I have a question about buffers.
Part A
So it starts with 20ml 0.1 sodium acetate and 25ml 0.1 acetic acid.
Calculate ph of buffer is 4.74 because the acid and conjugate base have the same molarity correct? So the Ph is just pKa (1.8e-5)?
Part B
So the next part is add 5ml of 0.1 HCl and this is the part where it messes me up.
First I use an ICE chart.
Equation is
CH3COO- + HCl --> CH3COOH + H2O
So I get confused here. Should I use moles or concentrations in this ice chart or does it make a difference?
If I use concentrations, would the initial concentration be from the calculated ones after the buffer is at equilibrium, as in part a? So the initial concentration of CH3COO- is 0.1 + 1.8e-5M? and CH3COOH is 0.1 - 1.8e-5M? (This is from buffer solution in part a)
It doesn't work to use molarity because the HCl is already 0.1 M! And subtracting that from the M of acetic acid and acetate ion would make the number negative or very small.
So I try to use mol. So the amount of mol HCl is 0.0005. What I'm confused about is the mol of CH3COO- and CH3COOH.
I know the concentrations, but what volume do I mulitply it by? Initially I have 45 ml since (20ml of sodium acetate and 25ml of acetic acid). So would I just multiply by 45 ml?
And after the change which is add 0.0005 mol of HCl, does that mean the concentration is 50 ml now? So when I have the final mol of CH3COO- and CH3COOH do I divide by 50ml to get the concentration??
Examples of this question seem to show that you just use the final volume throughout so I would multiply initial concentration by 50ml to get initial # of mol?
And then I use Pka + log (conjugate base/acid) = PH
But the PH I got here is much lower the the measured Ph. I got 4.65. Is that correct?
My final concentrations I got were CH3COO- = 0.09018M and CH3COOH = 0.109982. Is this correct???
Rather than go through each "problem" point by point, we would do a lot of back-tracking, and I think that would be even more confusing. I can see what your troubles are so let me just start from the beginning.
1. Yes, at the beginning, the pH = pKa because the concns of base and acid are the same.
2. Apparently you are ok until you add the 5 mL of 0.1 M HCl. So let me begin there. To make things a little simpler typing, I will call CH3COOH just HAc and the acetate ion, CH3COO^-, will be Ac^-. When HCl is added, the Ac^- grabs it (that's why it's a buffer--the Ac^- ties up what would have been free H^+) and makes HAc of it. Also I would like to work in millimoles so I don't have to type all those zero's. I think from your post that you're a pretty good student and can handle that. Instead of M x L = moles, we have mL x M = millimoles and mmoles/mL = M.
What did we have to start:
HAc = 25 mL x 0.1 M = 2.5 mmoles.
Ac^- = 20 mL x 0.1 M = 2.0 mmoles.
Now we add 5 mL x 0.1 M HCl.
Ac^- + H^+ ==> HAc
5 mL x 0.1 M = 0.5 mmoles. From your ICE chart, I think it is better to work in moles (millimoles).
So Ac^- is decreased by 0.5 and HAc is increased by 0.5.
After equilibrium is established,
new HAc = 2.5 + 0.5 = 3.0 mmoles.
new Ac^- = 2.0 - 0.5 = 1.5 mmoles.
new pH = pKa + log (base/acid).
You can substitute as well as I can but I get new pH of 4.44 if I didn't goof but you need to confirm that and let me know if I goofed.
Should you use concn or moles? I use moles (millimoles) to do the ICE chart because that's so much easier. TECHNICALLY, for substitution in the H-H equation, you should use concn BECAUSE the Henderson-Hasselbalch equation is pH = pKa + log [(base)/(acid)] and that is concn units in molarity. HOWEVER, the astute will notice that the volume is ALWAYS (and that's ALWAYS ALWAYS) the same since the acid, the salt, and the HCl all go into the same container. I always counted off if the student used moles and not concn BUT I always showed them that since the volume is always the same, there is no need to go through all the trouble to divide moles by liters to obtain molarity. We just write 1.5/V for concn Ac^- and 3/V = concn HAc because the V will cancel anyway and we can use any number we choose. The net effect is that we can use moles. I must admit that when the students aren't looking, I use millimoles, don't change to molarity by dividing by mL. It just makes for things a lot faster. After re-reading your post, I see I've addressed everything, I think, except the pH of 4.65. I don't know how you obtained that. If any of the above is confusing, post again with your questions. I THINK the 4.65 you calculated probably is the 4.44 I calculated above. It's a little strange to see buffers with more HAc than Ac^- at the beginning unless we want to add NaOH but I don't believe I have transposed those values. I hope this helps. I love these problems.
Oh ok thanks! I understand what I did wrong. I used the concentrations calculated from part A as my starting concentrations.
Isn't it impossible to use concentrations in part b though? Since HAc and Ac^- both start a 0.1 M and you are adding 0.1M HCl? Wouldn't that then make [] of HAc 2.0M and Ac^- 0M?
So what if the buffer in part A was made and then went to equilibrium. And then after a while HCl was added? Would that change the final pH since the initial mol of HAc and Ac^- would be different in this case right? What volume would you have to multiply the concentrations by to get the # of mol in this case?
So you're saying that in Intial change and Equlibrium states, the volume you multiply or divide by to get the #mol or concentration is the same?? But why is that since in the change part you add a certain # of mol and that obviously means a larger volume. Some questions I see are like that and that just confuses me. Thanks for all your help. Sorry if I'm making a simple question much more complicated than it really needs to be.
Oh ok thanks! I understand what I did wrong. I used the concentrations calculated from part A as my starting concentrations.
