A student found that the sulfate ion concentration in a solution of Al2(SO4)3 was 0.22 M. What was the concentration of Al2(SO4)3 in the solution?

The concentration of #Al_2(SO_4)_3# in solution will be #"0.073 M"#.

So, you're dealing with an aluminium sulfate solution.

![https://crispassinato.wordpress.com/2008/05/27/estrutura-do-al2so43-respondendo-ao-erivelton/]()

In aqueous solution, the compound will dissociate into aluminium cations and sulfate anions.

![http://www.chemicalbook.com/ProductChemicalPropertiesCB8435192_EN.htm]()

Write the equation for aluminium sulfate's dissociation (all the species are in aqueous solution)

#Al_(color(blue)(2))(SO_4)_(color(red)(3)) rightleftharpoons color(blue)(2)Al^(3+) + color(red)(3)SO_4^(2-)#

Notice the 1 mole of aluminium sulfate produces #color(blue)(2)# moles of aluminium cations and #color(red)(3)# moles of sulfate anions.

This means that, for a given volume of solution, the concentration of the sulfate anions will be three times larger than the concentration of aluminium sulfate.

Think of it like this - same volume, three times the number of moles in favor of the sulfate anion #-># 3 times the concentration.

Thus,

#C_(Al_2(SO_4)_3) = 1/3 * C_(SO_4^(2-)) = 1/3 * "0.22 M" = color(red)("0.073 M")#