# What is the concentration of the ions in 3 L of 0.50 M Al_2(SO_4)_3?

Jul 6, 2016

["Al"^(3+)] = "1.0 M"
["SO"_4^(2-)] = "1.5 M"

#### Explanation:

The first thing to notice here is that the problem asks for the concentration of the ions present in a $\text{0.50 M}$ solution of aluminium sulfate, "Al"_2("SO"_4)_3, so you don't need to worry about the volume of the solution.

All you need to focus on here is the dissociation of the salt in aqueous solution and the given concentration.

So, aluminium sulfate is a soluble ionic compound that dissociates completely in aqueous solution to form aluminium cations, ${\text{Al}}^{3 +}$, and sulfate anions, ${\text{SO}}_{4}^{2 -}$.

${\text{Al" _ color(red)(2)("SO"_ 4)_ (color(blue)(3)(aq)) -> color(red)(2)"Al"_ ((aq))^(3+) + color(blue)(3)"SO}}_{4 \left(a q\right)}^{3 -}$

As you can see, every mole of aluminium sulfate that dissociates in solution produces

• two moles of aluminium cations, $\textcolor{red}{2} \times {\text{Al}}^{3 +}$
• three moles of sulfate anions, $\textcolor{b l u e}{3} \times {\text{SO}}_{4}^{2 -}$

This means that for a given concentration of the salt, the solution will contain

["Al"^(3+)] = color(red)(2) xx ["Al"_2("SO"_4)_3] = color(red)(2) xx "0.50 M" = "1.0 M"

["SO"_4^(2-)] = color(blue)(3) xx ["Al"_2("SO"_4)_3] = color(blue)(3) xx "0.50 M" = "1.5 M"

Mind you, these are the concentration of the ions in a $\text{0.50 M}$ aluminium sulfate solution regardless of its volume!

In other words, if you dissolve enough aluminium sulfate in water to form a solution that is $\text{1.0 M}$ in aluminium cations and $\text{1.5 M}$ in sulfate anions, you will have these concentrations regardless of what volume of said solution you use.

For the given concentrations, the number of moles of ions depends on the volume of the solution.