Doubt on the process to determine the amount of precipitate of a salt, product of a reaction between two solutions
I am attempting to determine whether there is going to be the formation of a $\ce{AgCl}$ salt precipitate if the following two solutions were to be added together:
1. $50mL$ of $\ce{NaCl}$, $0.04M$,
2. $50mL$ of $\ce{AgNO3}$, $0.05M$.
Knowing the $K_{sp}$ of $\ce{AgCl}$ is $1.6 \times 10^{-10}$.
My thought process is the following:
* Determine the reaction that takes place: $\ce{NaCl + AgNO3 -> AgCl + NaNO3}$ (is this an equilibrium? How could I certainly say?),
* Determine the amount of $\ce{AgCl}$ that forms at the end of the reaction - noticing that $\ce{NaCl}$ limits the reaction and that, thus, there are going to be at most $n$ moles of $\ce{Cl-}$. 870
I would say the amount of $\ce{AgCl}$ that forms, at most, is $50mL \cdot 0.04M = 0.002mol$,
And at this point I'm stuck. If the solubility product, $K_{sp}$ of $\ce{AgCl}$, expresses the amount of ions that can be dissolved in the solution at the equilibrium, I would be able to say that, since the amount of silver chloride that forms ($0.02M$, in $100mL$) is greater than the value of $s$:
$$s = \sqrt{K_{sp}} = 1.265 \times 10^{-5}M = \ce{[Ag+]{eq} = [Cl-]{eq} = [AgCl]_{eq}}$$
from $K_{sp} = s \cdot s$, since there are going to be equal amounts of $\ce{Ag+}$ and $\ce{Cl-}$ ions dissolved in the solution, then I would assume that some $\ce{AgCl}$ would precipitate out: $\ce{AgCl}_{precipitate} = 0.002mol - 1.265 \times 10^{-5}mol = 0,0019987$, or $(107.9u + 35.45u) \cdot 0,0019987 = 0.286g$.
Are my assumptions on the solubility product correct?
Also, if I wanted to calculate the molar concentration of the ionic species ($\ce{Ag+}$ and $\ce{Cl-}$) in the saturated solution... wouldn't I take the previously found value of $s$ and divide it by the total volume of the solution ($100mL$)? That would be $1.265 \times 10^{-5}M$ for both $\ce{Ag+}$ and $\ce{Cl-}$?
Thank you! 870