# Question 750c8

Feb 25, 2016

Here's what I got.

#### Explanation:

The problem wants you to use the base dissociation constant, ${K}_{b}$, of ammonia, ${\text{NH}}_{3}$, to determine the percent of ammonia molecules that ionize to produce ammonium cations, ${\text{NH}}_{4}^{+}$, and hydroxide anions, ${\text{OH}}^{-}$.

As you know, ammonia is a weak base, which means that it does not ionize completely in aqueous solution. Simply put, some molecules of ammonia will accept a proton from water, but most will not.

So, the balanced chemical equation for the ionization of ammonia looks like this

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

Notice that you have a $1 : 1$ mole ratios between ammonia and the two ions it produces in aqueous solution.

This means that every molecule of ammonia that ionizes will produce one molecule of ammonium cations and one molecule of hydroxide anions.

In order to determine the percent ionization of the base, you need to know two things

• the initial concentration of the base
• either the equilibrium concentration of hydroxide anions or the concentration of ammonium cations

Use the solution's pH to help you find the equilibrium concentration of hydroxide anions. Determine the solution's pOH first

$\textcolor{b l u e}{\text{pH " + " pOH} = 14}$

$\text{pOH} = 14 - 11 = 3$

Then use it to find $\left[{\text{OH}}^{-}\right]$

color(blue)("pOH" = - log(["OH"^(-)]) implies ["OH"^(-)] = 10^(-"pOH"))

["OH"^(-)] = 10^(-3)"M"

So, you know that the initial $\text{0.5 M}$ concentration of ammonia produces a concentration of ${10}^{- 3} \text{M}$ of hydroxide anions and of ammonium cations (remember the $1 : 1$ mole ratio that exists between the two ions).

color(red)(!!!) At this point, you need to ignore the initial concentration of ammonia provided by the problem. color(red)(!!!)

Assuming that ${\left[{\text{NH}}_{3}\right]}_{0}$ is the initial concentration of ammonia.

If you were to write an ICE table for the ionization of the base, you would have

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

color(purple)("I")" " " "["NH"_3]_0" " " " " " " " " " " " " " " "0" " " " " " " " " " "0
color(purple)("C")" " " "(-x)" " " " " " " " " " " " " "(+x)" " " " " " "(+x)
color(purple)("E")" "["NH"_3]_0-x" " " " " " " " " " " " " "x" " " " " " " " " " "x

The base dissociation constant for this equilibrium reaction will be

$\textcolor{b l u e}{{K}_{b} = \left(\left[{\text{NH"_4^(+)] * ["OH"^(-)])/(["NH}}_{3}\right]\right)} \to$ uses equilibrium concentrations

In your case, you will have

${K}_{b} = \frac{x \cdot x}{{\left[{\text{NH}}_{3}\right]}_{0} - x} = 1.76 \cdot {10}^{- 5}$

But you know that $x = {10}^{- 3}$, since that represents the equilibrium concentration of hydroxide and ammonium ions. This means that you have

1.76 * 10^(-5) = (10^(-3))^2/(["NH"_3]_0 - 10^(-3)

Rearrange to solve for ${\left[{\text{NH}}_{3}\right]}_{0}$

${\left[{\text{NH}}_{3}\right]}_{0} + {10}^{- 3} = {10}^{- 6} / \left(1.76 \cdot {10}^{- 5}\right)$

["NH"_3]_0 = "0.05682 M" + 10^(-3)"M" = "0.05782 M"

So, what this means is that out of an initial concentration of $\text{0.05872 M}$ of ammonia, only ${10}^{- 3} \text{M}$ ionizes to form ammonium cations and hydroxide anions. The rest remains as ammonia molecules.

So, percent ionization will be equal to

$\textcolor{b l u e}{\text{% ionization" = "what ionizes"/"what you start with} \times 100}$

"% ionization" = (10^(-3)color(red)(cancel(color(black)("M"))))/(0.05782color(red)(cancel(color(black)("M")))) xx 100 = color(green)(1.73%)#
This tells you that out of ${10}^{4}$ ammonia molecules present in aqueous solution, only $173$ will ionize to form ammonium cations and hydroxide anions.
${10}^{- 3} / 0.05782 = 1.73 \cdot {10}^{- 2}$