# What molar quantity of magnesium ion is present in a 1000*mL volume of 0.50*mol*L^-1 MgSO_4(aq)?

Oct 25, 2016

You have $0.5$ $\text{mol}$ of $M {g}^{2 +} \left(a q\right)$.

#### Explanation:

You specify a $0.5 \cdot m o l \cdot {L}^{-} 1$ solution of $M g S {O}_{4} \left(a q\right)$.

Since $\text{Concentration}$ $=$ $\text{Moles of solute"/"Volume of solution}$,

$\text{Moles of solute"="Concentration"xx"Volume of solution}$.

If we have a $1000 \cdot m L$ volume, then.......

$\text{Moles of solute}$

$0.5 \cdot m o l \cdot \cancel{{L}^{-} 1} \times 1000 \cdot \cancel{m L} \times {10}^{-} 3 \cancel{L} \cdot \cancel{m {L}^{-} 1} = 0.5 \cdot m o l \text{ with respect to } M {g}^{2 +} \left(a q\right)$.

Dec 6, 2016

Here's what I got.

#### Explanation:

It's a good idea to keep in mind that equivalents have various definitions depending on the type of reaction you're working with.

In this context, the number of equivalents of a given ion is given by the number of moles of opposite unitary charge ions that can combine with $1$ mole of the given ion.

For example, ${\text{Mg}}^{2 +}$ has a charge of $2 +$, which means that it can combine with $2$ ions that carry a $1 -$ charge.

So if $1$ mole of ${\text{Mg}}^{2 +}$ can combine with $2$ moles of $1 -$ ions, it follows that you get $2$ equivalents of ${\text{Mg}}^{2 +}$ for every $1$ mole of ${\text{Mg}}^{2 +}$ added to the solution.

In this particular case, your solution will contain

1000 color(red)(cancel(color(black)("mL solution"))) * (0.5color(red)(cancel(color(black)("moles MgSO"_4))))/(1000color(red)(cancel(color(black)("mL solution")))) * (1 color(red)(cancel(color(black)("mole Mg"^(2+)))))/(1color(red)(cancel(color(black)("mole MgSO"_4)))) * "2 eq Mg"^(2+)/(1color(red)(cancel(color(black)("mole Mg"^(2+))

$= \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{1 eq Mg}}^{2 +}}}}$

Remember, this depends on what exactly you're about to do with your solution. If you're going to mix this solution with a solution that contains $1 -$ ions, then you can talk about it having $\text{2 eq}$ of ${\text{Mg}}^{2 +}$.

In and of itself, the solution simply contains $0.5$ moles of ${\text{Mg}}^{2 +}$ and $0.5$ moles of ${\text{SO}}_{4}^{2 -}$.

The equivalents come into play when dealing with reactions. Equivalents are defined as an ion's capacity to combine with opposite unitary charge ions, so you need an actual reaction in order for them to make sense at all.