At 25 °C, a 10.0-mL solution of 0.0500 M AGNO3 was titrated with 0.0250 M NaBr in the cell: S.C.E. | titration solution | Ag(s). Find the cell voltage for 0.1 mL and 30.0 mL of titrant if the Ksp of AgBr is 5.0 x 10-13. (ESCE = 0.244 V; Ag+ + e -> Ag(s) E° = 0.799 V).

At 25 °C, a 10.0-mL solution of 0.0500 M AGNO3 was titrated with 0.0250 M NaBr in the cell: S.C.E. | titration solution | Ag(s). Find the cell voltage for 0.1 mL and 30.0 mL of titrant if the Ksp of AgBr is 5.0 x 10-13. (ESCE = 0.244 V; Ag+ + e -> Ag(s) E° = 0.799 V).

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...

See similar textbooks

Similar questions

Calculate the equilibrium concentrations of Fe3+, Fe2+, Mn2+, MnO4 in a solution prepared by mixing 20.0 mL of 0.100 M KMnO4 and 20.0 mL of 0.100 M FeSO4 at pH 0.00. EMnO/Mn²+ = 1.51 V, Efe+/Fe²+ = 0.771 V MnO + 5Fe2+ + 8H+ - Mn²+ + 5Fe³++ 4H₂O 4
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…
A4.
Calculate MW from Average EM. MW (g/mol) = n (eq/mole) x EM (g/eq) where n is the number of H+ ions titrated per moelcule of acid TRIAL # moles of OH- # moles of H+ # equivalents H+ Equivalent Mass (g/eq) 1 0.00264 0.00264 0.00264 113.26 2 0.003295 0.003295 0.003295 91.05 113.26 + 91.05 = 204.31/2 = 102.155 g/eq average EM Acid is Sulfamic Acid
14- 12- 10- PH 8. 6- 2- 0- 0 20.0 40.0 Volume of Titrant Added (mL) 60.0 80.0 A solution of 0.100 M HCI and a solution of 0.100 M NaOH are prepared. A 40.0 mL sample of one of the solutions is added to a beaker which is then titrated using the other solution. A pH electrode is used to obtain the data plotted in the titration curve shown here. a. Identify the solution that was initially added to the beaker. Explain your reasoning. b. On the titration curve, circle the point which corresponds to the equivalence point.
The concentration of Pb2+ in a solution saturated with Pb(OH)2(s) is 6.69 ✕ 10−6 M. Calculate Ksp for Pb(OH)2.
A buffer solution is made using a weak acid, HA, with a pK, of 5.18. If the ratio of [A] to [HA] is 0.10, what is the pH of the buffer? Macmill pH = 14 IDI 4 at 1 R LL % mw 5 a T f6 G 30 6 & Y 7 H 18 4+ D 8 J fg 3 9 K f10 ▶II O L L f11 88 P 62°F 4) 112 { [ insert 4:48 PM 10/5/2022 prt sc ← backspace 1 pause delete hor enter
Please answer the following pertaining to analytical chemistry.