That's right. Part A is the starting point.
Isn't it impossible to use concentrations in part b though? Since HAc and Ac^- both start a 0.1 M and you are adding 0.1M HCl? Wouldn't that then make [] of HAc 2.0M and Ac^- 0M?
If you are asking if you can subtract 0.1 M from 0.1 M, no but you already know that. When you have 20 mL of 0.1 M Ac^- and you add 5 mL of 0.1 M HCl, it isn't realistic to think you can subtract 0.1 M - 0.1 M and make sense of it. You can't because you added DIFFERENT VOLUMES of the same concn and its obvious that the different volumes represent different amounts. You had 2.0 mmoles of Ac^-, you took away 0.5 mmole with HCl, which leaves you with 1.5 mmoles Ac^- and it is in 25 mL so the concn is 1.5 mmoles/25 mL = 0.06 M (later I use another example with a different volume which isn't 0.06 M so don't get those confused.
So what if the buffer in part A was made and then went to equilibrium.
I think you are making it much more complicated that it needs to be. I can guarantee you that when the HAc and Ac^- are mixed and stirred that the system is at equilibrium before you take the stirring rod out of the solution. Even before you can wink your eye. :-)
And then after a while HCl was added? Would that change the final pH since the initial mol of HAc and Ac^- would be different in this case right? What volume would you have to multiply the concentrations by to get the # of mol in this case?
The HAc and the Ac^- are not going to change concn unless you add more of one or the other OR if you add something else that reacts with one of them, such as the HCl. You have so many millimoles of each there (determined by mL x M) and that's in concrete. Those moles aren't going anywhere.
So you're saying that in Intial change and Equlibrium states, the volume you multiply or divide by to get the #mol or concentration is the same?? But why is that since in the change part you add a certain # of mol and that obviously means a larger volume. Some questions I see are like that and that just confuses me. Thanks for all your help. Sorry if I'm making a simple question much more complicated than it really needs to be.
I think you misunderstood me. I never said that the volumes didn't change. I said that the volume was the same for all parts of the solution. For example, we had 20 mL of 0.1 M Ac^- = 2.0 mmoles. We had 25 mL x 0.1 M HAc = 2.5 mmoles. When we added the 5 mL of 0.1 M HCl (yes the volume just changed by 5.0 mL), now we have a TOTAL volume of 20 + 25 + 5 = 50 mL.
We have 1.5 mmols Ac^-, we have 3.0 mmols HAc and we have a total volume of 50 mL. So the (Ac^-) = 1.5/50 = 0.030 M and (HAc) = 3.0/50 = 0.06 and the 5 mL of 0.1 M HCl is diluted instantly to 0.5 mmol/50 = 0.01 M. Note that the moles HCl did not change but the concn did. That's why I like to work in moles. 5 mL x 0.1 M = 0.5 moles. That's what reacts, not the concns.
Don't feel that you are a bother. This is important and it is important that you understand it, thoroughly. So don't hesitate to post again if needed.
To calculate the pH of a buffer solution, we need to consider the Henderson-Hasselbalch equation, which is pH = pKa + log([A-]/[HA]). Here, [A-] represents the concentration of the conjugate base (in this case, sodium acetate) and [HA] represents the concentration of the acid (in this case, acetic acid). It is important to note that both [A-] and [HA] should be in the same units, either molarity or moles.
Part A:
Given that the initial concentrations of sodium acetate and acetic acid are both 0.1 M, we can use these values directly in the Henderson-Hasselbalch equation. The pKa of acetic acid is indeed 1.8x10^-5. By substituting the values, we get pH = 4.74.
Part B:
In this case, you want to calculate the effect of adding 5 mL of 0.1 M HCl to the buffer solution from Part A. Let's walk through the steps:
1. Use an ICE (Initial, Change, Equilibrium) chart:
- Initial: The initial concentrations/moles of each component in the reaction.
- Change: The change in concentration/moles of each component after the reaction.
- Equilibrium: The equilibrium concentrations/moles of each component.
2. You mentioned that it doesn't work to use molarity because the HCl is already 0.1 M. Actually, you can still use molarity, but it is essential to account for the diluted volumes. Here's how you can approach it:
- Initial:
- Concentration of CH3COO- (acetate ion) = 0.1 M (from Part A)
- Concentration of CH3COOH (acetic acid) = 0.1 M (from Part A)
- Volume = 45 mL (20 mL + 25 mL)
- Change:
- You are adding 0.0005 mol of HCl to the solution.
- Since HCl is a strong acid, it fully dissociates, resulting in 0.0005 mol of H+ ions.
- Equilibrium:
- Concentration of CH3COO- (acetate ion) = 0.1 M - 0.0005 mol / 0.045 L (final volume in liters, correcting for the added HCl)
- Concentration of CH3COOH (acetic acid) = 0.1 M - 0.0005 mol / 0.045 L (final volume in liters)
- H+ ions can be ignored in the Henderson-Hasselbalch equation since their concentration is negligible compared to the acid and conjugate base concentrations.
3. Substitute the equilibrium concentrations into the Henderson-Hasselbalch equation:
- pH = pKa + log([A-]/[HA])
- pH = 1.8x10^-5 + log((0.1 - 0.0005)/(0.1 - 0.0005))
- Calculate this value to obtain the new pH.
The resulting pH you calculated was 4.65, which seems close to the measured pH. However, it is always recommended to double-check calculations and review the overall procedure to ensure accuracy.
Remember to consider units carefully, perform necessary conversions, and pay attention to the given information to obtain accurate results.