Determine the number of mL of .494 M ammonium sulfite solution that are required to prepare 85.4 mL of a solution that is .179 M in ammonium ion?
i am confused on the set up... i went .0854 L x .179 M which = .0152 and divided by .494... and i got an answer that was on the worksheet but it wasnt correct...can somebody tell me how to set it up correctly it would be muchly appriciated