mols AgNO3 = grams/molar mass = approx 0.06 but that's only an estimate as are all of the other calculations I do here.
mols NaCl = M x L = approx 5E-4
.......NaCl + AgNO3 ==> AgCl + NaNO3
I......5E-4......0........0......0
add...........0.06..................
C.....-5E-4...-5E-4..................
E.about 0.about 0.06....5E-4
So the final solution consists of about 5E-4 mols solid AgCl in a 50 mL solution that has 0.0595 (that's 0.06mol - 0.0005mol) mols AgNO3 in excess and none of the NaCl.
(AgCl) = saturated solution
(AgNO3) = 0.0595 mols/0.05L = 1.19M
You may ask about the (Ag^+) from the AgCl and that can be calculated as follows to see if there is enough there to be considered.
AgCl ==> Ag^+ + Cl^-
Ksp = (Ag^+)(Cl^-) = 1.8E-10
(Ag^+)total = x from AgCl + 1.19M from excess AgNO3. (Cl^-) = x
Substitute into Ksp to obtain
(x+1.19)(x) = 1.8E-10
x = Cl^- = Ag^+ from AgCl = 1.5E-10 and that is insignificant when compared to 0.0595. That is, total (Ag^+) in the solution is 1.19+1.5E-10 = for all practical purposes 1.19M.
calculate the concentration of the silver ion remaining in the solution when 10g of solid AgNO3 is added to 50 ml of 1x10-2 M NaCl solution.
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1 answer