# Question 2c896

Dec 1, 2015

["H"_3"O"^(+)] = 1.7 * 10^(-5)"M"

#### Explanation:

Ammonium chloride, $\text{NH"_4"Cl}$, will dissociate completely in aqueous solution to produce ammonium cations, ${\text{NH}}_{4}^{+}$, and chloride anions, ${\text{Cl}}^{-}$

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

Notice that the salt dissociates in a $1 : 1$ mole ratio with the ammonium ion. This means that you get one mole of the latter for every mole of the former that is dissolved in solution.

So, if $0.20$ moles of ammonium chloride are used to make this solution, it follows that the molarity of the ammonium ions will be

${n}_{N {H}_{4}^{+}} = {n}_{N {H}_{4} C l} = \text{0.20 moles}$

$c = \text{0.20 moles"/(400.0 * 10^(-3)"L") = "0.50 M}$

Once in aqueous solution, the ammonium ions will act as a weak acid and react with the water molecules to form ammonia, ${\text{NH}}_{3}$, a weak base, and hydronium ions, ${\text{H"_3"O}}^{+}$.

To determine the equilibrium concentration of the hydronium ions, use an ICE table

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

color(purple)("I")" " " "0.50" " " " " " " " " " " " " " " " " "0" " " " " " " " " " " " "0
color(purple)("C")" "color(white)(x)(-x)" " " " " " " " " " " " " " " "(+x)" " " " " " " " "(+x)
color(purple)("E")" "0.50-x" " " " " " " " " " " " " " " "x" " " " " " " " " " " " "x#

Now, notice that the poroblem provides you with the base dissociation constant, ${K}_{b}$, of ammonia. This means that you're going to have to use the equation

$\textcolor{b l u e}{{K}_{a} \cdot {K}_{b} = {K}_{W}} \text{ }$, where

${K}_{W}$ - the self-ionization constant of water, equal to ${10}^{- 14}$ at room temperature

to find the acid dissociation constant, ${K}_{a}$, of the ammonium ion.

${K}_{a} \cdot {K}_{b} = {K}_{W} \implies {K}_{a} = {K}_{W} / {K}_{a}$

${K}_{a} = {10}^{- 14} / \left(1.8 \cdot {10}^{- 5}\right) = 5.6 \cdot {10}^{- 10}$

By definition, the acid dissociation constant is equal to

${K}_{a} = \left(\left[{\text{NH"_3] * ["H"_3"O"^(+)])/(["NH}}_{4}^{+}\right]\right)$

In your case, this will be equal to

${K}_{a} = \frac{x \cdot x}{0.50 - x} = {x}^{2} / \left(0.50 - x\right)$

Since the value of ${K}_{a}$ is so small, you can use the approximation

$0.50 - x \approx 0.50$

to get

${K}_{a} = {x}^{2} / 0.50 = 5.6 \cdot {10}^{- 10}$

This means that you have

$x = \sqrt{0.50 \cdot 5.6 \cdot {10}^{- 10}} = 1.7 \cdot {10}^{- 5}$

Since $x$ represents the equilibrium concentrations of ammonia and hydronium ions, it follows that you have

$\left[\text{H"_3"O"^(+)] = color(green)(1.7 * 10^(-5)"M}\right)$