Question

Can anyone please help me with my lab report data? Desperate college freshman student looking for help with the topic of buffers.

Answer 1

@acxbox22 @andrewjordan833 @alurahenderson @badmood @chycora @cupcakerain @desiringeden @desiraem @ElonaSushchik @freckles @GeniousCreation @haleyelizabeth2017 @HackberryAbby18 @ikram002p @JoKeR0331 @khalilforthewin @Lady.Liv1776 @mathmate @Nnesha

Answer 2

I don't believe I know how to do this. What science is this?

Answer 3

Chemistry, if you know anyone who can. Please let me know!

Answer 4

@alurahenderson @background43 @jadaaaa

Answer 5

@abbi5169 @Brandonsnider222

Answer 6

I'll take a look at it if you post it in a pdf format (rather than word)

Answer 7

okay

Answer 8

@aaronq - Here you go. If I am not on. Email is jdalloo97@gmail.com.

Answer 9

you should stay on if you wanna work through it. I'm only going to tell you how to do it, but i'm not gonna do it for you.

Answer 10

Okay, sure. Whatever's best.

Answer 11

I need to know theoretical pH, also but I don't know how to do that. Finding the missing columns are the most important ones.

Answer 12

do you have other data regarding the composition of the buffers, or anything like that?

there isn't enough information on these sheets to complete this

Answer 13

Sodium Acetate and Acetic Acid are the substances that I used.

Answer 14

Okay cool. For the theoretical pH, we can use the Henderson-Hasselbalch equation:

\(\sf \large pH=pKa+log\dfrac{[A^-]}{[HA]}\)

so we need to know the initial concentrations of the components (sodium acetate and acetic acid) and how much acid or base you added.

Answer 15

1.0 M for each solution, which was labeled on the Erlenmeyer flask for the lab.

Answer 16

Regarding the amount added, it should be on the pdf file.

Answer 17

Is the data one second for the first table? it doesnt seem like it.

For the first table, beaker 1. you added equal amounts of Na acetate and acetic acid?

how much acid did you add, at what concentration?

Answer 18

What do you mean?

Answer 19

I could send you a pic of my lab scanned....is that fine?

Answer 20

On the first table, where it says "beaker, solution, measured pH..", what are the amounts of everything that you used?

Answer 21

Lets just work with the data for the first line, "Beaker 1" where you used buffer A + HCl

Answer 22

Okay, so first I added 25 mL of Buffer Solution "A" into Beakers 1 +3. I added 25 mL of distilled water to beakers 2 +4. Moreover, I transfered 1.0 mL of 1 M HCl to beakers 1+2, and 1 mL NaOH to beakers 3+4. I measured the new pH after that was done.

Answer 23

Hm nothing on the concentrations of the buffer? or the volumes?

i was working through it and can't go forth without those

Answer 24

No sir. That's why i am stuck.

Answer 25

the volumes of the buffers + additional total to 26 mL.

Answer 26

We could assume that they're equal (both at 0.5 M to add up to 1 M buffer), since the pH (4.41) is close to the pKa (4.75) and it could be off solely because of experimental error.

Answer 27

Okay sure. Would that work, or cause any error in the calculations?

Answer 28

Well it would be wrong if the concentrations of the components of the buffers are anything but 0.5 M each.

Answer 29

True.

Answer 30

I mean i can show you how to do it with 0.5 M and if you come across the actual concentrations you can just rework through it later?

Answer 31

That would be great. Anything that's close is good for me. I had a headache trying to figure this out all day.

Answer 32

okay cool.

we need to make an ICE table to keep things organized. It stands for Initial, Change and Equilibrium (concentrations).

The reaction is: \(NaCH_3COO^-+HCl\rightarrow NaCl+ CH_3COOH \)

we're only interested in the concentrations of sodium acetate and acetic acid.

\([NaCH_3COO^-]\) \([CH_3COOH]\)
I
C
E

Initially, we had 25 mL of \([NaCH_3COO^-]\)=0.5 M and \([CH_3COOH]\)=0.5 M

we have: \(moles=Molarity*volume=0.5~M*0.025~L=0.0125~moles\) for each

\([NaCH_3COO^-]\) \([CH_3COOH]\)
I 0.0125 moles 0.0125 moles
C
E

The "change" was adding acid, \(moles=1~M*0.001~L=0.001~moles \)


\([NaCH_3COO^-]\) \([CH_3COOH]\)
I 0.0125 moles 0.0125 moles
C -0.001 +0.001
E


this makes sense because you neutralized some base (sodium acetate and made the same amount of acetic acid)

\([NaCH_3COO^-]\) \([CH_3COOH]\)
I 0.0125 moles 0.0125 moles
C -0.001 +0.001
E 0.0125 -0.001 0.0125+0.001

The volume doesnt matter because of the math

\(\sf \large pH=4.75+log\dfrac{(0.0125 -0.001)}{(0.0125+0.001)}\)

this is the theoretical pH.

Answer 33

I don't be mean to sound super rude or anything, but I have to go in a little while. I apologize for the inconvenience.

Answer 34

I posted each step individually of the table so you could see how i filled it in

Answer 35

Thank you :)

Answer 36

This is for the buffers. for the water just use: \(\huge \sf pH=-log[H^+]\)

Answer 37

where \([H^+]\) is the concentration of HCl (in the new volume)

Answer 38

no problem! let me know if you have questions

Answer 39

Also, for the \(\Delta pH\) column, (which means change in pH.

you just find the difference (subtract) between the two measured pH's, before and after you added acid or base.

Answer 40

Sorry, to be a bother but I had a question about the concentrations of the acids and bases for the first table. Is that for the first step above.

Answer 41

hmm i'm not sure i understand which step you're referring to?

Answer 42

Where you displayed the ICE table, is that just for theoretical pH?

Answer 43

yeah, that's for the theoretical pH.

for \(\Delta pH\) just find the differences between the measured pH values

Answer 44

and for the [acids] and [bases],, how would i follow from previously mentioned steps.

Answer 45

So were talking about the second table right?

Answer 46

Yes.

Answer 47

Okay, so for that we use the value you mentioned before 1.0 M for each of the components.

So for A, we had 25 mL of each base and acid, that is sodium acetate and acetic acid.

We need to find the moles of each and the new concentration (molarity).

\(\sf moles_{Base}=1~M*0.025~L=0.025~moles\)

\(\sf Molarity=\dfrac{0.025~moles}{0.025~L+0.025~L}=0.5~M\)

Answer 48

Thanks!

Answer 49

no problem!