Answer: Molar solubility is [tex]9.1\times 10^{-9}moles/liter[/tex] and solubility in grams per liter is [tex]2.13\times 10^{-6}[/tex]
Explanation: The equation for the reaction will be as follows:
[tex]AgI\leftrightharpoons Ag^++I^-[/tex]
1 mole of [tex]AgI[/tex] gives 1 mole of [tex]Ag^{+}[/tex] and 1 mole of [tex]I^{-}[/tex].
Thus if solubility of [tex]AgI[/tex] is s moles/liter, solubility of [tex]Ag^{+}[/tex] is s moles\liter and solubility of [tex]I^{-}[/tex] is s moles/liter.
Therefore,
[tex]K_sp=[Ag^+][I^-][/tex]
[tex]8.3\times 10^{-17}=[s][s][/tex]
[tex]s^2=8.3\times 10^{-17}[/tex]
[tex]s=9.1\times 10^{-9}moles/liter[/tex]
Solubility in grams/liter=[tex]\text{solubility in moles/liter}\times Molar mass[/tex]
Solubility in grams/liter=[tex]9.1\times 10^{-9}moles/liter\times 234.77g/mol=2.13\times 10^{-6}grams/liter[/tex]
