Question

What is an example of buffering capacity?

Answer 1

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

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Suppose you want the buffer capacity after the FIRST addition of `"5.00 mL 0.15 M HCl"` into a `"30.0 mL"` buffer solution. The buffer capacity is given by:

`beta = (DeltaN)/(Delta"pH") > 0`

where:

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

Let's say the `"pH"` changed from `6.73` to `6.15` after the first addition of `"HCl"`, and the burette readings are `"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 xx 10^(-4) "mols HCl"`

So the numerator becomes:

`DeltaN = (7.50 xx 10^(-4) "mols HCl")/(underbrace(30.0 cancel"mL buffer soln")_"initial volume" xx "1 L"/(1000 cancel"mL")`

`=` `"0.0250 mols HCl/L buffer soln"`

And the denominator is:

`Delta"pH" = |6.15 - 6.73| = |-0.58| = "0.58 pH units"`

So, the buffer capacity after this first addition of `"0.15 M HCl"` is:

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

`= color(blue)("0.043 mol/L"cdot"pH")`