Question

Help finding H+ concentration

Answer 1

So, I have two buffers:
Buffer A is made from 100 mL of .75M NaAc and 100 mL of .22M acetic acid and finally 10 mL of a .2% starch solution.

All contents are mixed together and added to a 500 mL flask, and water is added to get the solution to the 500 mL mark.

Now, 65 mL of this buffer is added to another solution (65mL of buffer and rest of solution is 35 mL, equaling 100 mL total new solution).

What is the concentration of the H+ ion in the final solution? That is found by 1.753x10^-5 * ((acetic acid conc)/(Ac- conc)) * (1/(.16)^2).

I am using M1V1=M2V2, and getting .000201 M. Can someone check.

Answer 2

Where (Ac - conc) is the concentration of NaAC. Sorry for confusion.

Answer 3

the 35 ml that you add to the 65 of buffer is water or is something else?

Answer 4

A bunch of other stuff, but not a source of H+ ions. This is an iodine clock experiment.

Answer 5

KIO3 and KI and H3AsO3 to be specific. But the book says only worry about the buffer for source of H+.

Answer 6

are you familiar with the henderson hasselbalch equation

Answer 7

Oh yes! Very much so.

Answer 8

But don't I have to get the concentrations right using the m1v1=m2v2 before I can get H+ conc?

Answer 9

Because I have to use this: 1.753x10^-5 * ((acetic acid conc)/(Ac- conc)) * (1/(.16)^2).
for that deybe huckle theory.

Answer 10

yes you have to calculate the concentration of NaAc and AcH in the original dilution consider both diluted to 500 mL final

Answer 11

That is, 1.753x10^-5 * ((acetic acid conc)/(Ac- conc)) * (1/(.16)^2) = H+

Answer 12

Can you help me find those conc? Not sure If I did it right.

Answer 13

but technically because both are the same volumes and you dilute both to the same final volume the relationship of concentrations will be the same

Answer 14

Hum. That is true. So you are saying 1.753x10^-5 * ((.22)/(.75)) * (1/(.16)^2) = H+ ??

Answer 15

That is good, that is the answer I got.

Answer 16

Can you help me with one more? It is similar?

Answer 17

what is this 1.753x10^-5?

Answer 18

That is the K value.
P-chem lab book says:

H+ = K (HAC/Ac-)((1/(.16)^2))

Answer 19

Don't you know the Ka for acetic acid? LOL!!!

Answer 20

yes but the equation is an addition not a product

Answer 21

Yea, whoops. It may help if I type correctly...

Answer 22

My bad! DUH!

Answer 23

the concentration of the base is 0.75 and the concentration of the acid is 0.22M

Answer 24

and the dilution doesn't supposed to affect the pH, because it is a buffer.

Answer 25

what is the 0.16^2?

Answer 26

The reason I am asking all of this is because for my table I have to make to determine the reaction order, all of my H+ values are the same and it wants me to find the reaction order of it. I can't if they are all the same! That .16^2 is to take the Debye-Huckle theory into account.

Answer 27

For trial 4, I had buffer B, made of 50 mL of .75 M NaAc and 100 mL of .22 M acetic acid and the 10 mL of .2% starch. Contents are added to a 250 mL flask and water is added to get it up to 250 mL. I then add 65 mL of that solution to 35 mL of other "stuff", making 100 mL. Is the concentration of buffer B still the same or is it different since we are starting with 50 mL of .75 M NaAc instead of 100 mL of it as in buffer A or is it the same as buffer A?

Answer 28

I can scan you what I mean @Cuanchi please don't give up yet!

Answer 29

the concentration of the buffer change because you dilute the buffer, but because it is a buffer you supposed to have the same pH. If you have the same pH you will have the same [H+]

Answer 30

Ok. Well, that is what I thought. I'm glad we are on the same page! Then how and why does it even want me to find the reaction order of [H+]?

Answer 31

Now if you are doing the clock reaction with the I-. the buffer only is there to keep the pH constant and you have to calculate the kinetics of the reaction looking at the concentrations of I- and the time to take to change the color

Answer 32

Right! Thanks! That is what I thought. But this stupid lab manual says to find the rate for IO3-, I- and the H+. It does not make sense.

Answer 33

here is what the book says. Maybe I just need sleep @Cuanchi

Answer 34

Also says:

d/dt=k (io3-)^m (I-)^n (H+)^p

So I guess what I am really asking is...what would you say the order for H+ is? 0? 1? or what? @Cuanchi thanks so much for all your help.

Answer 35

@Cuanchi @Photon336

Answer 36

@KamiBug

Answer 37

I think that I need more details about the reaction that you are trying to find the rate. I am familiar with the

\[S _{2}O _{8}^{2-} + 2 I ^{-}\rightarrow I _{2} + 2SO _{4}^{2-}\]

I think you have other reaction

Answer 38

Let me find a copy of the lab for you.

Answer 39

Im fixing to post it.

Answer 40

And that is the whole thing. Take your time. I owe you so much!!!

Answer 41

let me read and organize that

Answer 42

If you need to take a break it is ok.

Answer 43

What kills me is it says "...varying concentrations of i03-, I-, and H+...
Take your time. This is just my input. From what I asked in the beginning, it looks like the H+ concentrations should not be the same like I calculated. Maybe it is from a volume issue we missed. I can find h+ easy if the concentration of at least one differs!

Answer 44

That is, I can find the rxn order easily of H+

Answer 45

the issue here you have 2 buffers, the buffer A and the buffer B, In the reactions that you use the buffer A (reactions 1, 2 and 3) the [H+] = 10^-5 M
the reaction 4 that you use the buffer B the [H+]= 2 x 10^-5 M

Answer 46

How did you get those two [H+}?

Answer 47

That makes me concerned my other concentrations are wrong.

Answer 48

You then to see if the concentration of H+ affect the rate of the reaction you have to compare the trial 1 with the trial 4, if the rate is the same the order for the [H+} is zero, if increase double is the first order and so on

Answer 49

Absolutely. Thanks a million. But can you show me how you got your two [H+}? I like to know the "why."

Answer 50

it is written with the buffer composition or you can calculate with the combination of acid and salt to prepare them

Answer 51

page 257

Answer 52

may I know the title and what is the author of the book

Answer 53

ARE YOU KIDDING ME? That was there the whole time and I missed it.

Answer 54

Experiments in Physical Chemistry by Shoemaker, et. al.

Answer 55

I am sorry!!

Answer 56

Well, you learn by your mistakes. I sure learned a lot today. I can never repay you enough.

Answer 57

Ho! is Physical Chemistry that is so difficult, I was thinking was general chemistry

Answer 58

Gen chem is not too bad, yeah. This semester I am doing the thermo/kinetics part of p-chem. It is all lab. No lab instructor...I have a book and access to all the supplies. So I am literally learning it all alone. Challenging but rewarding! I can't believe I missed that though.

Answer 59

You are very smart student I am glad to meet you and I wish you the best!! are you in chemical engineering?

Answer 60

Actually, I am a junior undergrad, pure chemistry major. When I go to grad school I will either get a p-chem phd or analytical. Not sure yet. I want to be a full professor. I actually am a TA for gen chem 1 and 2. That is how I knew the H-H equation!

Answer 61

see you around!! GTG now

Answer 62

Thanks so much!!!