CORRECT


(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


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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.

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