A 0.400-g sample of toothpaste was boiled with a 50-mL solution containing a citrate buffer and NaCl to extract the fluoride ion. After cooling, the solution was diluted to exactly 100 mL. The potential of a selective ion/calomel system in a 25.0-mL aliquot of the sample was found to be - 0.1823 V. Addition of 5.0 mL of a solution containing 0.00107 mg F-/mL caused the potential to change to 0.2446 V. Calculate the weight percent F- in the sample.

Calculate the cell potential when 50.00 mL of 0.100 M Ag+ solution is titrated with 50.00 mL of 0.150 M Cl- solution. Silver wire indicator electrode and Ag-AgCl electrode are used as reference electrodes. E0 (Ag/AgCl) = 0,197 V , E0 Ag+/Ag = 0,799 V, Ksp (AgCl) = 1,8 x10-8

Calculate the potential at the equivalence point for titration of 0.100MFe^2+ with 0.100M Ce^4+.Given the formal potential for the reduction E^0 of Fe^3+ to Fe^2+ is +0.77V, and the formal potential for the reduction of E^0 Ce^4+ to Ce^3+ is 1.45V.

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…

Consider the titration of 100.0 mL of 0.010 0 M Ce4+ in 1 M HClO4 with 0.040 0 M Cu+ to give Ce3+ and Cu2+. Calculate the potential of the indicator electrode after adding 24.5 mL of Cu+.

Consider the titration of 60.00 mL of 0.100 M Co3+ with 0.130 M MnO²- . What is the potential after 30.00 mL of 0.130 M MnO²- has been added in units of Volts? The potential is measured against the Ag|AgCl reference electrode (E(saturated KCI) = 0.197 V). Co 3+ + e =Co 2+ E°= 1.92 V MnO¯ + e ¯ = MNO2- E°= 0.56 V