# Calculate the "pH" of a solution that resulted when "40 mL" of a "0.1-M" ammonia solution was diluted to "60 mL" ?

## The ${K}_{a}$ of the ammonium cation is $5.7 \cdot {10}^{-} 10$

Feb 19, 2018

$\text{pH} = 11.0$

#### Explanation:

Start by calculating the molarity of the diluted ammonia solution.

You know that your stock solution contains

40 color(red)(cancel(color(black)("mL solution"))) * "0.1 moles NH"_3/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.0040 moles NH"_3

After you dilute this solution by adding enough water to increase its volume from $\text{40 mL}$ to $\text{60 mL}$, its molarity will be--remember to use the volume of the solution in liters!

["NH"_3] = "0.0040 moles"/(60 * 10^(-3) quad "L") = "0.0667 M"

Now, ammonia will act as a weak base in aqueous solution.

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

By definition, the expression of the base dissociation constant, ${K}_{b}$, is given by

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

As you know, an aqueous solution at ${25}^{\circ} \text{C}$ has

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{K}_{a} \cdot {K}_{b} = 1 \cdot {10}^{- 14}}}}$

Notice that the problem gives you the acid dissociation constant, ${K}_{a}$, of the ammonium cation, ammonia's conjugate acid. This means that the base dissociation constant for ammonia is equal to

${K}_{b} = \frac{1 \cdot {10}^{- 14}}{5.7 \cdot {10}^{- 10}} = 1.75 \cdot {10}^{- 5}$

Now, if you take $x$ $\text{M}$ to be the equilibrium concentration of the ammonium cations and of the hydroxide anions, you can say that, at equilibrium, the solution will also contain

["NH"_3] = (0.0667 - x) quad "M"

This basically means that in order for the reaction to produce $x$ $\text{M}$ of ammonium cations and $x$ $\text{M}$ of hydroxide anions, the concentration of ammonia must decrease by $x$ $\text{M}$.

Plug this back into the expression of the base dissociation constant to find

$1.75 \cdot {10}^{- 5} = \frac{x \cdot x}{0.0667 - x}$

$1.75 \cdot {10}^{- 5} = {x}^{2} / \left(0.0667 - x\right)$

The value of the base dissociation constant is small enough compared to the initial concentration of the acid to justify the approximation

$0.0667 - x \approx 0.0667$

This means that you have

$1.75 \cdot {10}^{- 5} = {x}^{2} / 0.0667$

which gets you

$x = \sqrt{0.0667 \cdot 1.75 \cdot {10}^{- 5}} = 1.08 \cdot {10}^{- 3}$

Since $x$ $\text{M}$ represents the equilibrium concentration of the hydroxide anions, you can say that your solution contains

["OH"^(-)] = 1.08 * 10^(-3) quad "M"

Consequently, the $\text{pH}$ of the solution, which can be found using the fact that at ${25}^{\circ} \text{C}$, an aqueous solution has

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{pH + pOH = 14}}}}$

will be equal to

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

Since

color(blue)(ul(color(black)("pOH" = - log(["OH"^(-)]))))

you can say that your solution has

$\text{pH} = 14 - \left[- \log \left(1.08 \cdot {10}^{- 3}\right)\right] = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{11.0}}}$

The answer is rounded to one decimal place, the number of sig figs you have for your values.