To determine the amount of silver ions present after the complete electrolysis, we need to first find the number of moles of electrons transferred during the electrolysis process.1. Calculate the total charge passed during electrolysis:Current I = 2.00 mA = 0.002 A converting milliamperes to amperes Time t = 25.0 minutes = 1500 seconds converting minutes to seconds Total charge Q = Current I Time t Q = 0.002 A 1500 s = 3.00 Coulombs2. Calculate the number of moles of electrons transferred:1 mole of electrons = 96,485 Coulombs Faraday's constant Moles of electrons n = Total charge Q / Faraday's constantn = 3.00 C / 96,485 C/mol = 3.11 10^-5 moles3. Determine the stoichiometry of the reaction:The balanced half-reaction for the reduction of silver ions is:Ag+ + e- AgFrom the stoichiometry, 1 mole of electrons reduces 1 mole of silver ions.4. Calculate the moles of silver ions reduced:Moles of silver ions reduced = Moles of electrons transferredMoles of silver ions reduced = 3.11 10^-5 moles5. Calculate the initial moles of silver ions in the 50.0 mL of 0.500 M silver nitrate solution:Initial moles of silver ions = Molarity Volume in liters Initial moles of silver ions = 0.500 mol/L 0.050 L = 0.025 moles6. Calculate the moles of silver ions remaining after electrolysis:Moles of silver ions remaining = Initial moles - Moles of silver ions reducedMoles of silver ions remaining = 0.025 moles - 3.11 10^-5 moles = 0.02497 moles7. Calculate the amount of silver ions present in the sample after electrolysis:Amount of silver ions = Moles of silver ions remaining Molar mass of silver Ag Amount of silver ions = 0.02497 moles 107.87 g/mol = 2.69 gSo, the amount of silver ions present in the sample after the complete electrolysis of 50.0 mL of 0.500 M silver nitrate solution using coulometric titration at a constant current of 2.00 mA for 25.0 minutes is approximately 2.69 grams.