What is an example of buffering capacity?

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

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")#