Hello, i've been going through some questions and i am stuck with this

What is the solubility product of silver chromate, Ag

_{2}CrO

_{4}, give that its solubility is 8.43x10

^{-5} mol dm

^{-3}.

My steps are,

K

_{sp}=

(Ag

^{+})

^{2}(CrO

_{4}^{2-}) =

(8.43x10

^{-5})

^{3} =

5.99 x 10

^{-13} mol

^{3}dm

^{-9}But the printed answer is 2.4x10

^{-12} mol

^{3}dm

^{-9}What did I do wrong? Please advise.