To determine the moles of Ag needed to decrease the concentration of Cl anion from 4×10^-5M to 10^-5M in a 100 ml solution, we need to consider the equilibrium expression for the dissolution of AgCl:

AgCl(s) ↔ Ag+(aq) + Cl-(aq)

The equilibrium constant for this reaction is known as the solubility product constant, Ksp. In this case, the Ksp for AgCl is given as 10^-10M at 25 degrees Celsius.

Let's break down the problem into steps:

1. Calculate the initial concentration of Ag+ and Cl- ions:
Since AgCl is a strong electrolyte, it fully dissociates into Ag+ and Cl- ions in solution. Therefore, the initial concentration of Ag+ and Cl- ions is equal to the initial concentration of Cl- ions, which is 4×10^-5M.

2. Calculate the final concentration of Cl- ions:
The final concentration of Cl- ions is given as 10^-5M.

3. Calculate the change in concentration of Cl- ions:
Δ[Cl-] = [Cl-]final - [Cl-]initial
Δ[Cl-] = (10^-5M) - (4×10^-5M)
Δ[Cl-] = -3×10^-5M

4. Use the stoichiometry of the reaction to determine the moles of AgCl that must dissolve:
From the balanced equation, we know that the stoichiometric ratio between Cl- ions and AgCl is 1:1. Therefore, the moles of AgCl that must dissolve to produce the change in concentration of Cl- ions is equal to the absolute value of the change in concentration of Cl- ions.

5. Calculate the molar mass of AgCl:
The molar mass of AgCl can be calculated using the atomic masses of silver (Ag) and chlorine (Cl). Ag has an atomic mass of 107.87 g/mol, and Cl has an atomic mass of 35.45 g/mol. Therefore, the molar mass of AgCl is 107.87 g/mol + 35.45 g/mol = 143.32 g/mol.

6. Convert the moles of AgCl to moles of Ag:
Since the stoichiometric ratio between AgCl and Ag is 1:1, the moles of AgCl that dissolve is equal to the moles of Ag that form.

7. Convert the moles of Ag to mass of Ag:
Using the molar mass of Ag (107.87 g/mol), calculate the mass of Ag formed by multiplying the moles of Ag by the molar mass.

By following these steps, you should be able to calculate the moles of Ag needed to decrease the concentration of Cl anion from 4×10^-5M to 10^-5M in a 100 ml solution.