H2S(aq) is analyzed by titration with coulometrically generated I2 in Reactions 17-3a and 17-3b. To 50.00 mL of unknown H2S sample were added 4 g of KI. Electrolysis required 812 s at 52.6 mA. Find the concentration of H2S (mg/mL)in the sample.
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H2S(aq) is analyzed by titration with coulometrically generated I2 in Reactions 17-3a and 17-3b. To 50.00 mL of unknown H2S sample were added 4 g of KI.
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- A 40.00-mL aliquot of 0.05000 M HNO2 is diluted to 75.00 mL and titrated with 0.0800 M Ce4+ . The pH of the solution is maintained at 1.00 throughout the titration; the formal potential of the cerium system is 1.44 V. Calculate the potential of the indicator electrode with respect to a saturated calomel reference electrode after the addition of 5.00 mL of cerium (IV). (Use a MW value in 4 decimal places)Traces of aniline can be determined by reaction with an excess of electrolytically generated Br2: C6H5NH2 + 3Br2 =====è C6H2Br3NH2 + 3H+ + 3Br- The polarity of the working electrode is then reversed, and the excess bromine is determined by a coulometric titration involving the generation of Cu(I): Br2 + 2Cu+ ===è 2Br- + 2Cu2+ Suitbale quantities of KBr and copper (II) sulfate were added to a 25.0 ml sample containing aniline (C6H5NH2). Calculate the mass in micrograms of C6H5NH2 in the sample from the accompanying data. Working electrode Functioning as Generation Time (min) with a constant current of 1.00 mA Anode 3.76 cathode 0.270One method for determining whether an individual has recently fired a gun is to look for traces of antimony in the residue collected from the individual’s hands. Anodic stripping voltammetry at a mercury film electrode is ideally suited for this analysis. In a typical analysis a sample is collected from a suspect with a cotton-tipped swab wetted with 5% v/v HNO3. After returning to thelab, the swab is placed in a vial containing 5.0 mL of 4 M HCl that is 0.02 Min hydrazine sulfate. After allowing the swab to soak overnight, a 4.0-mL portion of the solution is transferred to an electrochemical cell along with 100 mL of 0.01 M HgCl2. After depositing the thin film of mercury and the antimony, the stripping step gives a peak current of 0.38 μA. After adding a standard addition of 100 μL of 5.00×102 ppb Sb, the peak current increases to 1.14 μA. How many nanograms of Sb were collected from the suspect’s
- acid dissociation constants for H3PO4 (Ka1 = 7.11 × 10-3, Ka2 = 6.32 × 10-8, Ka3 = 7.1× 10-14), calculate standard reduction potential for the half reactionH2PO4- + H+ + 2e HPO32- + H2OLead in dry river sediment was extracted with 25 wt% HNO3 at incubation temperatures for 1 hr. Then 1 mL of filtered extract was combined with other reagents to a total of 4.60 mL. Pb+2 was measured by an electrochemical method by spiking with 2.50 ppm Pb2+. added vol. of 2.50ppm Pb2 signal 0 mL 1.10 0.025 mL 1.66 .050 mL 2.20 .075 mL 2.81 what is Sx based on SyUse the shorthand notation to describe a cell consisting of a saturated calomel reference electrode and a silver indicator electrode for the measurement of (a) pSCN. (b) pI. (c) pSO3 (d) pPO. Then, generate an equation that relates pAnion to Ecell for each of the cells. (For Ag2SO3 Ksp = 1.5 * 10-14; for Ag3PO4, K,p = 1.3 *10-20 ).
- A titration of 50.0 mL of 0.10 M Sn2+ with 0.2 M Fe3+ in 1 M HCl to give Fe2+ and Sn4+ using Pt and calomel electrodes. Assuming the standard potential for Sn2+/Sn4+ = 0.139V ; calomel, 0.241 V; Fe2+/Fe3+ = 0.732 V. This my cathode reaction: Fe3+ (aq) + 1 electron <-> Fe2+(s) and anode reaction: Sn2+(s) <-> Sn4+(aq) + 2 electrons. How will I write my Nernst equation for the cell voltage for the oxidation reaction?A 0.2653 g of pure potassium dichromate was dissolved in 15 mL of 1.5 M sulfuric acid and transferred to a 250.0 mL volumetric flask and diluted to the mark with distilled water. A 25.0 mL aliquot of this solution was transferred to another 250.0 mL volumetric flask and diluted to the mark with distilled water and this solution has an absorbance 0f 0.387 in a 2.00 cm cell. Calculate the molar absorptivity of potassium dichromate.(Please give clear handwritten answer) 0.1 M HNO3 solution at 298K was electrolyzed in Hittorf cell using platinum electrodes. After electrolysis, 34.2 mg of copper was deposited in coulometer which in series with the Hittorf cell. 50 mL of HNO3 solution was withdrawn from anode compartment after electrolysis and found to have a concentration of 0.0821 M. [Atomic mass Cu = 63]. (i) Find the transference number of H + and NO3 −. (ii) If 50 mL HNO3 solution was run off from cathode compartment, what would its concentration be?
- To find the copper content in 50 mL of sample, a titration is carried out with 0.01 M AEDT. A selective Cu electrode is used to determine the end point of the titration. Three assessments are performed that use 12.5 mL, 12.3 mL; 12.4 mL of 0.01 M EDTA.a) Indicate the equipment and material necessary to carry out the potentiometric assessment.b) Indicate the reagent that must be added to the sample for the correct performance of the titration. explain the reason.c) Calculate the Cu concentration of the sample (in mg/L).Data: atomic weight Cu = 63.5A 40.00-mL aliquot of 0.05000 M HNO2 is diluted to 75.00 mL and titrated with 0.0800 M Ce4+. The pH of the solution is maintained at 1.00 throughout the titration; the formal potential of the cerium system is 1.44V a. Calculate the potential of the indicator electrode with respect to a saturated calomel reference electrode after the addition of 5.00, 10.00, 15.00, 25.00, 40.00, 49.00, 49.50, 49.60, 49.70, 49.80, 49.90, 49.95, 49.99, 50.00, 50.01, 50.05, 50.10, 50.20, 50.30, 50.40, 50.50, 51.00, 60.00, 75.00 and 90.00 mL of Cerium (IV). b. Draw a titration curve for these data. c. Generate a first- and second-derivative curve for these data. Does the volume at which the second-derivative curve crosses zero correspond to the theoretical equivalence point? Why or why not?A 5.00 mL tap water sample was measured out with a volumetric pipette, and added to a 25 mL Erlenmeyer flask. It was then titrated with a 0.0100 M Na2EDTA.2H2O solution, and found to take 0.635 mL of the EDTA solution to reach the blue endpoint.