(3.c) [Ag+] = 1.8 x 10-9 M
AgCl(s) | <---> | Ag+(aq) | + | Cl-(aq) | |
Initial (I) | 0 | 0.10 M | |||
Change(C) | +x | +x |
Equilibrium (E) | x | (0.10 M + x) |
where x = [Ag+]
The molar solubility of AgCl is equal to [Ag+] since they came from dissolving AgCl only. It makes the math easier if you assume that x is small compared to 0.10M Cl- from the NaCl. Remember that in problem 1 the [Ag+] was about 10-5 M. Hence that (0.10 + x) is approximately equal to 0.10.
Ksp = [Ag+][Cl-]
1.8 x 10-10 = (x)(0.10)
1.8 x 10-9 M = x
Note that x << 0.10 so your assumption was correct.