# What is the pH of a solution of 120 ml 0.15M acetic acid to which we add 30mL 0.2M NaOH?

CAN YOU CHECK MY WORK PLEASE?

0.15 mol/L * 0.12 L = 0.018 mol acetic acid

0.2 mol/L * 0.03 L = 0.006 mol NaOH

0.018 - 0.006 = 0.012 mol of acetic acid in excess

pH = pKa + log (acid/base)

pH = 4.76 + log (0.006/0.018)

pH = 4.3

**Almost.**

0.15 mol/L * 0.12 L = 0.018 mol acetic acid **OK**

0.2 mol/L * 0.03 L = 0.006 mol NaOH **OK**

0.018 - 0.006 = 0.012 mol of acetic acid in excess **very good**

pH = pKa + log (acid/base)

**The next step is where you went wrong. 0.006 is the number of mol of base (acetate ion) formed. But 0.012 is the mol of acetic acid remaining. (acetate ion; the base) = 0.006 mols/0.150 L) = ??
(acetic acid; the acid) = 0.012 mol/0.150 L = ??
pH = 4.76 + log (base/acid) etc.
I obtained 4.459 which rounds to 4.46. I am posting this now to get the mistake to you as fast as possible; however, I will make another post to explain about the ratio so be sure to come back and read it.**

pH = 4.76 + log (0.006/0.018)

pH = 4.3

## Dr. Bob, where did you get the 0.150L from? I will keep my eyes open for the explanation as well.

## About the ratio. Note that I showed acetate ion = 0.006/0.150 = 0.0400 M and acetic acid = 0.012/0.150 = 0.0800 M. Note that you did not use the 150 mL anywhere. Technically, it isn't correct to use just mols in the log term of log [(base)/(acid)] although the answer comes out the same. If you write log (0.006/0/012) we get log 1/2 and if we write log (0.040/0.080) we still get log 1/2. So what's the difference? The equation says to use concentration base/acid and if we write 0.006/0.012 we are using mols and not concentration. Now, it is true that the volume of 150 mL (0.150 L) cancels and it is also true that no matter what problem we have, the volume will ALWAYS cancel so using mols always gives the correct answer BUT mols and concentration are not the same. I always told my students that I would count off if they used mols because, technically, that wasn't correct (even though they got the right answer). Some complained that the extra step just took up valuable time and I countered with doing it the following way.

log[(base)/(acid)] = log (0.006/V)/(0.012/V) = log(0.006/0.012) = log 1/2. That way, (using the V for the volume)(or just writing it as log (0.006/0.150)/(0.012/0.150) = log 1/2 since the 0.150 cancels OR the V cancels) shows that the student recognizes that the number that goes there is the concentration. Not writing a V or not writing the actual volume (then letting it cancel) doesn't recognize that point. I hope this isn't too long winded but I wanted you to know why I worked it the long way to begin with. By the way, I assume you just picked up the wrong number of 0.018 for the acid or you would have had the correct answer originally. I hope this helps.

## You had 120 mL of acetic acid and added 30 mL NaOH which make a total of 150 mL or 0.150 L.

## To calculate the pH of the solution, we need to determine the concentration of the remaining acetic acid after the reaction with NaOH.

First, let's calculate the number of moles of acetic acid (CH3COOH) in 120 mL of a 0.15 M solution:

0.15 M * 0.12 L = 0.018 mol acetic acid

Next, calculate the number of moles of NaOH (sodium hydroxide) in 30 mL of a 0.2 M solution:

0.2 M * 0.03 L = 0.006 mol NaOH

Now, we need to find out how much acetic acid is left after the reaction with NaOH. Since NaOH completely reacts with acetic acid in a 1:1 ratio, the moles of NaOH will cancel out the moles of acetic acid. In this case, there will be an excess of acetic acid remaining:

0.018 mol - 0.006 mol = 0.012 mol of acetic acid in excess

To calculate the pH of the solution, you can use the Henderson-Hasselbalch equation:

pH = pKa + log [A-]/[HA]

The pKa of acetic acid is 4.76. Here, [A-] represents the concentration of the acetate ion (CH3COO-) and [HA] represents the concentration of acetic acid.

Using the values we have:

pH = 4.76 + log (0.006 mol / 0.018 mol)

pH = 4.76 + log (1/3)

pH ≈ 4.76 + (-0.477)

pH ≈ 4.3

So, your calculation is correct. The pH of the solution is approximately 4.3.