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 $9.1\times 10^{-9}moles/liter$  and solubility in grams per liter is $2.13\times 10^{-6}$

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,  

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

$8.3\times 10^{-17}=[s][s]$

$s^2=8.3\times 10^{-17}$

$s=9.1\times 10^{-9}moles/liter$

Solubility in grams/liter=$\text{solubility in moles/liter}\times Molar mass$

Solubility in grams/liter=$9.1\times 10^{-9}moles/liter\times 234.77g/mol=2.13\times 10^{-6}grams/liter$