Question

AgCl <==> Ag^+ + Cl^-
NaCl ==> Na^+ + Cl^-

Ksp = (Ag^+)(Cl^-) = 1.8 x 10^-10
Let S = solubility of AgCl, then
(Ag^+) = S
(Cl^-) = S+0.01
Solve for S.
Note: A similar problem to this post (0.1 M NaCl instead of 0.01 M NaCl) was on a couple of days ago; the one who posted said that the answer came back incorrect. Check my work. Check myu thinking.


Should the continuing work look like this?
S = 1.8 x 10^-10 + 1
= 1.00?

Please help I am very confused

Answer 1

I don't understand how you can be confused.
(S)(S + 0.01) = 1.8 x 10^-10
You can solve the quadratic, OR you can make the simplifying assumption that S + 0.01 is about 0.01 (that is, that S is very small in comparison so 0.01 is not really different than S + 0.01).
If we make the assumption, then
S(0.01) = 1.8 x 10-^-10
and S = 1.8 x 10^-8 M = (AgCl).