Solubilty prob?

Question

How many mL of a 0.001 M chloride solution must be added to a 100mL solution of 7.2 x 10^ -5 M Ag+ solution for AgCl to precipitate? Ksp (AgCl)= 1.8 x 10^ -10.

how do you get .25mL

Answer 1

Ksp = 1.8x10^-10 = [Ag+][Cl-]

[Ag+] = 7.2x10^-5

1.8x10^-10 = (7.2x10^-5) * [Cl-]

[Cl-] = 0.0000025 M (2.5x10^-6 M)

100 mL solution requires 2.5x10^-7 moles Cl- (2.5x10^-7/0.100 L = 2.5x10^-6 M)

liters x M = moles

liters x 0.001 M = 2.5x10^-7 moles

liters = 0.00025 (0.25 mL)