of 0.04333 M EDTA. How many milliliters of 0.02071 M Zn²+ is required for the back titration of the Ni²+ solution? volume: mL

A 1.000 ml. aliquot of a solution containing Cu?* and Ni* is treated with 25.00 ml, of a 0.03468 M EDTA solution. The solution is then back titrated with 0.02308 M Znt solution at a pH of 5. A volume of 20.28 ml. of the Zn+ solution was needed to reach the xylenol orange end point. A 2.000 ml. aliquot of the Cu* and Ni* solution is fed through an ion- exchange column that retains Ni*. The Cu+ that passed through the column is treated with 25.00 ml. of 0.03468 M EDTA. This solution required 23.18 ml. of 0.02308 M Zn* for back titration. The Ni" extracted from the column was treated witn 25.00 mL of 0.03468 M EDTA. How many milliliters of 0.02308 M Zn?* is required for the back titration of the NP solution? volume: ml.

A 1.000 mL aliquot of a solution containing Cu²+ and Ni²+ is treated with 25.00 mL of a 0.03703 M EDTA solution. The solution is then back titrated with 0.02138 M Zn²+ solution at a pH of 5. A volume of 22.13 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 0.03703 M EDTA. This solution required 18.53 mL of 0.02138 M Zn²+ for back titration. The Ni²+ extracted from the column was treated witn 25.00 mL of 0.03703 M EDTA. How many milliliters of 0.02138 M Zn²+ is required for the back titration of the Ni²+ solution? volume: mL

Chromel is an alloy composed of nickel, iron, and chromium. A 0.6418-g sample was dissolved and diluted to 250.0 mL. When a 50.00-mL aliquot of 0.05173 M EDTA was mixed with an equal volume of the diluted sample, all three ions were chelated, and a 5.27-mL back-titration with 0.06139 M copper(II) was required. The chromium in a second 50.0-mL aliquot was masked through the addition of hexamethylenetetramine; titration of the Fe and Ni required 35.81 mL of 0.05173 M EDTA. Iron and chromium were masked with pyrophosphate in a third 50.0-mL aliquot, and the nickel was titrated with 25.77 mL of the EDTA solution. Calculate the percentages of nickel, chromium, and iron in the alloy. Percentage of nickel = % Percentage of iron = Percentage of chromium = % %

2. A sample of mineral water containing Ca and Mg was analyzed by a complexation titration using EDTA as the titrant. A 50 ml aliquot of the sample was rendered strongly alkaline to precipitate the Mg as Mg(OH)2. Titration with a calcium specific indicator required 20 ml of 0.02M EDTA. A second 50 ml aliquot was adjusted to pH 10 titrating with 0.02M EDTA required 32.50 ml of 0.02M EDTA. Calculate the ppm CaCO3 (MW=100g/mol) and ppm M9CO3 (MW=84 g/mol) in the mineral water.

A solution containing 60 mL of a 0.025 mol / L metal ion (Mn +) buffer buffered to pH 7.0 was titrated with a 0.05 mol / L EDTA solution. Data: Conditional formation constant (Kf)=10^15. Determine: i) The concentration of the free metal when ½ of the equivalence volume is added. ii) the concentration of free metal in the equivalence volume. iii) the concentration of free metal with an excess of 2 mL of EDTA? (Data - EDTA constants: K1=0.01; K2=2.19.10^-3; K3=6.92.10^-7; K4=5.75.10^-11)

Standard oxylate solution used to standardize KMnO4 was prepared by dissolving 3.299g of Na2C2O4 in 1.000L of water containing H2SO4. add 25mL aliquot of the Na2C2O4 solution required 28.39 mL of KMnO4 for titration. A blank titration of 25mL of water containing H2SO4 required 0.03 mL of KMnO4. what is the molarity of the KMnO4 solution?   5C2O42-+2MnO4-+16H+ → 10CO2+2Mn2++8H2O

A 0.2420 g sample cntg. Calcium is dissolver and the metal precipitated as CaC2O4. The ppt. is filtered, washed and redissolved in acid. The pH is adjusted, 25 ml of 0.0400 M EDTA added and the xcss EDTA titrated with 33.28 ml of 0.012102 M MG2+. Calculate the % Ca (40.078) in the sample.