-the solubility of AgCl in water at 25c is 1.274*10^2, on the assumption that the Debye–Hückel limiting law apply.

a-calculate ∆G for the process of AgCl gives Ag+ + Cl-

b- the solubility for the AgCl in 0.005 K2SO4 solution.

I really appreciate the help.

for the solubility of an ionic compound where the cation to anion ratio is 1 to 1 the formula is

Ksp = x

^{2 }where x is the molar solbility∆G = -RT ln(Ksp)

for b you have to see if any of the ions in K2SO4 interact with either Ag+ or Cl

^{-}if they do not then nothing changes if they do, you have to take that equilibrium into account. In other words will either KCl or Ag2SO4 be formed?