1o A 5 cm sample of 0.05 mol dm3 sodium chloride is mixed with a 5 cm3 sample of 0.05moldm potassium iodide. 10 cm3 of acidified 0.05 moldm 3 silver nitrate is then added, followed by concentrated ammonia solution. What is seen after the addition of an excess of concentrated ammonia solution? a cream precipitate a white precipitate C a yellow precipitate no precipitate

1o A 5 cm sample of 0.05 mol dm3 sodium chloride is mixed with a 5 cm3 sample of 0.05moldm potassium iodide. 10 cm3 of acidified 0.05 moldm 3 silver nitrate is then added, followed by concentrated ammonia solution. What is seen after the addition of an excess of concentrated ammonia solution? a cream precipitate a white precipitate C a yellow precipitate no precipitate

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter26: Molecular Absorption Spectrometry

Section: Chapter Questions

Problem 26.14QAP

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Given the Information that Fe3++Y4FeYKi=1.01025Cu2++Y4CuY2Ki=6.31018 and the further information that, among the several reactants and products, only CuY2- absorbs radiationat 750 nm, describe how Cu(II) could be used as an indicator for the photometric titration of Fe(III) with H2Y2-. Reaction:+ Fe3++ H2Y2- FeY- + 2H+.
A 1.000-g sample containing bromide was dissolved in sufficient water to give 100.0 mL. A 50.00 mL aliquot was measured and after acidification, silver nitrate was introduced to precipitate AgBr, which was filtered, washed, and then dissolved in an ammoniacal solution of potassium tetracyanonickelate(II): Ni(CN)42- + 2AgBr(s) → 2Ag(CN)2- + Ni2+ + 2Br- The liberated nickel ion required 11.70 mL of 0.002146 M EDTA. Calculate mmol Ni in the 50.00 mL aliquot. Calculate mmol Br- in the 50.00 mL aliquot. Calculate the percentage of NaBr (102.894) in the 1.000 g sample.
A 1.000-g sample containing bromide was dissolved in sufficient water to give 100.0 mL. A 50.00 mL aliquot was measured and after acidification, silver nitrate was introduced to precipitate AgBr, which was filtered, washed, and then dissolved in an ammoniacal solution of potassium tetracyanonickelate(II): Ni(CN)42- + 2AgBr(s) → 2Ag(CN)2- + Ni2+ + 2Br- The liberated nickel ion required 11.70 mL of 0.002146 M EDTA The other 50.00 mL remaining solution was also analyzed for its Br- content by potentiometry using a metallic electrode of the second kind. Ecell recorded in running the solution using the potentiometric set-up was Ecell = 0.0286 V. (E0Ag/AgBr = 0.095 V) g) Compute for Eind. h) Compute pBr in the 50.00 mL aliquot. i) Compute for % NaBr ( in the potentiometric technique). j) Calculate the error between the obtained % NaBr fr the EDTA titration technique and the % NaBr from the potentiometric technique.
A 1.000-g sample containing bromide was dissolved in sufficient water to give 100.0 mL. A 50.00 mL aliquot was measured and after acidification, silver nitrate was introduced to precipitate AgBr, which was filtered, washed, and then dissolved in an ammoniacal solution of potassium tetracyanonickelate(II): Ni(CN)42- + 2AgBr(s) → 2Ag(CN)2- + Ni2+ + 2Br-Nickel ion required 11.70 mL of 0.002146 M EDTAWhat is the percentage of NaBr (102.894) in the 1.000 g sample?
A 1.000-g sample containing bromide was dissolved in sufficient water to give 100.0 mL. A 50.00 mL aliquot was measured and after acidification, silver nitrate was introduced to precipitate AgBr, which was filtered, washed, and then dissolved in an ammoniacal solution of potassium tetracyanonickelate(II): Ni(CN)42- + 2AgBr(s) → 2Ag(CN)2- + Ni2+ + 2Br-50.00 mL remaining solution was analyzed for its Br- content by potentiometry using a metallic electrode of the second kind. a) Write the cell notation of the potentiometric set-up with SCE as the reference electrode. b) Write the Nernst equation that describes the indicator electrode set-up. Ecell recorded in running the solution using the potentiometric set-up was Ecell = 0.0286 V. (E0Ag/AgBr = 0.095 V) c) Compute for Eind. d) Compute pBr in the 50.00 mL aliquot. e) Compute for % NaBr ( in the potentiometric technique).
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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.
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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?
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