# What is an example of buffering capacity?

May 25, 2018

Here is an example using one of my students' data.

Suppose you want the buffer capacity after the FIRST addition of $\text{5.00 mL 0.15 M HCl}$ into a $\text{30.0 mL}$ buffer solution. The buffer capacity is given by:

$\beta = \frac{\Delta N}{\Delta \text{pH}} > 0$

where:

• $\Delta N$ is the mols of $\text{HCl}$ (or $\text{NaOH}$) added to the buffer, divided by the volume of the buffer you actually used (not the $\text{HCl"//"NaOH}$).
• $\Delta \text{pH}$ is simply the change in $\text{pH}$ you got due to the first addition. Treat this as a positive quantity.

Let's say the $\text{pH}$ changed from $6.73$ to $6.15$ after the first addition of $\text{HCl}$, and the burette readings are $\text{0.23 mL" -> "5.23 mL}$.

We get mols of:

n_"HCl" = overbrace("0.15 mol HCl"/cancel"1 L")^"conc." xx [overbrace((5.23 cancel"mL" - 0.23 cancel"mL"))^"volume added via burette" xx cancel"1 L"/(1000 cancel"mL")]

$= 7.50 \times {10}^{- 4} \text{mols HCl}$

So the numerator becomes:

$\Delta N = \left(7.50 \times {10}^{- 4} \text{mols HCl")/(underbrace(30.0 cancel"mL buffer soln")_"initial volume" xx "1 L"/(1000 cancel"mL}\right)$

$=$ $\text{0.0250 mols HCl/L buffer soln}$

And the denominator is:

$\Delta \text{pH" = |6.15 - 6.73| = |-0.58| = "0.58 pH units}$

So, the buffer capacity after this first addition of $\text{0.15 M HCl}$ is:

color(blue)(beta) = "0.0250 mols HCl"/("L buffer soln" cdot "0.58 pH")

$= \textcolor{b l u e}{\text{0.043 mol/L"cdot"pH}}$