Answer:
[tex]6.5X10^-^5~\frac{mol}{L}[/tex]
Explanation:
For this question, we have to start with the ionization equation for [tex]Ag_2CrO_4_(_s_)[/tex], so:
[tex]Ag_2CrO_4_(_s_)~<->~2Ag^+~_(_a_q_)~+~CrO_4^-^2_(_a_q_)[/tex]
With this in mind we can write the Ksp expression:
[tex]Kps~=~[Ag^+]^2[CrO_4^-^2][/tex]
Additionally, for every mole of [tex]CrO_4^-^2[/tex] formed, 2 moles of [tex]Ag^+[/tex] are formed. We can use "X" for the unknown concentration of each ion, so:
[tex][CrO_4^-^2]~=~X[/tex] and [tex][Ag^+]~=~2X[/tex]
Now, we can plug the values into the Ksp expression:
[tex]1.12x10^-^1^2~=~(2X)^2(X)[/tex]
Now we can solve for "X" :
[tex]1.12x10^-^1^2~=~4X^3[/tex]
[tex]X^3=\frac{1.12X10^-^1^2~}{4}[/tex]
[tex]X=(\frac{1.12X10^-^1^2~}{4})^(^1^/^3^)[/tex]
[tex]X=6.5X10^-^5~\frac{mol}{L}[/tex]
I hope it helps!
