# How many grams CH3COONa do I need to create this buffer?

1. Apr 18, 2010

### 4LeafClover

Problem:
I need 750mL of an acetic acid-sodium acetate buffer with pH = 4.3. Solid sodium acetate (CH3COONa) and glacial acetic acid are available. Glacial acetic acid is 99% CH3COOH by mass and has a density of 1.05 g/mL. If the buffer is to be 0.25M in CH3COOH, how many grams of CH3COONa must be used?

My work:
I looked up the Ka value of CH3COOH and found it to be 1.8*10^-5. I also took the pH and solved for [H+] by using the mathematical process of 10^-4.3 and found [H+] to be 5.0*10^-5. Because I am told that the buffer is 0.25M CH3COOH and I found the concentration of [H+] I went ahead and set up an equilibrium expression which looks like Ka = [H+][CH3COO-]/[CH3COOH] --> 1.8*10^-5 = (5.0*10^-5)(x)/(0.25) and I found x to be 0.09M. Since I only have .750L of solution, I went ahead and multiplied 0.09 by 0.750 and found that there are 0.0675 moles of CH3COONa in the solution. Now, when I multiply the number of moles of CH3COONa in solution by the molecular weight of CH3COONa, I come up with 6 grams, but for some reason this is incorrect. Where did I go wrong?

Additional Question: How many milliliters of glacial acetic acid must be used?2

2. Apr 18, 2010

### Staff: Mentor

What equation describes pH of the buffer?

--
methods

3. Apr 18, 2010

### 4LeafClover

The henderson hasselbach equation.

So I have 4.3 = 4.74 + log ([CH3COO-]/[CH3COOH])
--> -0.44 = log ([CH3COO-]/[CH3COOH])
--> 0.904 = [CH3COO-]/[CH3COOH]
--> [CH3COO-] = 1 - [CH3COOH]
-->1 - [CH3COOH]/[CH3COOH] = 0.904
-->[CH3COOH] = 0.525M and [CH3COO-] = 0.475M

So I have (0.475M) * (0.75L) = 0.356 moles of CH3COONa. The molecular weight of CH3COONa is 89 g/mol. 89 * 0.356 = 31.68 grams of CH3COONa.

For the second part to this problem (how many mL of glacial acetic acid should be used?), I need to calculate grams of CH3COOH used. The molecular weight of CH3COOH is 60 g/mol. This means that I have 0.525*60 = 31.5 grams of CH3COOH. Since the density of glacial acetic acid is 1.05 g/mL, I have 31.5/1.05=30mL of glacial acetic acid. But since glacial acetic acid is only 99% CH3COOH, does this mean that I actually only have 29.7mL? (30mL*.99=29.7mL)

4. Apr 18, 2010

### Staff: Mentor

Why 1?

--

5. Apr 18, 2010

### 4LeafClover

I chose 1 because since this is an acid and its conjugate base, for some reason I thought that together they must = 1. But I think for some reason I was thinking of mole fractions and their parts adding up to = 1. Since this is NOT a mole fraction, then the ratio of conjugate base to conjugate acid would have to equal the molarity of the solution??? And since I don't have the molarity of the solution, I can't perform the calculations above. I do however have the molarity of the acid though. Which gives me these calculations:

4.3 = 4.74 + log (x/0.25)
--> -0.44 = log (x/0.25)
--> 0.363 = x/0.25
--> x = 0.907M
--> 0.907M * 0.750L = 0.680 moles

The molecular weight of CH3COONa is 89 g/mol. 89 * 0.0680 = 6.06 grams of CH3COONa.

So now, for the second part, I have 0.25M *0.75L = 0.1875 moles CH3COOH. 0.1875moles * 60g/mol = 11.25 grams. 11.25g / 1.05 g/mL = 10.7 mL of CH3COOH needed. Since glacial acetic acid is only 99% CH3COOH though, I am really only going to need 10.7 * .99 = 10.6mL?

6. Apr 18, 2010

### Staff: Mentor

Much better, but it is obvious at first sight that there is something wrong with your math again.

--

7. Apr 18, 2010

### 4LeafClover

Shoot, I just went back and looked at this, and it looks like doing the calculations with the henderson hasselbach equation give me almost exactly the same answer I came up with when I did the calculations using the equilibrium equation...so obviously it's wrong. This fact is really confusing me though - I am coming up with the same answer using two different equations. Grr...

8. Apr 18, 2010

### 4LeafClover

Ok, is this better if I split the log function up?
4.3 = 4.74 + log (x/0.25)
-0.44 = log x - log 0.25
0.363 = x - .25
x = 0.113M
0.113M * 0.750L = 0.848 moles
0.848 moles * 89 = 7.54 grams

9. Apr 18, 2010

### Staff: Mentor

No, because you obviously don't know how to do it.

This is incorrect, it has nothing to do with properties of logaritimc function.

You were much closer previously, you just did math error, almost in the same place.

--
methods

10. Apr 18, 2010

### 4LeafClover

You're right, I don't know how to do it. That's why I am asking the questions I am, so that I can understand. I know it is late where you are, so I will look through everything we've talked about again (for both my questions) and try to see if I can synthesize the information and understand what is going on. I will get back to you (prob tomorrow) after I've pondered and see if I can get a better grasp of what is going on.

11. Apr 18, 2010

### Staff: Mentor

Just check your math, don't try to invent something fancy.

--
methods

12. Apr 19, 2010

### 4LeafClover

Must've been a typo, because as I go back and look at my math in the other one, a couple steps down, it is corrected and I return to 0.0680 moles....so I am getting the same answer of roughly 6 grams....

0.363 = x/0.25
--> x = 0.0908M
0.0908M * 0.75L = 0.0681 moles
0.0681 moles * 89 g/mol = 6.06 grams CH3COONa

This answer also gives me the same calculations for the second part where I get 10.7 mL of CH3COOH needed, but since glacial acetic acid is only 99% CH3COOH, I really only need 10.7 * .99 = 10.6 mL.

So if 6 grams if wrong, why do I keep getting it? Either my professor made a typo in the answer, which is entirely possible, or I am missing something, which is probably even more likely.

13. Apr 19, 2010

### Staff: Mentor

Could be it was a typo, I haven't checked exact result. I usually trace the solution to the first error and it was obvious that 0.25*0.363 can't be larger than both multiplied numbers.

What is molar mass of sodium acetate that you used for calculations?

Think again. Imagine you need 100 mL of pure acetic acid for something. You have just stated that as the acetic acid is 99% you need only 99 mL. That means that using 5% table vinegar you need 5 mL. Following this line of thinking if you will use pure water 0 mL will contain exactly the same amount of acetic acid as 100 mL of pure acetic acid.

You may have a hard time convincing me that's the way it works.

--
methods

14. Apr 19, 2010

### 4LeafClover

I was using 89 g/mol. My being sick right now must be making me foggy in the head because I was using 30 g/mol for Na! I redid calculations and found it to be 5.5 grams which is the correct answer....which would have been correct for my first calculations if I had just been a little more careful.

I also worked through what you said about the volume of acetic acid needed and got the answer right. Thanks for working me through this the last couple of days!