Question

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?

Answer 1

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.