# When the ionic compound NH_4Cl dissolves in water, it breaks into one ammonium, NH_4^+ and one chloride Cl^-. If you dissolved 10.7 g of NH_4Cl in water, how many moles of ions would be in the solution?

Dec 19, 2015

$\text{0.400 moles}$

#### Explanation:

Well, you know that one mole of ammonium chloride, $\text{NH"_4"Cl}$, dissociates completely in aqueous solution to form one mole of ammonium cations, ${\text{NH}}_{4}^{+}$, and one mole of chloride anions, ${\text{Cl}}^{-}$.

${\text{NH"_4"Cl"_text((aq]) -> "NH"_text(4(aq])^(+) + "Cl}}_{\textrm{\left(a q\right]}}^{-}$

This tells you that regardless of how many moles of ammonium chloride you dissolve in water, you will always get twice as many moles of ions.

All you have to do now is to figure out how many moles of ammonium chloride you get in that $\text{10.7-g}$ sample. To do that, use the compound's molar mass

10.7 color(red)(cancel(color(black)("g"))) * ("1 mole NH"_4"Cl")/(53.49 color(red)(cancel(color(black)("g")))) = "0.200 moles NH"_4"Cl"

So, if $0.200$ moles of ammonium chloride are dissociated in aqueous solution, you will get $0.200$ moles of ammonium cations and $0.200$ moles of chloride anions.

Therefore, the resulting solution will contain a total of

n_"total" = 0.200 + 0.200 = color(green)("0.400 moles ions")

SIDE NOTE The ammonium cations act as a weak acid in aqueous solution, meaning that they react with water to form ammonia, ${\text{NH}}_{3}$, and hydronium ions, ${\text{H"_3"O}}^{+}$

${\text{NH"_text(4(aq])^(+) + "H"_2"O"_text((l]) rightleftharpoons "NH"_text(3(aq]) + "H"_3"O}}_{\textrm{\left(a q\right]}}^{+}$