$ K_{SP} of AgCl is 4.0 \times 10 ^ {-10} at 298 k.solubility of AgCl is 0.04 M CaCl_2 $ होगा
यदि x, 0.04 M $ CaCl_2 $ में AgCl की विलेयता है, तो
$ [Ag^+] = x mol L ^ {-1} $
$ [Cl^1] = (0.04 \times 2 ) + x = 0.08 M $
$ 0.08 (x) = 4 \times 10 ^ {-10} $
$ x = 5.0 \times 10 ^ {-9} M $