To determine the amount of chloride ions Cl- present in the 75 mL sample, you need to perform the potentiometric titration and record the volume of the 0.1 M silver nitrate AgNO3 solution used to reach the endpoint. The endpoint is the point at which the potential difference voltage between the reference electrode and the working electrode changes significantly, indicating that all chloride ions have reacted with the silver ions.Once you have the volume of the 0.1 M AgNO3 solution used, you can calculate the moles of chloride ions using the stoichiometry of the reaction:AgNO3 + NaCl AgCl + NaNO3In this reaction, one mole of silver nitrate reacts with one mole of sodium chloride, producing one mole of silver chloride and one mole of sodium nitrate. Since the stoichiometry is 1:1, the moles of chloride ions will be equal to the moles of silver nitrate used.To calculate the moles of AgNO3 used, use the formula:moles = molarity volumeAssuming you have the volume of AgNO3 used in liters convert mL to L by dividing by 1000 , you can calculate the moles of Cl-:moles of Cl- = 0.1 M volume of AgNO3 in L Once you have the moles of Cl-, you can calculate the concentration of Cl- in the original 75 mL sample:concentration of Cl- = moles of Cl- / 0.075 L Finally, to find the amount of chloride ions in the 75 mL sample, you can use the formula:amount of Cl- in grams = moles of Cl- molar mass of Cl- The molar mass of Cl- is approximately 35.45 g/mol. Multiply the moles of Cl- by the molar mass to get the amount of chloride ions in grams.