Assuming the volume is unchanged in a 0.5 M acetic acid/sodium acetate buffer at a starting pH of 4.76, if 1 mL of 0.1 M HCl is added, what does the pH become?

In assuming the volume is "unchanged", we could be saying that we start with a large volume and add a very small volume of #"HCl"#.

BACKGROUND INFORMATION

The starting buffer has the following equation (with sodium as the unstated counterion):

#"CH"_3"COOH"(aq) + "H"_2"O"(l) rightleftharpoons "CH"_3"COO"^(-)(aq) + "H"_3"O"^(+)(aq)#

ACQUIRING INITIAL QUANTITIES OF ACID AND CONJUGATE BASE

Next, we need to determine the initial concentration of acid and conjugate base (before adding #"HCl"#).

This can be done by using the Henderson-Hasselbalch equation:

#\mathbf("pH" = "pKa" + log\frac(["A"^(-)])(["HA"]))#

where:

• #"pH" = 4.76#
• #"pKa" = -log(K_a) = -log(1.75xx10^(-5)) = 4.757#; higher #"pKa"# = weaker acid = dissociates LESS.
• #["HA"]# is the concentration of generic Bronsted acid #"HA"# in #"M"#. In this case, #"HA" = "CH"_3"COOH"#.
• #["A"^(-)]# is the concentration of generic conjugate base #"A"^(-)# in #"M"#. In this case, #"A"^(-) = "CH"_3"COO"^(-)#.

First, let's say we had a #\mathbf("1 L")# solution. (It can really be any large volume, based on the assumptions we'll make.)

Now, a nice trick with this equation is that since both substances are in the same solution, and since the amount of #"HA"# neutralized is the amount of #"A"^(-)# produced #x#, we can treat this in terms of mols,

#\frac(["A"^(-)]_(n ew))(["HA"]_(n ew)) = (n_(A^(-))-x)/(n_(HA)+x),#

with #x# being the mols of added #"H"^(+)#.

ADDING THE NEW ACID TO SOLUTION

Now, let's see how the concentration of protons changes after adding #"0.1 M"# of #"HCl"#, a strong acid, from #"HCl"#"without" changing the volume of the solution. Since it is a strong acid, we saw that #color(green)(["HCl"] = ["H"^(+)])#.

We can do this by supposing that we have #"0.0001 mol"# of #"HCl"# in #"0.001 L"# (which is #"1 mL"# of #"0.1 M"# #"HCl"#).

In doing so, we reacted #"0.0001 mol"#s of #"CH"_3"COO"^(-)#, the BASE, to neutralize that ACID, decreasing the #"mol"#s of conjugate base, increasing the #"mol"#s of Bronsted acid. Thus, #["H"^(+)]uarr# relative to the starting buffer.

This also increases the volume to #"1.001 L"#, by the way (not that it matters much in this case). This gives:

#n_(A^(-))^"new" = "0.5000 mols" - "0.0001 mols H"^(+) = "0.4999 mols A"^(-)#

#n_(HA)^"new" = "0.5000 mols" + "0.0001 mols H"^(+) = "0.5001 mols HA"#

Plugging these values back into the Henderson-Hasselbalch equation, we would get:

#color(blue)("pH") = 4.757 + log\frac(0.4999)(0.5001)#

#= 4.757 + log\frac(0.4999)(0.5001)#

#~~ color(blue)(4.76)#

which makes sense because:

• It's a buffer. It's supposed to resist #"pH"# change.
• You added some concentration of #"HCl"# such that the volume didn't change significantly, i.e. you barely touched the solution.