Answer:

Step-by-step explanation:
For this question, we have to start with the ionization equation for
, so:

With this in mind we can write the Ksp expression:
![Kps~=~[Ag^+]^2[CrO_4^-^2]](https://img.qammunity.org/2021/formulas/chemistry/college/hhub69nf03juovai5dmjulvww19ywdeoki.png)
Additionally, for every mole of
formed, 2 moles of
are formed. We can use "X" for the unknown concentration of each ion, so:
and
Now, we can plug the values into the Ksp expression:

Now we can solve for "X" :




I hope it helps!