# Question a8da2

Apr 12, 2015

The pH of the solution is $11.22$.

#### Explanation:

To solve this problem, you need the value of the base dissociation constant of ammonia, ${K}_{b}$, which is listed as being equal to $1.8 \cdot {10}^{- 5}$ $\to$ see here.

Since ammonia is a weak base, it will increase the concentration of ${\text{OH}}^{-}$ ions in the solution, so you would expect the solution to have a pH greater than $7$.

Use the ICE table (more here) to solve for the concentration of ${\text{OH}}^{-}$ ions, which are labeled here as $x$

${\text{ " " ""NH"_ (3(aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "NH" _ (4(aq))^(+) + "OH}}_{\left(a q\right)}^{-}$

"I"color(white)(aaaaacolor(black)(0.15)aaaaaaaaaaaaaaaa color(black)(0) aaaaaaa color(black)(0)
"C"color(white)(aaaaacolor(black)(-x)aaaaaaaaaaaaaaa color(black)(+x) aaaaa color(black)(+x)
"E"color(white)(aaacolor(black)(0.15-x)aaaaaaaaaaaaaa color(black)(x) aaaaaaa color(black)(x)

Use the definition of the base dissociation constant

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

$1.8 \cdot {10}^{- 5} = \frac{x \cdot x}{0.15 - x} = {x}^{2} / \left(0.15 - x\right)$

Because ${K}_{b}$ is so small compared with the initial concentration of ammonia, you can use the approximation

$0.15 - x \approx 0.15$

You will have

$1.8 \cdot {10}^{- 5} = {x}^{2} / 0.15$

$x = \sqrt{0.15 \cdot 1.8 \cdot {10}^{- 5}}$

$x = 0.001643$

This means that you have

["OH"^(-)] = "0.001643 M"

The solution's $\text{pOH}$ will be

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

$\text{pOH} = - \log \left(0.001643\right) = 2.78$

Therefore, the $\text{pH}$ of the solution will be

$\text{pH" = 14 - "pOH}$

$\text{pH} = 14 - 2.78 = \textcolor{g r e e n}{11.22}$