Notifications
Clear all
Equilibrium
1
Posts
2
Users
1
Likes
302
Views
0
24/10/2021 11:41 am
Topic starter
The solubility (in mol L-1) of AgCl (Ksp = 1.0 × 10-10) in a 0.1 M KCl solution will be
(a) 1.0 × 10-9
(b) 1.0 × 10-10
(c) 1.0 × 10-5
(b) 1.0 × 10-11
1 Answer
1
24/10/2021 11:46 am
Correct answer: (a) 1.0 × 10-9
Explanation:
Let solubility of AgCl = x mole/L
AgCl ⇌ Ag+ + Cl-
i.e., Ksp(Agcl) = \(x \times x\)
KCl → K+ + Cl-
0.1
[Cl-] from KCl = 0.1 m
Total [Cl-] in solution = x + 0.1
Ksp(AgCl) = [Ag+] [Cl-] = x (x + 0.1)
1.0 × 10-10 = x(x + 0.1)
1.0 × 10-10 = x2 + 0.1x
1.0 × 10-10 = 0.1x (as x2 << 1)
x = 1.0 × 10-9 mol/L