Question

Calculate the molar solubility of AgI in the following. Ksp for AgI is 8.30 ×10−17.(a) pure water×10MEnter your answer in scientific notation.(b) 0.0400 M NaI×10MEnter your answer in scientific notation.

Answer 1

(a) solubility in pure water =9.11 x 10⁻⁹ (M)

(b) solubility in 0.08 (M) NaI = 1.04 x 10⁻¹⁵ (M)

Step-by-step explanation:

(1)

AgI (s) + H₂O ⇄ Ag⁺(aq) + I⁻ (aq)

s s

Solubility product of AgI

Ksp = [Ag⁺] [I⁻]

⇒Ksp = s²

⇒s
`S=√(Ksp)\\\\=\sqrt{8.3X10^(-17) } \\\\=9.11 X10x^(-9)(M)`

solubility in pure water

(2)

Ksp = [Ag⁺] [I⁻]

⇒ 8.3 X 10⁻¹⁷ = S x (0.08+S)

and assuming S is small relative to 0.08, due to common ion effect (I⁻) solubility of salt decreases

S + 0.08 = 0.08

⇒S X 0.08 = 8.3 X 10⁻¹⁷

⇒S = 1.04 X 10⁻¹⁵ (M) solubility in 0.08 (M) NaI