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#
