So would [Ag+]= Ksp/0.25Reaction: NaCl + AgNO3 --> AgCl + NaNO3
You know that NaCl is the limiting reagent, so use that to find the moles of AgCl, ie .25 mol. Therefore, moles of Cl^- in AgCl will also be 0.25 moles, therefore the conc of the Cl^-1 is 0.25 molL^-1.
sub the values in, and you get conc of Ag+
But what's the concentration for Ag+ ?We start with [Ag+] = 1.00 mol L-1 and [Cl-] = 0.250 mol L-1
The Ksp equilibrium is
AgCl(s) <---> Ag+(aq) + Cl-(aq)
and it lies far to the left given the small value of Ksp = [Ag+][Cl-] = 1.76 x 10-10
So, at equilibrium, we will have [Ag+] = 1.00 - x mol L-1 and [Cl-] = 0.250 - x mol L-1
and for the equation (1.00 - x)(0.250 - x) = 1.76 x 10-10 to be solvable, noting x is the extent to which the system moved left (and hence x < 0.250), we will have [Cl-] = being very close to 0, x very close to 0.250, and so [Ag+] very close to 0.750 mol L-1.
At equilibrium, [Ag+] = 0.750 mol L-1 (3 sig. fig.)
[Cl-] = Ksp / [Ag+] = 1.76 x 10-10 / 0.750 = 2.34 x 10-10 mol L-1 (3 sig. fig.)
x = 0.250 - [[Cl-] = 0.249 999 999 76... mol L-1
and so actually [Ag+] = 0.750 000 000 23... mol L-1
which fits above results.
The 0.25 mol of chloride ions must be split between the solid AgCl and the small amount in solution - they can't both be 0.25 mol.So "TheShy" answer is incorrect I presume?
Ahhh I see tyvmThe 0.25 mol of chloride ions must be split between the solid AgCl and the small amount in solution - they can't both be 0.25 mol.
As my answer shows, the [Cl-] in the solution is around 2.34 x 10-10 mol in the 1 L of solution, and the n(AgCl) precipitated as solid is 0.249 999 999 76... mol, which add to 0.25 mol