[Zeratul]

I attempted the problem and feel that I did most of it correct, but I seem to keep making a mistake somewhere along the line.


30ml of .20M NaCl is added to 20ml of .2M AgNO3.  How many moles and grams of ppt are formed? and what is the final concentrations of each of the remaining ions in the final solution?

NaCl + AgNO3 -> NaNO3 + AgCl

.20M/1000*30ml = .006 moles NaCl
.20M/1000*20ml = .004 moles AgCl

NaNO3 is limited by the NO3 which gives an excess of .002 moles of Na
AgCl is limited by the Ag which gives an excess of .002 moles of Cl

This leaves us with .004 moles of both NaNO3 and AgCl, AgCl is the ppt

.004 moles AgCl
g=moles*mw=.004moles x 143(AgCl) = .57g AgCl

That answers the first part of the question, but for the remaining part, "What is the final concentration of each of the remaining ions in the final solution".
Would it just be the concentration of the Na, the NO3, and the excess Na and Cl? If so, then i would end up with 2 different numbers for the concentration of Na, one number for NaNO3 and one for the excess Na

I'm not really too good at explaining things, so if something is not clear please let me know, thanks.

[Borek]

.004moles x 143(AgCl) = .57g AgCl

Check molar mass of AgCl. Otherwise you are OK.

Would it just be the concentration of the Na, the NO3, and the excess Na and Cl? If so, then i would end up with 2 different numbers for the concentration of Na, one number for NaNO3 and one for the excess Na

No - you are asked about concentration of each indivdual ion - so no such thing as NaNO₃. Na⁺ was just a spectator, as well as NO₃^-.

[LQ43]

try setting up a table of moles of each ion to keep track

moles	Na+		NO3-		Ag+		Cl-
before
reacted
after

remember you mixed the volumes of two solutions - keep this in mind when you calculate the final concentration