An unknown solution containing 25.00 mL of Ni2* in dilute HCI is treated with a known excess of 25.00 mL of 0.05382 M EDTA solution. The solution turns yellow when a few drops of indicator are added. The excess unreacted EDTA back titrated with 0.02299 M Zn2*, which requires 17.00 mL to reach the red end point. What is the molarity of Ni2* in the unknown solution? (b)

An unknown solution containing 25.00 mL of Ni2* in dilute HCI is treated with a known excess of 25.00 mL of 0.05382 M EDTA solution. The solution turns yellow when a few drops of indicator are added. The excess unreacted EDTA back titrated with 0.02299 M Zn2, which requires 17.00 mL to reach the red end point. What is the molarity of Ni2 in the unknown solution? (b)

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter17: Complexation And Precipitation Reactions And Titrations

Section: Chapter Questions

Problem 17.31QAP

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The concentration of ammonia in a cleaning product was determined by back titration.Firstly, 10.00 cm3 of the cleaning product was pipetted into a large conical flask,containing 250.00cm3 of 0.50 mol/l HCl to give Solution A.Following a period of reaction and shaking, 50.00cm3 of Solution A was removed anddiluted to 250 cm3 with water in a volumetric flask to give Solution B.20 cm3 samples of Solution B were titrated against 0.05 mol/l Na2CO3 solution, givingan average titre of 12.45 cm3. i) Write equations for the reactions that have taken place.ii) Determine the concentration of NH3 in the original cleaning product in mol/l,g/l, ppm, and % w/v.
Chromel is an alloy composed of nickel, iron and chromium.A 0.6472 g sample was dissolved and diluted to 250 mL. When a50 mL aliquot of 0.05182 M EDTA was mixed with an equal volumeof the diluted sample and all the three ions were chelated, a 5.11 mLback titration with 0.06241 M copper (II) was required.The chromium in a second 50 mL aliquot was masked through theaddition of hexamethylenetetramine, titration of the Fe and Nirequired 36.28 mL of 0.05182 M EDTA.Iron and chromium were masked with pyrophosphate in a third50 mL aliquot and the nickel was titrated with 25.91 mL of theEDTA solution.Calculate the percentage of nickel, chromium and iron in thealloy.
A 0.3045 g of CaCO3 primary standard was dissolved using concentrated HCl, evaporated to incipient dryness and quantitatively transferred to a 250 mL volumetric flask, and diluted to mark. A 10 mL of aliquot was then transferred to an Erlenmeyer flask, together with 5mL buffer and 5 drops of EBT indicator, and was used to standardize the EDTA titrant. The solution turned blue after the addition of 24.10 mL of EDTA. A blank correction was performed and was determined to be 0.4 mL. What is the concentration of EDTA obtained (MW CaCO3 = 100.0869 g/mol)?
A 0.3045 g of CaCO3 primary standard was dissolved using concentrated HCl, evaporated to incipient dryness, and quantitatively transferred to a 250 mL volumetric flask and diluted to mark. A 10 mL of aliquot was then transferred to an Erlenmeyer flask, together with 5mL buffer and 5 drops of EBT indicator, and was used to standardize the EDTA titrant. The solution turned blue after the addition of 24.10 mL of EDTA. A blank correction was made and was determined to be 0.4 mL. [Analysis] A 30 mL unknown water sample was treated with 37.6 mL of EDTA. A blank correction was made and was determined to be 0.4 mL. What is the concentration of Ca2+ (40.0780 g/mol) in ppm?
A 0.9352g sample of ore containing Fe³+, Al³+ and Sr²+ was dissolved and made up to 500.00 mL. The analysis of metals was performed using complexation volumetry. Initially, an aliquot of 50.00 mL had its pH adjusted to 1.0 and titrated with a standard 0.03145 mol/L EDTA solution, requiring 6.95 mL to reach the end point. Subsequently, another 25.00 mL aliquot was buffered at pH=5 and titrated with the same EDTA solution, requiring 6.24 mL to reach the end point. Finally, a third aliquot of 25.00 mL was titrated at pH=11, requiring 11.10 mL of the same EDTA solution to complete the titration. Given the molar masses: Fe=55.845 g/mol; Al-26.982 g/mol and Sr-87.620 g/mol. a) Determine the percentage of each of the metals in the sample. b) Explain why the change in pH allows the determination of the three ions in this sample.
The amount of iron in a meteorite was determined by a redox titration using KMnO4 as the titrant. A 0.4185 -g sample was dissolved in acid and the liberated Fe3+ quantitatively reduced to Fe2+, using a reductor column. Titrating with 0.0044 M KMnO4 requires 28.76 mL to reach the end point. Determine the %w/w Fe2O3 in the sample of meteorite. (Fe = 55.845 amu, Fe2O3 = 159.69 g/mol
A foot powder sample containing Zn was dissolved on 50.00 mL water and was titrated to the end point color with 22.57 mL of 0.01639 M EDTA at pH=4. (α= 3.61x109, KKzny2 = 3.2 x 1016). The pZn in the sample is:
A student withdraws 3.00Ml of supernate from a saturated solution of KHC8H4O4 at room temp (22C). This sample was titrated to the henolphthalein endpoint and 12.85mL of 0.0997 M NaOH was required A similare tiration was done at 22C for a saturated solution of KHP in 0.500 M KCl. The withgrawn supernant was 5.00mL. The volume of titrant (0.0997M NaOH) used was 11.75. a) calculate the % decrease in solubility by using this equation %decrease = [HC8H4O4]water - [HC8H4O4]sat'd KCl /[HC8H4O4]water
When I was a boy, I watched Uncle Wilbur measure the iron content of runoff from his banana ranch. He acidified a 25.0-mL sample with HNO3 and treated it with excess KSCN to form a red complex. (KSCN itself is colorless.) He then diluted the solution to 100.0 mL and put it in a variablepathlength cell. For comparison, he treated a 10.0-mL reference sample of 6.80X 1024 M Fe31 with HNO3 and KSCN and diluted it to 50.0 mL. The reference was placed in a cell with a 1.00-cm pathlength. Runoff had the same absorbance as the reference when the pathlength of the runoff cell was 2.48 cm. What was the concentration of iron in Uncle Wilbur’s runoff ?
Chromel is an alloy composed of nickel, iron and chromium. A 0.6553-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 5.34-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 36.98 mL of 0.05173M EDTA. Iron and chromium were masked with pyrophosphate in a third 50.00-mL aliquot, and the nickel was titrated with 24.53 mL of the EDTA solution. Calculate the percentage of Cr in the alloy. Express your answer in 2 decimal places.
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.
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