Question

what is the solubility of AgI in a 0.274 molar solution of NaI. (Ksp of AgI =8.52×10^-17)

Answer 1

Answer:

1) Dissociation equation:

AgI (s) ⇌ Ag+ (aq) + I¯ (aq)

2) Ksp expression:

Ksp = [Ag+] [I¯]

3) Let us substitue into the Ksp expression:

8.52 x 10¯17 = (s) (0.274 + s)

4) The answer (after neglecting the +s in 0.274 + s:

[Ag+] = 3.11 x 10¯16 M

By the 1:1 stoichiometry between silver ion and AgI, the solubility of AgI in the solution is 3.11 x 10¯16 M

5) By the way, the solubility of AgI in pure water is this:

8.52 x 10¯17 = (s) (s)

x = 9.23 x 10-9 M

The solubility of the AgI has been depressed by a factor of a bit less than 30 million times.

Explanation: