Question

0.125 L of 0.02157 M AgNO₃ is added to a 50.00 ml sample that contains the chloride ion (Cl-). A PPT forms which has a mass of 211.6 mg. What was the concentration of the chloride ion in that original 50.0 ml sample? [Cl-] = 0.0295M

Answer 1

The concentration of the chloride ion in the original 50.0 ml sample is 0.002696 M.

Step-by-step explanation:

To find the concentration of the chloride ion in the original 50.0 ml sample, we can use the stoichiometry of the reaction. The balanced equation is: AgNO3(aq) + Cl-(aq) -> AgCl(s) + NO3-(aq). From the equation, we can see that one mole of AgNO3 reacts with one mole of Cl-. The number of moles of AgNO3 is given by (0.125 L) x (0.02157 M) = 0.002696 mol. Since the stoichiometry is one-to-one, the concentration of Cl- in the original sample is also 0.002696 M.