How do I find Equilibrium [FeSCN2+] with given values. See image. PART A-Table 1Soln (Flask) No.Vol of NaSCN Soln (mL)Initial [SCN-]Eqm [FeSCN2+]AbsorbanceConcentration of Fe(NO3), in 0.10 M HNO₂: .Flask No. 3Flask No.4Flask No.1 Flask No.2000.00114.00E-050.038.00E-050.1631.20E-040.2344.00E-02 Concentration of NaSCN in 0.10 M HMFlask No.6Flask No.51.60E-040.30352.00E-040.362Eqn

According to Beer's law, the absorbance for any solution is directly proportional to its initial…

ART A-Table 1 Soln (Flask) No. Vol of NaSCN Soln (mL) Initial [SCN-] Eqm [FeSCN2+] Absorbance Concentration of Fe(NO3)3 in 0.10 M HNO,:. Flask No. 3 Flask No.4 Flask No.1 Flask No.2 0 0 0 0.001 1 4.00E-05 0.03 2 8.00E-05 0.16 3 1.20E-04 0.234 4.00E-02 Concentration of NaSCN in 0.10 M HI Flask No.6 Flask No.5 4 1.60E-04 0.303 5 2.00E-04 0.362 E

ART A-Table 1 Soln (Flask) No. Vol of NaSCN Soln (mL) Initial [SCN-] Eqm [FeSCN2+] Absorbance Concentration of Fe(NO3)3 in 0.10 M HNO,:. Flask No. 3 Flask No.4 Flask No.1 Flask No.2 0 0 0 0.001 1 4.00E-05 0.03 2 8.00E-05 0.16 3 1.20E-04 0.234 4.00E-02 Concentration of NaSCN in 0.10 M HI Flask No.6 Flask No.5 4 1.60E-04 0.303 5 2.00E-04 0.362 E

Principles of Instrumental Analysis

7th Edition

ISBN:9781305577213

Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch

Publisher:Douglas A. Skoog, F. James Holler, Stanley R. Crouch

Chapter1: Introduction

Section: Chapter Questions

Problem 1.9QAP

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The acid-base indicator HIn undergoes the following reaction in dilute aqueous solution: HIncolor1H++Incolor2 The following absorbance data were obtained for a 5.00 I0-4 M solution of HIn in 0.1 M NaOH and 0.1 M HC1. Measurements were made at wavelengths of 485 nm and 625 nm with 1.00-cm cells. 0.1 M NaOH A485 = 0.075 A625 = 0.904 0.1 M HC1 A485 = 0.487 A625 = 0.181 In the NaOH solution, essentially all of the indicator is present as In-; in the acidic solution, it is essentially all in the form of HIn. (a) Calculate molar absorptivities for In- and HIn at 485 and 625 nm. (b) Calculate the acid dissociation constant for the indicator ¡fa pH 5.00 buffer containing a small amount of the indicator exhibits an absorbance of 0.567 at 485 nm and 0.395 at 625 nm (1.00-cm cells). (c) What is the pH of a solution containing a small amount of the indicator that exhibits an absorbance of0.492 at 485 nm and 0.245 at 635 nm (1.00-cm cells)? (d) A 25.00-mL aliquot of a solution of purified weak organic acid HX required exactly 24.20 mL of a standard solution of a strong base to reach a phenolphthalein end point. When exactly 12.10 mL of the base was added to a second 25.00-mL aliquot of the acid, which contained a small amount of the Indicator under consideration, the absorbance was found to be 0.333 at 485 nm and 0.655 at 625 nm (1.00-cmcells). Calculate the pH of the solution and Ka for the weak acid. (e) What would be the absorbance of a solution at 485 and 625 nm (1.50-cm cells) that was 2.00 10-4 M in the indicator and was buffered to a pH of 6.000?
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+.
The chromium in an aqueous sample was determined by pipetting 10.0 ml. of the unknown into each of five 50.0-mL volumetric flasks. Various volumes of a standard containing 12.2 ppm Cr were added to the flasks, following which the solutions were diluted to volume. Unknown,mLStandard, mLAbsorbancc 10.00.00.201 10.010.0 0.292 10.020.0 0.378 10.030.0 0.467 10.040.0 0.554 (a) Plot the data using a spreadsheet. (b) Determine an equation for the relationship between absorbance and volume of standard. (c) Calculate the statistics for the least-squares relationship in (b). (d) I)ctcrmine the conccnt ration oÍCr in ppm in the sample. (e) Find the standard deviation of the result in (d).
6. Blue Blue dye stock solution 0.293 M Absorbance at 630 nm 0.00265 Calibration curve y = 0.0833x A solution is prepared by diluting 2.79 mL of the blue dye stock solution to 25.00 mL. The measured absorbance for the prepared solution is listed in the data table. (a) What is the theoretical molar concentration? [Blue]theoretical x 10 |M (b) What is the experimental molar concentration? [Blue]experimental x 10 M (c) What is the percent error? Percent error (blue) = %
Activities and Assessments: 1. Determine the Ecell values for the following: a. TI (s)|Tl* (aq) || Sn²+ (aq) | Sn(s) b. Zn(s)|Zn²*(aq) || Fe³*(aq), Fe²*(aq)| Pt c. Cd(s)| Cd²+(aq)|| H* (aq)|H2(g) | Pt d. Fe2+ + Mn0; → Fe3+ + Mn²+ e. Sn + NOz → Sn02 + NO2 f. Zn + NO, → Zn²* + NH g. The oxidation of iodide ion I- by permanganate ion Mno, in basic solution to yield molecular iodine l2 and manganese (IV) oxide Mn02. 170
*13-7_ A solution containing the complex formed between Bi(III) and thiourea has a molar absorptivity of 9.32 X 10³ L mol¹ cm¹ at 470 nm. 00.5 701 X 00.A 1910 919 juloa wot What is the absorbance of a 4.25 × 10-5 M : I solution of the complex at 470 nm in a 1.00-cm cell? (a) W (b) (alios mo-00.1) What is the percent transmittance of the solution described in (a)? (c) 200 What is the molar concentration of the complex in a solution that has the absorbance described in Torito or tovo (a) when measured at 470 nm in a 2.50-cm cell? (non?+ 1rd
A 2.50 mL aliquot of a solution that contains 3.85 ppm Fe(III) is treated with an appropriate excess of KSCN and diluted to 100.0 mL. What is the absorbance of the resulting solution at 580 nm in a 0.50 cm cell? (Molar absorptivity of Fe (SCN)2+is 7.0 x 105L mol-1cm-1). AW: K =39.098 S=32.08 C=12.011 N=14.007 Fe=55.845
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A 2.50 mL aliquot of a solution that contains 3.85 ppm Fe(III) is treated with an appropriate excess of KSCN and diluted to 100.0 mL. What is the absorbance of the resulting solution at 580 nm in a 0.50 cm cell? (Molar absorptivity of Fe (SCN)2+is 7.0 x 105L mol-1cm-1).
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1) Pd (II) with Au (III) methiomeprazine (C19 H24 N2 S2) complex. A mixture of their complexes is being analyzed. Pd (II) complex at 480 nm, The measurement results for Au (II) with a maximum peak at 635 nm are as follows. Of this mixture What is its concentration? (b = 1) Pd(II) ɛ Au(III)ɛ 2.96x 103 A 480 nm 3.55x 103 0.533 635 nm 5.64x 102 1.45x 104 0.590