Question

'24.6 mL of 0.488 M NaCl is added to 24.6 mL of 0.312 M AgNO3. How many moles of AgCl would precipitate? What would be the concentrations of each of the ions in the reaction mixture after the reaction? Ion Concentration (M) Ag+ M NO3- M Na+ M Cl- M'

Answer 1

millimols NaCl = mL x M = 24.6 x 0.488 = estimated 12
mmols AgNO3 = estd 7.7
You need to redo these more accurately as well as all of the other numbers below. How advanced is this class? Basically, you have this but the final results depend upon just how complicated you want to make it. Remember these numbers are estimates.
..........NaCl + AgNO3 ==> AgCl + NaNO3
I.........12......7.7.......0........0
C........-7.7....-7.7......7.7.....
E.........4.3.....0?........7.7(solid)..

So Na^+ is a spectator ion. It is 12 initially and 12 at the end. M = mmols/total mL where total mL = 24.6+24.6 = ?
Cl was 12 and reduced by 7.7 = 4.3 left. M = mmols/mL
NO3^- is a spectator ion. 7.7 initially and 7.7 at the end. M = mmols/mL.
The (Ag^+) depends upon how advanced this calss is. If low level then (Ag^+) = 0 since all of it was pptd.
If the class is somewhat advanced the Ag^+ is what is left due to the solubility of AgCl in an excess of Cl^-
That is Ksp = (Ag^+)(Cl^-). Plug in Ksp from tables in your text/notes, plug in Cl from above and solve for Ag^+. Post your work if you get stuck.