Please help me prepare for finals?

Question

What's the concentration of the sulfate ion SO4-2, in a .50 M solution of Al2(SO4)3? Explain in a step by step process in a easy way.

Answer 1

First, you should write down the ionization eq. and balance it:


Al2(SO4)3--------->2 Al^3+ + 3 SO4 ^2-

This can be read as: for every mole of Aluminum sulfate, you have 2 moles of Ion aluminum, and three moles of ion sulfate. Or for every molecule of substance you have two atoms of Al2-, and three ions of SO4.

The concentration 0.5 M means that for every liter of solution you have half a mole of Sulfate aluminum. You can see from the ionization that from every mole of sulfate aluminum you get 3 moles of ion sulfate.

Soy you must write down the relation, suppose you are working with one liter of solution, it means you have 0.5 mol of Al2(SO4)3:

1mol of Al2(SO4)3-------------3 mol of SO4 ^2-
0.5 mol of Al2(SO4)3----------x mol of SO4 ^2-

and x= (3 x 0.5 )/1

that's it. Since I CHOSE to work with one liter from the beginning, x is the molar concentration of ion SO4 ^2-

Answer 2

The 3 at the end tells you that the concentration of sulphate will be 3 x the concentration of the compound as a whole: 1.5M.