The Telluride ion was determined by indirect titration with EDTA. To a solution containing50.00 mL of 0.08664M Zn(CIOA)2 plus 15 mL of 1 M acetate buffer (pH 4.5) were added 50.00mL of unknown telluride solution with vigorous stirring. The ZnTe precipitate was filteredand washed with hot water. Then NH3 was added to the filtrate (which contained excess Zn2+)until the blue colour of Zn(NH3)42+ was observed. Titration of the filtrate with 0.07854 M EDTArequired 24.22 mL to reach the murexide end point.Calculate the molarity of telluride ion in the unknown.

Given Concentration of Zn(ClO4)2= 0.08664 M Given Telluride Solutions = 50 ml We have to calculate…

The Telluride ion was determined by indirect titration with EDTA. To a solution conta 50.00 mL of 0.08664M Zn(CIO4)2 plus 15 mL of 1 M acetate buffer (pH 4.5) were added mL of unknown telluride solution with vigorous stirring. The ZnTe precipitate was fi and washed with hot water. Then NH3 was added to the filtrate (which contained excess until the blue colour of Zn(NH3)42+ was observed. Titration of the filtrate with 0.07854 M I required 24.22 mL to reach the murexide end point. Calculate the molarity of telluride jon in the unknown

The Telluride ion was determined by indirect titration with EDTA. To a solution conta 50.00 mL of 0.08664M Zn(CIO4)2 plus 15 mL of 1 M acetate buffer (pH 4.5) were added mL of unknown telluride solution with vigorous stirring. The ZnTe precipitate was fi and washed with hot water. Then NH3 was added to the filtrate (which contained excess until the blue colour of Zn(NH3)42+ was observed. Titration of the filtrate with 0.07854 M I required 24.22 mL to reach the murexide end point. Calculate the molarity of telluride jon in the unknown

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

A 1.000 mL aliquot of a solution containing Cu²+ and Ni²+ is treated with 25.00 mL of a 0.03169 M EDTA solution. The solution is then back titrated with 0.02152 M Zn²+ solution at a pH of 5. A volume of 15.93 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 0.03169 M EDTA. This solution required 15.86 mL of 0.02152 M Zn²+ for back titration. The Ni²+ extracted from the column was treated witn 25.00 mL of 0.03169 M EDTA. How many milliliters of 0.02152 M Zn²+ is required for the back titration of the Ni²+ solution? volume: mL
A 50.00 ml aliquot of a solution containing Ca2+ and Mg2+ was buffered at pH 10 and titrated with 0.0474 M EDTA. The endpoint volume was 40.99 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 0.0474 M EDTA and the endpoint volume was found to be 30.04 ml.
2. Titration of Ca²+ and Mg²+ in a 50mL sample of hard water required 23.65mL of 0.01205M EDTA. A second 50mL aliquot was made strongly basic with NaOH to precipitate Mg2+ as Mg(OH)2. The supernatant liquid was titrated with 14.53mL of the EDTA solution. Calculate a. concentration in ppm of CaCO3 in the sample. b. concentration in ppm of MgCO3 in the sample.
Calculate the concentration of an EDTA solution of which 24.22 mL were needed to titrate the Ca2+ions present in a 50.00 mL solution containing 246.7 mg CaCO3. Express your answer in terms of (a) molar concentration of EDTA (b) Ca Titer (mg Ca per mL of EDTA) MM Ca= 40.078
A 1.509-g samle of a Pb/Cd alloy was dissolved in acid and diluted to exactly 250.0 mL in a volumetric flask. A 50.00-mL aliquot of the diluted solution was brought to a pH of 10.0 with an HCN/NaCN buffer, which also served to mask the Cd2+; 11.56mL of the EDTA solution were needed to titrate the Pb2+. Calculate the percentage of Pb and Cd in the sample.
1. A 1.000 g sample containing Na,C,O, (MM=134 mg/mmol) is titrated with 40.00 mL of 0.0200 M KMNO, in acid solution. Calculate the percentage of Na,C2O, in the sample. Rxn: 5C20,2 + 2MnO, + 16H* analyte - 2MN2+ + 10 CỐ, + 8H2O titrant
The Cl– content of a solution was analyzed via an EDTA titration. 2.00 g of AgNO3 was added to 25.00 mL of the sample solution. A titration with 0.096 53 M EDTA was then carried out, reaching an endpoint at 38.25 mL. What’s the concentration of Cl – (M)?
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 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 solution containing Mg2+ was buffered to pH 9.75 and titrated with EDTA. Eriochrome black T, with pKa1 = 6.3 and pKa2 = 11.6 was used as the indicator. It is a diprotic acid, H2E What is the fraction of the predominant form of eriochrome black T in solution at this pH? (To answer this, report the alpha fraction of the indicator's most prevalent species at this pH) Review text sections 10.4 and 10.5 for help with this!
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 14.92 g sample of bacon is pureed in a blender with 100.0 mL of water. The suspension is filtered and the clear solution containing dissolved sodium nitrite is adjusted to pH 2. The solution is treated with 25.0 mL of 0.0150 M KMnO4 to oxidize the nitrite to nitrate and after standing for 10 minutes, the excess KMnO4 is back titrated with 15.0 mL of 0.0818 M FeSO4. What is the concentration of nitrite in the bacon as parts per million?