Question

Calculate the molar solubility and the solubility in g / l of agi at 25°c. the ksp of agi is 8.3 × 10−17.

Answer 1

Answer is: solubility of silver iodide is 9.11·10⁻⁹ M..
silver iodide in water: AgI(s) → Ag⁺ + I⁻.
Ksp = 8.3·10⁻¹⁷.
[Ag⁺] = [I⁻] = x.
Ksp = [Ag⁺] · [I⁻].
8.3·10⁻¹⁷ = x².
x= √8.3·10⁻¹⁷.
x = [Ag⁺] = [I⁻] = 9.11·10⁻⁹ mol/L.

Answer 2

Answer: Molar solubility is
`9.1* 10^(-9)moles/liter` and solubility in grams per liter is
`2.13* 10^(-6)`

Explanation: The equation for the reaction will be as follows:

`AgI\leftrightharpoons Ag^++I^-`

1 mole of
`AgI` gives 1 mole of
`Ag^(+)` and 1 mole of
`I^(-)`.

Thus if solubility of
`AgI` is s moles/liter, solubility of
`Ag^(+)` is s moles\liter and solubility of
`I^(-)` is s moles/liter.

Therefore,

`K_sp=[Ag^+][I^-]`

`8.3* 10^(-17)=[s][s]`

`s^2=8.3* 10^(-17)`

`s=9.1* 10^(-9)moles/liter`

Solubility in grams/liter=
`\text{solubility in moles/liter}* Molar mass`

Solubility in grams/liter=
`9.1* 10^(-9)moles/liter* 234.77g/mol=2.13* 10^(-6)grams/liter`