# If a buffer contains "0.110 M" weak base and "0.440 M" of the weak conjugate acid, what is the "pH"? The "pK"_b is 4.96.

Jul 30, 2017

$\text{pH} = 8.44$

#### Explanation:

This looks like a job for the Henderson - Hasselbalch equation, which for a weak base/conjugate acid buffer looks like this

"pH" = 14 - overbrace(["p"K_b + log( (["conjugate acid"])/(["weak base"]))])^(color(blue)("the pOH of the buffer solution"))

As you know, you have

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

In your case, you know that the buffer contains $\text{0.110 M}$ of the weak base and $\text{0.440 M}$ of its conjugate acid, so even without doing any calculations, you should be able to say that the $\text{pOH}$ of the buffer will be higher than the $\text{p} {K}_{b}$ of the weak base.

In other words, the $\text{pH}$ of the buffer will be lower than $14 - \text{p} {K}_{b}$, what you would get for a $\text{pOH}$ equal to the $\text{p} {K}_{b}$ of the weak base.

Plug in your values into the Henderson - Hasselbalch equation to find

"pH" = 14 - [-log(1.1 * 10^(-5)) + log((0.440 color(red)(cancel(color(black)("M"))))/(0.110color(red)(cancel(color(black)("M")))))]

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{pH} = 8.44}}}$

The answer is rounded to two decimal places, the number of sig figs you have for the base dissociation constant.

Jul 30, 2017

Stefan has a good answer, but I thought I'd give another approach to this. For buffers (i.e. weak acid + conjugate base, weak base + conjugate acid), the Henderson-Hasselbalch equation applies.

To make it so I only have to know one Henderson-Hasselbalch equation, I use the ${\text{pK}}_{a}$ one and recall the relationships to interconvert between ${\text{pK}}_{a}$, ${\text{pK}}_{b}$, $\text{pH}$, and $\text{pOH}$.

"pH" = "pK"_a + log\frac(["A"^(-)])(["HA"])

Using the idea that ${\text{pK"_a + "pK}}_{b} = 14$, we get:

$- \log \left({K}_{b}\right) = {\text{pK}}_{b} = - \log \left(1.1 \times {10}^{- 5}\right) = 4.96$

$\implies {\text{pK}}_{a} = 14 - 4.96 = 9.04$

And thus, noting the difference in notation (treating the base as ${\text{A}}^{-}$ or $\text{B}$ and the conjugate acid as $\text{HA}$ or ${\text{BH}}^{+}$), the $\text{pH}$ is alternatively found as:

color(blue)("pH") = 9.04 + log (("0.110 M")/("0.440 M"))

$= 9.04 - \log 4 = \textcolor{b l u e}{8.44}$

And this makes physical sense, as we started with a weak base, whose conjugate acid dissociates less (${K}_{a} < {K}_{b}$ if ${K}_{b} > {10}^{- 7}$ at ${25}^{\circ} \text{C}$ and $\text{1 atm}$).

Furthermore, there is a higher concentration of conjugate acid than the weak base. So, we should expect the $\text{pH}$ to be basic, but also more acidic than the ${\text{pK}}_{a}$ of the conjugate acid.

APPENDIX

And just so you see, this gives the same equation Stefan has. Recall that:

• At ${25}^{\circ} \text{C}$ and $\text{1 atm}$, ${\text{pK"_a + "pK"_b = 14 = "pK}}_{w}$
• $\log \left(\frac{a}{b}\right) = - \log \left(\frac{b}{a}\right)$

Therefore:

"pH" = (14 - "pK"_b) + log\frac(["A"^(-)])(["HA"])

= (14 - "pK"_b) - log\frac(["HA"])(["A"^(-)])

= 14 - ["pK"_b + log\frac(["HA"])(["A"^(-)])]

which is what Stefan used.

With slightly changed notation, and knowing that $\text{pH" + "pOH} = 14$ at ${25}^{\circ} \text{C}$ and $\text{1 atm}$, we can get the other form of the Henderson-Hasselbalch equation:

"pH" = 14 - overbrace(["pK"_b + log\frac(["BH"^(+)])(["B"])])^("pOH")

$= 14 - \text{pOH}$

Thus,

$\overline{\underline{| \stackrel{\text{ ")(" " "pOH" = "pK"_b + log\frac(["BH"^(+)])(["B"])" }}{|}}}$