Respuesta :
Answer: [tex]1.31\times 10^4moles/L[/tex]
Explanation:
Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as [tex]K_{sp}[/tex]
The equation for the ionization of the silver chromate is given as:
[tex]Ag_2CrO_4\leftrightharpoons 2Ag^++CrO_4^{2-}[/tex]
We are given:
Solubility of [tex]Ag_2CrO_4[/tex] = S mol/L
By stoichiometry of the reaction:
1 mole of [tex]Ag_2CrO_4[/tex] gives 2 moles of [tex]Ag^{+}[/tex] and 1 mole of [tex]CrO_4^{2-}[/tex].
When the solubility of [tex]Ag_2CrO_4[/tex] is S moles/liter, then the solubility of [tex]Ag^{+}[/tex] will be 2S moles\liter and solubility of [tex]CrO_4^{2-}[/tex] will be S moles/liter.
Expression for the equilibrium constant of [tex]Ag_2CrO_4[/tex] will be:
[tex]K_{sp}=[Ag^+]^2[CrO_4^{2-}][/tex]
[tex]9.00\times 10^{12}=[2s]^2[s]=4s^3[/tex]
[tex]s=1.31\times 10^4M[/tex]
Hence, the solubility of [tex]Ag_2CrO_4[/tex] is [tex]1.31\times 10^4moles/L[/tex]
