(1.b) 1.3 x 10-5
Ksp = [Ag+] ][Cl-] and since [Ag+] = [Cl-] = x when they are produced by AgCl dissolving
1.8 x 10-10 = x2
x = 1.3 x 10-5
x = [Ag+] = [Cl-] = 1.3 x 10-5
| AgCl(s) | <---> | Ag+ (aq) | + | Cl- (aq) | |
| Initial (I) | 0 | 0 | |||
| Change(C) | +x | +x |
| Equilibrium (E) | x | x | |||
Reminder: The solid AgCl must be present for equilibrium to be established and its activity is constant and equals 1. If no AgCl(s) is present the activity is 0 and the equilibrium does not apply.