Answer:
d. 0.25.
Explanation:
Hello,
In this case, since the equilibrium repression for the considered chemical reaction is:
[tex]Ksp=[Ag^+]^2[CrO_4^{2-}][/tex]
For a concentration of silver of 6.0x10⁻⁶ we need a concentration of chromate anion that makes the reaction quotient greater than the solubility product, thus, we write:
[tex][CrO_4^{2-}]=\frac{Ksp}{[Ag^+]^2} =\frac{9.0x10^{-12}}{(6.0x10^{-6})^2}[/tex]
[tex][CrO_4^{2-}]=0.25M[/tex]
It means that at concentrations of chromate anion of 0.25 M or more, the reaction quotient Q becomes greater than the solubility product which means that precipitation will begin occurring, therefore, answer is d. 0.25.
Best regards.
