A foot powder samplecontaining Zn was dissolvedon 200.00 mL water andwas titrated to the end pointcolor with 22.57 mL of 0.01639M EDTA at pH=4.(a4= 3.61x10-9, K ZnY²- = 3.2 ×1016). The pZn in the sample is:%3DPlease fill in the spacewith a numerical valuewith two digits ornumerical values withoutunitsAnswer:

The question is based on the concept of chemical equilibrium. we have to calculate pZn value for the…

A foot powder sample containing Zn was dissolved on 200.00 mL water and was titrated to the end point color with 22.57 mL of 0.01639 M EDTA at pH=4. (a4= 3.61×10-9, K ZnY²- = 3.2 x 1016). The pZn in the sample is: Please fill in the space with a numerical value with two digits or numerical values without units Answer:

A foot powder sample containing Zn was dissolved on 200.00 mL water and was titrated to the end point color with 22.57 mL of 0.01639 M EDTA at pH=4. (a4= 3.61×10-9, K ZnY²- = 3.2 x 1016). The pZn in the sample is: Please fill in the space with a numerical value with two digits or numerical values without units Answer:

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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a primary standard of (COOK)2.H2O was prepared up to 0.05M. This primary standard was then titred against an unknown concentration of 50cm3 of KMnO4. An average titre value of 13 was obtained. The balanced equation is as follows: 5(COOK)2.H2O + 2KMnO4 + 8H2SO4 = 6K2SO4 +2MnSO4 + 10CO2 +18H2O. Find the concentration of the KMnO4.
A 1.000 ml. aliquot of a solution containing Cu?* and Ni* is treated with 25.00 ml, of a 0.03468 M EDTA solution. The solution is then back titrated with 0.02308 M Znt solution at a pH of 5. A volume of 20.28 ml. of the Zn+ solution was needed to reach the xylenol orange end point. A 2.000 ml. aliquot of the Cu* and Ni* solution is fed through an ion- exchange column that retains Ni*. The Cu+ that passed through the column is treated with 25.00 ml. of 0.03468 M EDTA. This solution required 23.18 ml. of 0.02308 M Zn* for back titration. The Ni" extracted from the column was treated witn 25.00 mL of 0.03468 M EDTA. How many milliliters of 0.02308 M Zn?* is required for the back titration of the NP solution? volume: ml.
COMPLEXOMETRIC TITRATION A deep well sample of 100 mL was titrated with 0.010 M EDTA at pH=10 (with Mg-EDTA-PH10 reagent) and consumed 31.95 mL using EBT indicator. Titration of another 100 mL of the same deep well sample with the same titrant but the pH was adjusted to 12.10 and murrexide as indicator consumed 24.25 ml to reach end point. Calculate: А. Total Hardness of water as CaCO3 В. ppm of Ca and Mg 76
Calculate the equilibrium concentrations of all species when 10 mL of 0.90M Fe2+ is mixed with 10 mL of 0.10M Fe3+, 10 mL of 0.20M Cr2072-and 10 mL of 0.2M Cr3*all buffered to a pH = 3. 6FE2++ Cr2072-+ 14H* - 6FE3++ 2Cr3++ 7H2O K =1×1059
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8) 25.00 mL of a halide solution, which contains approximately 0.025 M iodide and approximately 0.05 M chloride, is titrated with 0.0500 M AgNO3. Ksp (Agl) = 8.3 x 10-17 and Ksp (AgCl) = 1.8 x 10-10./25.00 mL van 'n haliedoplossing, wat ongeveer 0.025 M jodied en ongeveer 0.05 M chloried bevat, word getitreer met 0.0500 M AgNO3. Ksp (Agl) = 8.3 x 10-17 en Ksp (AgCl) = 1.8 × 10-10. a) Sketch a rough pAg* vs volume AgNO3 added curve of the titration of 25 mL of the halide solution with the AgNO3 solution. Your sketch should be consistent with the information provided, including that the pAg* scale should be an approximate representation of the true value at different points of the titration. Your curve should also include realistic estimates of the equivalence points. / Skets 'n rowwe benoemde pAg* vs volume AgNO3 bygevoeg kurwe van die titrasie van 25 mL van die halied oplossing met die AgNO3 oplossing. Jou skets moet ooreenstem met die inligting wat verskaf word, insluitend dat die…
If a 25.00 ml aliquot of a solution containing sulfide ions was added to 25.00 ml of 0.0108 M Cu2+ solution. The CuS precipitate was washed and collected. The pH of the combined filtrate and washings was then adjusted by the addition of ammonia and the solution was titrated with 0.01453 M EDTA, using a suitable indicator. The endpoint volume was found to be 12.12ml. Find the molarity of the sulfide in the sample solution.
A 340.00 mL solution of 0.00175 M AB2 is added to a 140.00 mL solution of 0.00235 M CD4. What is pQsp for AD2?
08:37 * E 6 • @ 3 53% PREPARATION%200F%20A%20BUI fx D E 37 TABLE 2: 38 Volume of Titrant pH of 0.05M Na-Phosphate pH of 0.05M Na-phophate buffer with HCI (0.05M) (ml) 0,0 buffer with NaOH (0.05M) 7,73 39 40 7.73 41 0,5 7,54 6.78 5.91 1,0 1,5 2,0 2,5 42 7.12 43 3.26 2.45 2.25 7.52 44 8.22 45 10.50 46 3,0 2.15 3,5 4,0 4,5 5,0 5,5 6.0 47 2.02 48 49 50 51 52 6,0 53 6,5 54 7,0 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87
DATA AND RESULTS: Titration Data (Standardization of EDTA) Example II Mass, g Caco, 0.0250 0.0251 0.0246 0.0254 Final reading, mL 25.00 24.91 24.87 25.31 Initial Reading, mL 0.00 0.02 0.02 0.02 Volume, mL EDTA 25.00 Molarity of EDTA, M 0.009992 B. Analysis Data Example II Volume of Sample, mL 25.00 24.90 25.10 25.0 Final reading, mL 27.46 25.08 25.06 24.99 Initial Reading, mL 0.00 0.02 0.02 0.02 Volume of EDTA, mL 27.46 Molarity of EDTA, M 0.009992 Ca" concentration, ppm 440 STATISTICAL EVALUATION OF RESULTS Standardization of EDTA Mean, M EDTA Median, M EDTA Standard Deviation Ci for u at 95% Prob. Level Reported value: M EDTA Ca" Concentration Mean, Ca" ppm Median, Ca" ppm Standard deviation Cl for u at 95% Prob. Level Reported value: Ca" ppm
A 1.00 mL aliquot of a solution containing Cu and Ni is treated with 25.00 ml. of a 0.03468 M EDTA solution. The solution is then back titrated with 0.02308 M Zn+ solution at a pH of 5. A volume of 20.28 ml. of the Zn solution was needed to reach the xylenol orange end point. A 2.000 mL aliquot of the Cu* and Ni* solution is fed through an ion- exchange column that retains Ni?*. The Cu?+ that passed through the column is treated with 25.00 mL of 0.03468 M EDTA. This solution required 23.18 mL of 0.02308 M Zn* for back titration. The Ni extracted from the column was treated witn 25.00 mL of 0.03468 M EDTA. How many milliliters of 0.02308 M Zn+ is required for the back titration of the Ni solution? volume: ml.