Question

Suppose that a 10-mL sample of a solution is to be tested for Cl− ion by addition of 1 drop (0.2 mL) of 0.10 M AgNO3.

What is the minimum number of grams of Cl−that must be present in order for AgCl(s) to form?

Answer 1

Ksp AgCl = 1.77×10–10

AgCl(s) <-> Ag+ + Cl-

Ksp = [Ag+][Cl-]

1.77*10^-10 = [Ag+][Cl-]

Total V = 10 mL + 0.2 mL = 10.2 mL

[Ag+] in solution = 10*0.1 / 10.2 = 0.098039 M of Ag+

then

if Q > Ksp, there will be precipitate

1.77*10^-10 = [Ag+][Cl-]

1.77*10^-10 = (0.098039)(Cl-)

[Cl-]= (1.77*10^-10)/0.098039 =1.80540*10^-9

V = 10.2 mL = 10.2*10^-3 L

mol = MV = (1.80540*10^-9)(10.2*10^-3) = 1.84150*10^-11 mol of Cl-

mass = mol*MW = (1.84150*10^-11)(35.5) = 6.537*10^-10 g of Cl required