5 A 0.400-g sample of toothpaste was boiled with a 50.00 mL solution containing citrate buffer and NaCl to extract the fluoride ion. After cooling, the solution was diluted to exactly 100.00 mL. The potential of a selective ion/calomel system in a 25.00 mL aliquot of the sample was found to be -0.1823 V. Addition of 5.00 mL of a solution containing 0.00107 mg F/mL caused the potential to change to -0.2446V. Calculate the weight-percent F in the sample

5 A 0.400-g sample of toothpaste was boiled with a 50.00 mL solution containing citrate buffer and NaCl to extract the fluoride ion. After cooling, the solution was diluted to exactly 100.00 mL. The potential of a selective ion/calomel system in a 25.00 mL aliquot of the sample was found to be -0.1823 V. Addition of 5.00 mL of a solution containing 0.00107 mg F/mL caused the potential to change to -0.2446V. Calculate the weight-percent F in the sample

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

Chapter24: Coulometry

Section: Chapter Questions

Problem 24.4QAP: Halide ions can he deposited at a silver anode, the reaction being Ag(s) + X- AgX(s) +e- Suppose...

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Halide ions can he deposited at a silver anode, the reaction being Ag(s) + X- AgX(s) +e- Suppose that a cell was formed by immersing a silver anode in an analyte solution that was 0.0250 M Cl-,Br-, and I -ions and connecting the half-cell to a saturated calomel cathode via a salt bridge. (a) Which halide would form first and at what potential? Is the cell galvanic or electrolytic? (b) Could I- and Br- be separated quantitatively? (Take 1.00 l0-5 M as the criterion for quantitative removal of an ion.) If a separation is feasible, what range of cell potential could he used? (c) Repeat part (b) for I- and Cl-. (d) Repeat part (b) for Br- and Cl-.
The measurement of pH using a glass electrode obeys the Nernst equation. The typical response of a pH meter at 25 00C is given by the equation where contains the potential of the reference electrode and all other potentials that arise in the cell that are not related to the hydrogen ion concentration. Assume that = 0.250 V and that a. What is the uncertainty in the values of pH and [H+] if the uncertainty in the measured potential is 1 m V ( 0.001 V)? b. To what precision must the potential be measured for the uncertainty in pH to be 0.02 pH unit?
What is the cell potential of the following cell at 25C? Ni(s)Ni2+(1.0M)Sn2(1.5104M)Sn(s)
From the standard potentials Ag2SeO4(s)+2e2Ag(s)+SeO42-E0=0.355V Ag++2eAg(s)E0=0.799V calculate the solubility product constant for Ag2SeO4.
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A soluion of 10.0 mL of 0.0500 mol/L AgNO3 is titrated with 0.0250 mol/L NaBr in the cell: VKE||onbekende oplossing||Ag(s) Calculate the potential of the cell after adding 0.1 mL, 10.0 mL, 20.0 mL and 30.0 ml of titrant. E\deg (Ag+, Ag = 0.799 V; E(VKE = 0.244 V; Ksp(ABr) = 5.0*10-13
A pH meter is calibrated using a pH 4.00 buffer with 174.48 mV, pH 7.00 with 0 mV and pH 10.0 buffer with (-)174.48 mV. Calculate the Molar concentration of a HCl solution with electrode potential of 396.31 mV.
In the analysis at 25 ° C of chloride using an ion selective electrode (ISE), two 100 mL solutions were prepared. These contain 1.0129x10-2 M and 1.01239x10-3 M Cl- and yielded a potential of 20.3 mV and 79.5 mV, respectively. The value of the electrode slope in mV is:
A solution prepared by mixing 60.0 mL of 0.360 M AGNO, and 60.0 mL of 0.360 M TINO, was titrated with 0.720 M NaBr in a cell containing a silver indicator electrode and a reference electrode with a constant potential of 0.175 V. The reference electrode is attached to the positive terminal of the potentiometer, and the silver electrode is attached to the negative terminal. The solubility constant of TIBr is Kp = 3.6 × 10-6 and the solubility constant of AgBr is = 5.0 x 10-13. Ksp Which precipitate forms first? O TIBI O AgBr Which of the expressions shows how the cell potential, E, depends on [Ag*]? E = 0.175 V – [0.799 V – 0.05916log ([Ag+])] 1 E = |0.799 V – 0.05916log - 0.175 V [Ag*) E = [0.799 V – 0.05916log ([Ag*]D] – 0.175 V E = 0.175 V - |0.799 V – 0.05916log [Ag* * (ল) Calculate the first and second equivalence points of the titration. first equivalence point: mL second equivalence point: mL What is the cell potential after each of the given volumes of 0.720 M NaBr have been…