The mass percent of Cl− in a seawater sample is determined by titrating 25.00 mL of seawater with AgNO3 solution, causing a precipitation reaction....

Question

An indicator is used to detect the endpoint, which occurs when a free Ag+ ion is present in a solution after all the Cl− has reacted. If 57.50 mL of 0.2036 M AgNO3 is required to reach the endpoint, what is the mass percent of Cl− in the seawater (d of seawater = 1.024 g/mL)?

Answer 1

Let's look at the reaction taking place:

AgNO₃(aq) + Cl^-(aq) ==> AgCl(s) + NO₃^-(aq)

moles AgNO₃ needed to reach endpoint = 57.50 ml x 1 L/1000 ml x 0.2036 mol/L = 0.011707 moles

moles Cl^- present in the 25.00 ml = 0.011707 mol AgNO₃ x 1 mol Cl^- / mol AgNO₃ = 0.011707 mol Cl^-

mass of Cl^- present in 25.00 ml = 0.011707 mol Cl^- x 35.45 g/mol = 0.4150 g Cl^-

Total mass of 25.00 ml = 25.00 ml x 1.024 g/ml = 25.6 g

Mass % Cl^- = 0.4150 g / 25.6 g (x100%) = 1.621 %