# 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?

May 17, 2016

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

BACKGROUND INFORMATION

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

${\text{CH"_3"COOH"(aq) + "H"_2"O"(l) rightleftharpoons "CH"_3"COO"^(-)(aq) + "H"_3"O}}^{+} \left(a q\right)$

ACQUIRING INITIAL QUANTITIES OF ACID AND CONJUGATE BASE

Next, we need to determine the initial concentration of acid and conjugate base (before adding $\text{HCl}$).

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

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

where:

• $\text{pH} = 4.76$
• $\text{pKa} = - \log \left({K}_{a}\right) = - \log \left(1.75 \times {10}^{- 5}\right) = 4.757$; higher $\text{pKa}$ = weaker acid = dissociates LESS.
• $\left[\text{HA}\right]$ is the concentration of generic Bronsted acid $\text{HA}$ in $\text{M}$. In this case, $\text{HA" = "CH"_3"COOH}$.
• $\left[{\text{A}}^{-}\right]$ is the concentration of generic conjugate base ${\text{A}}^{-}$ in $\text{M}$. In this case, ${\text{A"^(-) = "CH"_3"COO}}^{-}$.

First, let's say we had a $\setminus m a t h b f \left(\text{1 L}\right)$ 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 $\text{HA}$ neutralized is the amount of ${\text{A}}^{-}$ produced $x$, we can treat this in terms of mols,

$\setminus \frac{{\left[\text{A"^(-)]_(n ew))(["HA}\right]}_{n e w}}{=} \frac{{n}_{{A}^{-}} - x}{{n}_{H A} + x} ,$

with $x$ being the mols of added ${\text{H}}^{+}$.

ADDING THE NEW ACID TO SOLUTION

Now, let's see how the concentration of protons changes after adding $\text{0.1 M}$ of $\text{HCl}$, a strong acid, from $\text{HCl}$"without" changing the volume of the solution. Since it is a strong acid, we saw that $\textcolor{g r e e n}{\left[{\text{HCl"] = ["H}}^{+}\right]}$.

We can do this by supposing that we have $\text{0.0001 mol}$ of $\text{HCl}$ in $\text{0.001 L}$ (which is $\text{1 mL}$ of $\text{0.1 M}$ $\text{HCl}$).

In doing so, we reacted $\text{0.0001 mol}$s of ${\text{CH"_3"COO}}^{-}$, the BASE, to neutralize that ACID, decreasing the $\text{mol}$s of conjugate base, increasing the $\text{mol}$s of Bronsted acid. Thus, $\left[{\text{H}}^{+}\right] \uparrow$ relative to the starting buffer.

This also increases the volume to $\text{1.001 L}$, by the way (not that it matters much in this case). This gives:

${n}_{{A}^{-}}^{\text{new" = "0.5000 mols" - "0.0001 mols H"^(+) = "0.4999 mols A}} ^ \left(-\right)$

${n}_{H A}^{\text{new" = "0.5000 mols" + "0.0001 mols H"^(+) = "0.5001 mols HA}}$

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

$\textcolor{b l u e}{\text{pH}} = 4.757 + \log \setminus \frac{0.4999}{0.5001}$

$= 4.757 + \log \setminus \frac{0.4999}{0.5001}$

$\approx \textcolor{b l u e}{4.76}$

which makes sense because:

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