# Question 62d6f

Oct 20, 2015

$p \text{H} = 11.13$

#### Explanation:

The idea here is to use the ammonium ion's acid dissociation constant, ${K}_{a}$, to find the value of ammonia's base dissociation constant, ${K}_{b}$.

The relationship between the base dissocaition constant, the acid dissociation constant, and water's self-ionization constant, ${K}_{W}$, is given by the equation

${K}_{W} = {K}_{a} \times {K}_{b}$

In your case, the base dissociation constant of ammonia will be

${K}_{b} = {K}_{W} / {K}_{a} = {10}^{- 14} / \left(5.62 \cdot {10}^{- 10}\right) = 1.78 \cdot {10}^{- 5}$

To determine the pH of the solution, you first need to know the $p \text{OH}$ of the solution, which in turn requires the cocnentration of hydroxide ions, ${\text{OH}}^{-}$.

Ammonia will react with water to form ammonium and hydroxide ions. Use an ICE table to determine the equilibrium concentration of hydroxide ions

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

color(purple)("I")" " " " 0.1" " " " " " " " " " " " " " " " " " " "0" " " " " " " " " " " "0
color(purple)("C")" " color(white)(x)-x" " " " " " " " " " " " " " " " " "(+x)" " " " " " " " "(+x)
color(purple)("E")" " color(white)(x)0.1-x" " " " " " " " " " " " " " " "color(white)(xx)x" " " " " " " " " "color(white)(x)x

By definition, the base dissociation constant will be

${K}_{b} = \left(\left[{\text{NH"_4^(+)] * ["OH"^(-)])/(["NH}}_{3}\right]\right) = \frac{x \cdot x}{0.1 - x}$

Because the value of ${K}_{b}$ is so small, you can say that

$\left(0.1 - x\right) \approx 0.1$

This means that you have

${K}_{b} = {x}^{2} / 0.1 = 1.78 \cdot {10}^{- 5}$

$x = \sqrt{0.1 \cdot 1.78 \cdot {10}^{- 5}} = 0.001334$

This means that you have

["OH"^(-)] = x = "0.001334 M"

The $p \text{OH}$ of the solution will be

p"OH" = -log(["OH"^(-)])#

$p \text{OH} = - \log \left(0.001334\right) = 2.87$

The pH of the solution will thus be

$p \text{H" = 14 - p"OH} = 14 - 2.87 = \textcolor{g r e e n}{11.13}$