Solubility equilbrium problem

Question

A weighed amount of solid AgCl is ground up and thoroughly shaken with a liter of water until saturation. When the saturated solution is filtered, the residual solid AgCl is dried and weighed. If the loss in mass of AgCl is found to be 0.0019 g, what will be the Ksp of AgCl at the same temperature?

If a saturated solution is why is there residual solid ?

Answer 1

The loss of mass, 0.0019 g is that amount of AgCl that went into solution, or it is the "solubility" of AgCl.  This is in a volume of 1 liter, so we can convert to moles and then to molar (M).  Thus ...

1.9x10^-3 g AgCl x 1 mole AgCl/143.3 g = 1.33x10^-5 moles AgCl = 1.33x10^-5 M since it is in 1 liter.

AgCl ==> Ag⁺ + Cl^-

Ksp = [Ag⁺][Cl^-] = (1.33x10^-5)²

Ksp = 1.77x10^-10

 

To answer the second part of your question, the residual solid is that AgCl that did NOT dissolve.  A saturated solution only means that the liter of solution holds as much AgCl as possible, the rest will settle as a solid.