Calculate the molar solubility of AgCl in 0.05M NaCl. (Ksp = 1.77 x 10-10)

Answer: 3.54 x 10^{-9}

Explanation:

For this Case we have keep in mind the effect of the Common Ion, which is Cl.

You can find the amount of silver ions at equilibrium from Ksp since You already know the chloride ion concentration. Keep in mind that only a few amount of silver chloride will be dissolved. This means the chloride concentration will stay in 0.05 M, Then I can use Ksp to solve for the silver ion concentration. So you can use the following equation.

AgCl Ksp = Ksp/[Cl^-] = (1.77 x 10^{-10}) / (0.05) = 3.54 x 10^{-9}