Final answer:
The concentration of chloride ions in the final solution, after mixing 50.0 mL of 0.245M NH₄Cl with 75.5 mL of 0.165M FeCl₃, is 0.3955 M.
Step-by-step explanation:
In an experiment, 50.0 mL of a 0.245M NH₄Cl solution is added to 75.5 mL of a 0.165M FeCl₃ solution. To calculate the concentration of chloride ions in the final solution, we need to consider the total moles of chloride (Cl-) ions coming from both solutions and the total volume of the mixture.
NH₄Cl dissociates into NH₄+ and Cl- ions, contributing one Cl- per molecule. FeCl₃ dissociates into Fe3+ and 3 Cl- ions, contributing three Cl- ions per molecule. Therefore, the number of moles of chloride from NH₄Cl is 50.0 mL × 0.245 mol/L = 0.01225 mol, and from FeCl₃, it is 75.5 mL × 0.165 mol/L × 3 = 0.0373875 mol. The total moles of Cl- is the sum of these two figures: 0.0496375 mol.
The total volume of the mixed solution is 50.0 mL + 75.5 mL = 125.5 mL = 0.1255 L. To find the molarity of chloride ions in the final mixture:
Molarity (M) = Total moles of solute / Total volume of solution = 0.0496375 mol / 0.1255 L = 0.3955 M.