Ap Chemistry: Building (and Breaking) Buffers lab

[jeybeyb]

Homework Statement

Overall goals: Your first task will be to make a buffer solution of defined pH. Part of your grade will be based on how close your solution comes to the desired pH. Your second task will be to titrate your buffer using 0.5 M NaOH or 0.5 M HCl until the buffer “breaks”. You will predict what quantity of acid or base is needed to break the buffer (as judged by a color change of a chosen indicator). Y

Materials: You will use one of the three conjugate acid-base pairs listed below to build your buffer. Make sure you choose the right system for the pH you are shooting for.

Homework Equations

glacial acetic acid (pure CH₃COOH) Ka = 1.8 x 10^-5
sodium acetate NaCH₃COO

pH = pKa + log [base]/[acid]

The Attempt at a Solution

CH3COOH --> CH3COO + H

Ka= [H+][CH3COO-]/[CH3COOH] = 1.8 x10^-5
[H+]= 4.85

pH = pKa + log [base]/[acid]
5.0 (chosen pH of buffer) = 4.85 + log [base]/[acid]

.15 = log [base]/[acid]

10^0.15 = [1.41 M]/[1 M]

Homework Statement

Homework Equations

pH = pKa + log [base]/[acid]

The Attempt at a Solution

CH3COOH --> CH3COO + H

Ka= [H+][CH3COO-]/[CH3COOH] = 1.8 x10^-5
[H+]= 4.85

pH = pKa + log [base]/[acid]
5.0 (chosen pH of buffer) = 4.85 + log [base]/[acid]

.15 = log [base]/[acid]

10^0.15 = [1.41 M]/[1 M]

We're doing a lab in my class and we're supposed to make a buffer at a certain pH.

I have decided to make a buffer with a pH of 5.0 using glacial acetic acid (pure CH3COOH) and sodium acetate.

I found that the ratio of acid to base is 1.41 M: 1.0 M and now I'm stuck.

My teacher only has 17.4 M glacial acetic acid and we basically have to dilute it in 75 mL of water. And I'm confused at this point. Help me please with a step by step description on how to do this?

 

[Borek]

Do I understand correctly you don't know how to dilute the acid?

http://www.chembuddy.com/?left=concentration&right=dilution-mixing

 

[jeybeyb]

Yes. That's correct.

 

[symbolipoint]

You say you have for glacial acetic acid, only that form of the unnuetratlized acid, and that you have its concentration as 17.4 M. You have what you need. You have something of 17.4 M and you want to dilute it to 1.0 M. What's the problem?

If you use numeric symbols, C for concentration in moles per liter, V for volume in liters, and you want to start with one value of C and find how much volume to add (water in this case) to get a different value for C, this is solvable. Let C₁ be starting concentration and C₂ be the desired concentration. Also, you want V₁ and V₂ for volume of starting acid and volume of WATER to add... your way of assigning variables might vary from mine but essential algebra expressions and equations are the same.

Note that in practice you should add the acid TO the water; not the other way around:

[tex]\frac{C_{1} V_{1}}{V_{1}+ V_{2}}[/tex] = C₂

 

[Borek]

[tex]\frac{C_{1} V_{1}}{V_{1}+ V_{2}}[/tex] = C₂

There is a problem here. V₁ plus V₂ doesn't have to be what you want, as volumes are not additive. It is much better to fill up to the needed volume.

It is described on the page linked to from my first.

 

[symbolipoint]

There is a problem here. V₁ plus V₂ doesn't have to be what you want, as volumes are not additive. It is much better to fill up to the needed volume.

It is described on the page linked to from my first.

Very good, Borek! Yes, the volume may need to be adjusted with monitoring to finish. The student may also choose to decide on his preparation using weight & volume percentage, needing the use of density of the glacial acetic acid. The use of just molarity and literes seemed more straightforward.