Question #09240

1 Answer
Dec 12, 2015

Answer:

#"20 mEq Mg"^(2+)#

Explanation:

An equivalent is defined as the product between the number of moles of a substance and its valence.

When it comes to ions, you can think of an equivalent as being a measure of that ion's capacity to combine with opposite-charge ions.

In your case, magnesium chloride, #"MgCl"_2#, is an ionic compound that dissociates in aqueous solution to give magnesium cations, #"Mg"^(2+)#, and chloride anions, #"Cl"^(-)#

#"MgCl"_text(2(aq]) -> "Mg"_text((aq])^(2+) + color(red)(2)"Cl"_text((aq])^(-)#

Notice that every mole of magnesium chloride produces #1# mole of magnesium cations and #color(red)(2)# moles of chloride anions.

This tells you that in a magnesium chloride solution, every magnesium cation can combine with #color(red)(2)# chloride anions. As a result, for every mole of magnesium chloride, you will have two equivalents of magnesium cations.

The concentration of magnesium cations will be equal to that of the solution, since it forms in a #1:1# mole ratio from magnesium chloride.

This means that every liter of this solution will contain

#10 color(red)(cancel(color(black)("mM"))) * "2 mEq Mg"^(2+)/(1color(red)(cancel(color(black)("mM")))) = color(green)("20 mEq Mg"^(2+)#

Here #"mEq"# represent milliequivalents, the #1/1000"th"# part of an equivalent.

This is what the product of the number of moles and the valence of the ion is all about.

Now, in any solution, the number of equivalents of cations must be equal to the number of equivalents of anions.

In this case, the solution will contain #"20 mM"# of chloride anions, since we've established that every mole of magnesium chloride produces #color(red)(2)# moles of chloride anions.

Chlorine's valence is equal to #1#, since it's located in group 17 and forms #1-# anions. This means that you will have

#20 color(red)(cancel(color(black)("mM"))) * "1 mEq Cl"^(-)/(1color(red)(cancel(color(black)("mM")))) = color(green)("20 mEq Cl"^(-)#

Here you get one equivalent per mole of chloride anions, but you have twice as many present in solution.