The reaction that will occur here is a precipitation reaction with the net ionic equation
Ag^+(aq)+Cl-(aq) rarr AgCl (s)
The barium and nitrate ions play no part in it, and can be ignored for simplicity.
We must calculate the quantity of AgNO_3 and of BaCl_2 though, so that we can determine the amount of each ion present in the original solutions.
moles AgNO_3 = (1.5 "mol"/L xx 5L) = 7.5 "mol"
This means the quantity of Ag^+ ion present is also 7.5 mol.
moles BaCl_2 = (2.0 "mol"/Lxx5 L) = 10 "mol"
Note however, that each mole of BaCl_2 produces two moles of Cl^- when it dissolves.
So, the quantity of Cl^- present is 20 mol.
As a first approximation, 7.5 mol Ag^+ will consume 7.5 mol of Cl^- leaving 12.5 mol of Cl^- still in the solution.
If we stop at this point, it seems the concentration of Ag^+ would be zero (as it has all been used up in the reaction).
However, if your question wishes you to consider the equilibrium nature of this solubility problem, we must go on...
The K_(sp) value for AgCl is 1.6xx10^(-10)
Therefore [Ag^+][Cl^-] = 1.6 xx 10^(-10)
From the above, we see we have 12.5 mol of Cl^- in a total volume of 10 L ( the mixture of the two 5L solutions), meaning that the concentration of [Cl^-] is
(12.5 "mol")/(10 L)= 1.25 "mol"/L
Therefore, from the K_(sp) relation.
[Ag^+] = (1.6xx10^(-10))/1.25 = 1.28xx10^(-10) "mol"/L
