lol
AgNO3 + xcess KCL <-----> KNO3 (ppt) + AgCl
Given: 33.5 mL AgNO3, 0.680g KNO3
MW AgNO3 = 170g/mol, MW KNO3 = 101g/mol
(I usually carry sig figs through calculations)
First multiply through using the 1:1 mol rat to get the number of moles of reactant used. All molar and ionic quantities are 1:1, so that simplifies things:
0.680g KNO3 * (mol KNO3/ 101g KNO3) * (mol AgNO3/ mol KNO3) = 6.71x10^-3 mol of reactant AgNO3 used.
Then, using the definition M = mol/L:
M = mol/L = (6.71x10^-3)/(0.335 L) = 0.2M AgNO3 sol'n
Now, since the ionic compound Ag+ NO3- is in a 1:1 ratio, the molarity of Ag+ ions is also 0.2M.
It's been been a while since I've done this, so you might wanna verify the work. It seems kinda basic, so hopefully thier isn't something I overlooked. It's not a K(sp) problem, is it? It seems to be a basic 1:1 ppt ionicy-type problem.