Answer to Question #286128 in Chemistry for eli

Question

The chloride ion content of a 250.0-mL seawater sample was titrated against 0.1102 M silver nitrate, requiring 13.56 mL to reach end point. What is the molarity of the chloride ion in the seawater sample?

Answer 1

Solution:

Cl⁻(aq) + AgNO₃(aq) → AgCl(s) + NO₃⁻(aq)

According to stoichiometry:

1 mol of chloride ion (Cl⁻) reacts with 1 mol of silver nitrate (AgNO₃)

Therefore,

Molarity of Cl⁻ × Volume of seawater = Molarity of AgNO₃ × Volume of AgNO₃

Molarity of Cl⁻ × (250.0 mL) = (0.1102 M) × (13.56 mL)

Molarity of Cl⁻ = (0.1102 M × 13.56 mL) / (250.0 mL) = 0.005977 M Cl⁻ = 0.0060 M Cl⁻

Answer: The molarity of the chloride ion in the seawater sample is 0.0060 M