# 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?

##### 1 Answer

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

**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

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"# =weakeracid = dissociatesLESS.#["HA"]# is the concentration of genericBronsted acid#"HA"# in#"M"# . In this case,#"HA" = "CH"_3"COOH"# .#["A"^(-)]# is the concentration of genericconjugate base#"A"^(-)# in#"M"# . In this case,#"A"^(-) = "CH"_3"COO"^(-)# .

First, let's say we had a **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 **neutralized** is the amount of **produced**

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

with

**ADDING THE NEW ACID TO SOLUTION**

Now, let's see how the concentration of protons changes after adding **strong acid**, we saw that

We can do this by supposing that we have

In doing so, we reacted **neutralize** that ACID, **decreasing** the *conjugate base*, **increasing** the *Bronsted acid*. Thus,

This also *increases* the volume to

#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.