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A 1.000-mL aliquot of a solution containing Cu2* and Ni2+ is treated with 25.00 mL of...
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of...
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.04328 M EDTA solution. The solution is then back titrated with 0.02246 M Zn2 solution at a pH of 5. A volume of 21.60 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...
A 1.000-mL aliquot of a solution containing Cu and Ni2 is treated with 25.00 mL of...
A 1.000-mL aliquot of a solution containing Cu and Ni2 is treated with 25.00 mL of a 0.05182 MEDTA solution. The solution is then back titrated with 0.02198 M Zn2 solution at a pH of 5. A volume of 21.82 mL of the Zn2 solution was needed to reach the xylenol orange end point. A 2.000-mL aliquot of the Cu and Ni2 solution is fed through an ion-exchange column that retains Ni2. The Cu2 that passed through the column is...
A titration was performed to standardize an EDTA solution. a) A 25.00 mL aliquot of a...
A titration was performed to standardize an EDTA solution. a) A 25.00 mL aliquot of a standard solution containing 0.01500 M Ca2+ required 42.87 mL of EDTA to reach the endpoint. The molarity of the EDTA solution is: ________ M b) That same EDTA solution was then used to titrate an unknown hard water sample. A 40.00 mL aliquot of unknown hard water required 34.21 mL of EDTA solution to reach a distinct endpoint. The concentration of Ca2+ ions (assuming...
An unknown solution was analyzed for Ni by an EDTA titration. A 50.00 mL sample of...
An unknown solution was analyzed for Ni by an EDTA titration. A 50.00 mL sample of the unknown was treated with 25.00 mL of 0.2404 M EDTA. The excess EDTA was then back titrated with 8.52 mL of 0.0694 M Zn2+ to reach the equivalence point. What was the concentration of Ni (in unit of M) in the 50.00 mL sample? Please keep your answer to three decimal places.
A 25.0 mL aliquot of 0.0440 M EDTA was added to a 49.0 mL solution containing...
A 25.0 mL aliquot of 0.0440 M EDTA was added to a 49.0 mL solution containing an unknown concentration of V3+. All of the V3+ present in the solution formed a complex with EDTA, leaving an excess of EDTA in solution. This solution was back-titrated with a 0.0340 M Ga3+ solution until all of the EDTA reacted, requiring 11.0 mL of the Ga3+ solution. What was the original concentration of the V3+ solution?
A 25.00 mL aliquot of a solution containing phenol was treated with 5 mL of 1 M KBr and 10 mL of ...
A 25.00 mL aliquot of a solution containing phenol was treated with 5 mL of 1 M KBr and 10 mL of 1 MHC A platinum generator electrode was made the anode and Br was oxidized to Br. A constant current of 8.00 mA was passed for 2 min and 38 s (not 2.38 min) to reach the potentiometric endpoint using an ORP electrode. Calculate the concentration of phenol in molL in the original sample, given 2Br-→Br2+22 Br Br 3Bra...
Please help! Will rate. A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 56.5...
Please help! Will rate.
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 56.5 mL of 0.0600 M EDTA. Titration of the excess unreacted EDTA required 13.3 mL of 0.0100 M Ca. The Cd2t was displaced from EDTA by the addition of an excess of CN. Titration of the newly freed EDTA required 11.2 mL of 0.0100 M Ca2+. What are the concentrations of Cd2+ and Mn2+ in the original solution? M Mn2+ concentration: M Cd2 concentration: A...
This is for an analytical chemistry. Please show all work and excel sheets. Cu Determination (5...
This is for an analytical chemistry. Please show all work and excel
sheets.
Cu Determination (5 Points: 5 Points titration, 10 points error/statistics) Copper ions in solution were directly determined with an EDTA titration at a pH of 5.50 using glycinecresol red indicator. Assume a 25.00 mL aliquot of tap water in this titration. Cu Titration Trial VEDTA (mL) 16.98 16.48 16.92 16.29 18.12 16.41 17.12 Pb2 Determination (15 Points: 5 Points Tiration, 10 Points Erroristatistics) Lead was determined through...
