# Question 49dfd

Feb 19, 2016

["Cl"^(-)] = 5.84 * 10^(-2)"M"

#### Explanation:

You're dealing with thee ionic compounds that are soluble in aqueous solution, which means that they exist as cations and anions in solution.

Simply put, these compounds will dissociate completely to form cations and anions in aqueous solution.

${\text{NaCl"_text((aq]) -> "Na"_text((aq])^(+) + "Cl}}_{\textrm{\left(a q\right]}}^{-}$

${\text{CaCl"_(color(red)(2)(aq]) -> "Ca"_text((aq])^(2+) + color(red)(2)"Cl}}_{\textrm{\left(a q\right]}}^{-}$

${\text{Li"_2"SO"_text(4(aq]) -> 2"Li"_text((aq])^(+) + "SO}}_{\textrm{4 \left(a q\right]}}^{2 -}$

Now, you're interested in find the molarity of the chloride anions, ${\text{Cl}}^{-}$, which means that you can ignore the dissociation of lithium sulfate, ${\text{Li"_2"SO}}_{4}$, completely.

Now take a look at the dissociation of sodium chloride, $\text{NaCl}$, and calcium chloride,${\text{CaCl}}_{2}$.

Notice that every mole of sodium chloride added to the solution will dissociate to form one mole of sodium cations and one mole of chloride anions, ${\text{Cl}}^{-}$.

This means that you'll have

$\left[\text{Cl"^(-)] = 1 xx ["NaCl}\right]$

In your case, this will get you

["Cl"^(-)]_1 = "15.6 mM"

On the other hand, every mole of calcium chloride added to the solution will dissociate to form one mole of calcium cations and two moles of chloride anions.

This means that you'll have

$\left[{\text{Cl"^(-)] = color(red)(2) xx ["CaCl}}_{2}\right]$

This will get you

["Cl"^(-)]_2 = color(red)(2) xx "21.4 mM" = "42.8 mM"

The total concentration of chloride anions in solution will be

${\left[{\text{Cl"^(-)]_"total" = ["Cl"^(-)]_1 + ["Cl}}^{-}\right]}_{2}$

["Cl"^(-)]_"total" = "15.6 mM" + "42.8 mM" = "58.4 mM"

To convert this to molar, use the conversion factor

$\text{1 M" = 10^3"mM}$

You will thus have

58.4 color(red)(cancel(color(black)("mM"))) * "1 M"/(10^3color(red)(cancel(color(black)("mM")))) = color(green)(5.84 * 10^(-2)"M")#