To solve this problem, we will use the solubility product constant Ksp expression for the reaction between AgNO3 and NaCl, which is:AgNO3 aq + NaCl aq AgCl s + NaNO3 aq The Ksp expression for the formation of AgCl is:Ksp = [Ag+] [Cl-]We are given the initial concentration of Ag+ and NO3- ions, as well as the Ksp value for AgCl. We want to find out how much NaCl is needed to decrease the [Ag+] concentration by a factor of 10, which means the new [Ag+] will be 0.001 M.First, we need to find the initial concentration of Cl- ions in the solution. Since there is no NaCl initially, [Cl-] = 0 M.Now, let's denote the amount of NaCl added as 'x' moles. When NaCl is added to the solution, it will dissociate completely into Na+ and Cl- ions. So, the new concentration of Cl- ions will be x moles.At the new equilibrium, the Ksp expression will be:Ksp = 0.001 x We can now plug in the given Ksp value:1.8 x 10^-10 = 0.001 x Now, we can solve for x:x = 1.8 x 10^-10 / 0.001x = 1.8 x 10^-7 molesNow that we have the moles of NaCl needed, we can convert it to grams using the molar mass of NaCl 58.44 g/mol :mass of NaCl = 1.8 x 10^-7 moles 58.44 g/mol mass of NaCl = 1.05192 x 10^-5 gSo, approximately 1.05 x 10^-5 grams of NaCl is needed to be added to the solution in order to decrease the [Ag+] concentration by a factor of 10.