# Question bf38f

May 4, 2017

$\text{pH} = 9.31$

#### Explanation:

For starters, you know that your buffer contains ammonia, ${\text{NH}}_{3}$, a weak base, and the ammonium cation, ${\text{NH}}_{4}^{+}$, its conjugate acid.

When you add sodium hydroxide, a strong base, the hydroxide anions delivered to the solution by the sodium hydroxide will react with the ammonium cations to form ammonia and water

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

As you can see, the buffer will convert a strong base to a weak base, which is why you should expect the to see a small increase, much smaller than what you see when sodium hydroxide is added to pure water, in the $\text{pH}$ of the solution.

Now, you're working with a $\text{1.00-L}$ buffer, so you can treat molarity and number of moles interchangeably.

This means that the initial solution contained $0.15$ moles ${\text{NH}}_{3}$ and $0.15$ moles ${\text{NH}}_{4}^{+}$.

When you add the sodium hydroxide, the hydroxide anions will react with the ammonium cations in a $1 : 1$ mole ratio. The reaction will also produce ammonia in a $1 : 1$ mole ratio.

This means that after the reaction takes place, the buffer will contain

n_ ("OH"^(-)) = "0 moles " -> completely consumed

n_ ("NH"_ 4^(+)) = "0.15 moles" - "0.010 moles" = "0.014 moles NH"_4^(+)

n_ ("NH"_ 3) = "0.15 moles" + "0.010 moles" = "0.16 moles NH"_3

In other words, every mole of hydroxide anions added to the buffer will consume $1$ mole of ammonium cations and produce $1$ mole of ammonia.

If you assume that the volume of the buffer does not change, you will end up with

["NH"_3] = "0.16 M" " " and $\text{ " ["NH"_4^(+)] = "0.14 M}$

Now, you can use the Henderson-Hasselbalch equation to calculate the $\text{pOH}$ of the buffer

"pOH" = "p"K_b + log(( ["NH"_4^(+)])/(["NH"_3]))

This would make the $\text{pH}$ of the solution equal to

"pH" = 14 - ["p"K_b + log(( ["NH"_4^(+)])/(["NH"_3]))]

The $\text{p} {K}_{b}$ of the weak base is defined as

$\text{p} {K}_{b} = - \log \left({K}_{b}\right)$

In your case, this will be equal to

$\text{p} {K}_{b} = - \log \left(1.76 \cdot {10}^{- 5}\right) = 4.75$

Therefore, you can say that the $\text{pH}$ of the solution after the addition of the strong base is equal to

"pH" = 14 - [4.75 + log((0.14 color(red)(cancel(color(black)("M"))))/(0.16color(red)(cancel(color(black)("M")))))] = color(darkgreen)(ul(color(black)(9.31)))

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

Notice that the $\text{pH}$ of the solution before adding the strong base was lower than $9.31$

"pH"_ "before adding NaOH" = 14 - [4.75 + log ( color(red)(cancel(color(black)("0.15 M")))/color(red)(cancel(color(black)("0.15 M"))))]#

$\text{pH"_ "before adding NaOH} = 14 - 4.75 = 9.25$