A 24.34 mL aliquot of a Pb2 solution, containing excess Pb2+, was added to 11.50 mL...
A 24.34 mL aliquot of a Pb2 solution, containing excess Pb2+, was added to 11.50 mL of a 2,3-dimercapto-1-propanol (BAL) solution of unknown concentration, forming the 1:1 Pb2+-BAL complex. The excess Pb2+ was titrated with 0.0157 M EDTA, requiring 9.88 mL to reach the equivalence point. Separately, 39.89 mL of the EDTA solution was required to titrate 32.75 mL of the Pb2+ solution. Calculate the BAL concentration, in molarity, of the original 11.50 mL solution concentration:
10.6 10.00 mL of a sample of a solution containing Feand Co are added to 100...
10.6 10.00 mL of a sample of a solution containing Feand Co are added to 100 mL of pH 4.5 buffer and then titrated to a Cu-PAN endpoint with 18.82 mL of 0.05106 M EDTA. Then, 25.00 mL of the same of the same solution are added to 100 mL of pH 4.5 buffer along with 10 mL of 1 M potassium fluoride, which is added to mask the Fe'. This solution was titrated to a Cu-PAN endpoint with 13.40...
A 1.000-mL aliquot of a solution containing Cu and Ni2 is treated with 25.00 mL of a 0.05182 MEDTA solution. The solution is then back titrated with 0.02198 M Zn2 solution at a pH of 5. A volume of 21.82 mL of the Zn2 solution was needed to reach the xylenol orange end point. A 2.000-mL aliquot of the Cu and Ni2 solution is fed through an ion-exchange column that retains Ni2. The Cu2 that passed through the column is...
A 25.00 mL aliquot of a solution containing phenol was treated with 5 mL of 1 M KBr and 10 mL of 1 MHC A platinum generator electrode was made the anode and Br was oxidized to Br. A constant current of 8.00 mA was passed for 2 min and 38 s (not 2.38 min) to reach the potentiometric endpoint using an ORP electrode. Calculate the concentration of phenol in molL in the original sample, given 2Br-→Br2+22 Br Br 3Bra...
Please help! Will rate.
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 56.5 mL of 0.0600 M EDTA. Titration of the excess unreacted EDTA required 13.3 mL of 0.0100 M Ca. The Cd2t was displaced from EDTA by the addition of an excess of CN. Titration of the newly freed EDTA required 11.2 mL of 0.0100 M Ca2+. What are the concentrations of Cd2+ and Mn2+ in the original solution? M Mn2+ concentration: M Cd2 concentration: A...
This is for an analytical chemistry. Please show all work and excel
sheets.
Cu Determination (5 Points: 5 Points titration, 10 points error/statistics) Copper ions in solution were directly determined with an EDTA titration at a pH of 5.50 using glycinecresol red indicator. Assume a 25.00 mL aliquot of tap water in this titration. Cu Titration Trial VEDTA (mL) 16.98 16.48 16.92 16.29 18.12 16.41 17.12 Pb2 Determination (15 Points: 5 Points Tiration, 10 Points Erroristatistics) Lead was determined through...
A 24.34 mL aliquot of a Pb2 solution, containing excess Pb2+, was added to 11.50 mL of a 2,3-dimercapto-1-propanol (BAL) solution of unknown concentration, forming the 1:1 Pb2+-BAL complex. The excess Pb2+ was titrated with 0.0157 M EDTA, requiring 9.88 mL to reach the equivalence point. Separately, 39.89 mL of the EDTA solution was required to titrate 32.75 mL of the Pb2+ solution. Calculate the BAL concentration, in molarity, of the original 11.50 mL solution concentration:
10.6 10.00 mL of a sample of a solution containing Feand Co are added to 100 mL of pH 4.5 buffer and then titrated to a Cu-PAN endpoint with 18.82 mL of 0.05106 M EDTA. Then, 25.00 mL of the same of the same solution are added to 100 mL of pH 4.5 buffer along with 10 mL of 1 M potassium fluoride, which is added to mask the Fe'. This solution was titrated to a Cu-PAN endpoint with 13.40...
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