The solubility of AgCl(s) with solubility product 1.6 10-10 in 0.1 M NaCl solution would be
- 1.25 10-5 M
- 1.6 10-9 M
- 1.6 10-11 M
- zero
Answer
(2) 1.6 10-9 M
Solution
Let the solubility of AgCl in 0.1 M NaCl be s.
[Ag+] = s ; [Cl-] = s + 0.1 0.1 (common ion)
Ksp = [Ag+][Cl-] ⇒ 1.6 10-10 = s 0.1
⇒ s = 1.6 10-9 M
Answer: (2)