Question

The solubility product of AgI at 25ºC is 1.0×10–16 mol2L–2. The solubility of AgI in 10–4N solution of KI at 25ºC is approximately (in mol L–1):

• A. 1.0×10–16
• B. 1.0×10–12
• C. 1.0×10–10
• D. 1.0×10–8

Answer 1

Answer: (B)

1.0×10–12

Answer 2

Using the solubility product expression for AgI:
Ksp = [Ag+][I-]
We know that
Ksp = 1.0 x 10-16 mol2 L-2
[I-] = 10-4 M
Now,
[Ag+] = $\frac {K_{sp}}{[I^-]}$
[Ag+] = $\frac {1.0 \times 10^{-16 }}{10^{-4}}$
[Ag+] = 1.0 x 10-12 M

So, the solubility of AgI in a 10⁻⁴ N solution of KI at 25ºC is approximately 1.0×10−12 M, which corresponds to option (B).