# How do I find the theoretical pH of a buffer solution after HCl and NaOH were added, separately?

This question is about the theoretical pH of a buffer solution.

Calculate the theoretical pH values expected for a $\pu{200mL}$ buffer solution containing a 1:1 ratio of acetic acid and sodium acetate (their concentrations are $\pu{0.05M}$), following the addition of;

1. $\pu{10mL}$ of $\pu{0.2M}$ $\ce{HCl}$ and
2. $\pu{10mL}$ of $\pu{0.2M}$ $\ce{NaOH}$.

$K_\mathrm{a}$ of acetic acid is given.

So I just found the $\mathrm{p}K_\mathrm{a}$ for the Henderson equation.

Attempt for (1):

$\ce{[HCl]}$ = $\pu{0.2M}$ and $V=\pu{10mL}$/$\pu{0.01L}$ thus $n = 0.002$

$0.002 \times \pu{0.2L} = \pu{0.01M}$

ICE table:

$$\begin{array}{cccc} &\ce{H3O+ + &CH3COONa <=> &CH3COOH + &H2O}\\ \text{I}& x& 0.05&0.05\\ \text{C}& -0.01&-0.01&+0.01\\ \text{E}& x - 0.0&0.04&0.06\\ \end{array}$$

$\mathrm{pH} = 4.74 + \log (0.06/0.04) = 4.926$

Although in our experiments the actual pH was 4.17 after $\pu{10mL}$ of $\ce{HCl}$ was added and the pH after 10 mL of NaOH was 4.98

Attempt for b:

$\ce{[NaOH]} = \pu{0.2M}$ and $V = \pu{10mL}/\pu{0.01L}$ thus $n= 0.002$

$0.002 \times \pu{0.2L} = \pu{0.01M}$

Is this how I would properly set it out?

$$\begin{array}{cccc} &\ce{CH3COOH + &OH- &<=> &CH3COONa + H2O}\\ \text{I}& 0.05&x&&0.05\\ \text{C}& -0.01&x-0.01&&+0.01\\ \text{E}& -&0.04&&0.06\\ \end{array}$$

$\mathrm{pH} = 4.74 + \log(0.04/0.06)= 3.74$

So I feel like I have done these the wrong way because it makes more sense but I don't know why or how.

• The Wikipedia page on buffer solutions has an example ICE calculation similar to yours towards the bottom of the page. May 1 '18 at 14:24

First I want to point out a couple things.

1. When you add more liquid to a solution, the concentration of the solute is going to decrease. Therefore, after the addition of 10 mL of HCl, the concentration of sodium acetate and acetic acid will no longer be 0.5 M since you diluted the solution a little bit. When calculating the concentration after dilutions, keep in mind the amount of moles is still the same, so just use the formula $\ce{M1V1 = M2V2}$.

2. Also, sodium acetate is a soluble salt (see solubility rules), so in the net ionic equation, don't write out the sodium as it is a spectator ion. The same applies for HCl. It is a strong acid, so it completely dissociates in water.

3. For this problem, using the Henderson-Hasselbalch equation is not really the right way to do it. When you are using this formula, keep in mind that it is only good for calculating pH when you already know the equilibrium concentrations of the acid/conjugate base. Here's a derivation of the formula if you're interested that will probably help clear things up a bit.

I think you mostly understand this point last point since you did made an ICE table before you applied the equation. However, in the context of this problem, making an ICE table will actually lead you to the pH value without requiring the HH equation. You'll see why.

Alright; now we can actually get to how to approach the problem. First, you're going to want to look out for a reaction. When solving these problems, being able to identify a reaction will make it 10 times easier.

How do you identify a reaction? Well, look at the reagants in this problem: $\ce{HCl}$, which is essentially just hydronium and chloride, sodium acetate, which as mentioned earlier is essentially just sodium and acetate, and acetic acid. Acids react with bases, so using that, it isn't too hard to determine what exactly happens in the solution.

You actually got the net ionic equation for the reaction correct (without the Na, however), but you didn't solve the ICE table correctly.

Actually, we're going to start with simple stoichiometry as a reaction is occurring. To simplify the process, we're going to assume the reaction reacts to completion, and then reverses until it reaches equilibrium.

For this, just use limiting reactants and such to calculate the final concentrations of $\ce{CH3COOH}$ and $\ce{CH3COO-}$.

As I said earlier, we're going to first assume that the reaction reacts to completion, and then goes back on itself and reaches equilibrium. We're at the second part, the equilibrium part. This is where an ICE table comes in.

First, write out the equilibrium reaction. This is pretty simple; it is just the acid dissociation reaction. It's going to look something like this: $$\ce{HA<->H+ +A-}$$

For your ICE table, you didn't really set it up properly. Remember these three things:

1. You usually won't be writing any unknowns for your initial concentrations. Everything should be known.

2. You usually won't be writing any numbers for your change. It's an equilibrium. You don't know the exact amount it is going to react to reach equilibrium. That's why we write out the ICE table, so we can calculate how much needs to react for the system to reach equilibrium.

3. Don't use the Henderson-Hasselbalch equation. Use the $K_a$ and its definition to calculate the concentration of $\ce{H+}$, which leads to your pH.

I know I pretty much just wrote a wall of text, but hopefully that helps nonetheless.

You seem to be on the right track. You have the correct number of moles of HCl added (0.002 moles H+) and the correct number of moles of acetate (0.01 moles Ac-). This reaction will result in an increase in the moles of HAc and a concomitant decrease in the moles of Ac- since H+ + Ac- => HAc. After addition, moles of Ac- = 0.01 - 0.002 = 0.008
moles HAc = 0.01 + 0.002 = 0.012
In a final volume of 210 ml (0.210 L), you can calculate final concentrations:
[Ac-] = 0.008/0.210 = 0.038 M
[HAc] = 0.012/0.210 = 0.057 M
You can then use the HH equation as you did, substituting these concentrations. Do similar calculation for addition of NaOH where HAc will decrease and Ac- will increase.
NOTE: The experimentally determined pH will not always be the same as the calculated pH. There are several reasons why this might be the case. Use your imagination.