science( chem)
posted by christina .
When a 0.1M solution of weak acid HA was titrated with a 0.1M NaOH. The pH measured when Vol_{base}=1/2Vol_{eq point} was 4.62. Using the activity coeficents, calculate pKa.
f_{A}=0.854 and f_{HA}=1.00
For this I'm confused once again. I know that using the activity coeficents I can get the activity which would then be used as concentration right?
So..
a_{x}=[A]^{x} * f^{x}
a_{NaA}=[NaA]*f_{NaA}
a_{NaA}= [0.1][0.0854]= 0.0854
a_{HA}=[HA]*f_{HA}
a_{HA}= [0.1][1.0]= 0.1
[H30+]= 10^pH= 10^4.62= 2.40e5
But I'm not sure what equation to use but since it is a buffer I, assuming it would be using a buffer eqzn
BUT since it is a titration wouldn't I have to go and subtract the moles of HA  moles NaOH? This is where I think that the 1/2 vol of equivalence point comes into play but I'm not sure. So do I go and assume that since it is 0.1M that it is cHA= 0.10.05 ? but what would the cNaA be then? would it be 0.05moles ? I guess it would but I'm not sure how to apply that here with the activity coeficents since I've only seen it used with the Ksp.
anyways not using that this is what I did..
[H30+]=Ka(cHA/cA)
2.40e5 = Ka (0.1/0.0854)
Ka=2.04e5
pKa= log(Ka)= 4.69
I really don't know if that is correct..
Someone PLEASE help me on this...
I didn't check any of your work but I can start by taking care of part of your confusion on where to begin. Yes, it is a buffer. And you can use the equation you have been using for Ka. But the trick here is to remember that when the solution is just half titrated, then the acid remaining is equal to the salt formed, so they are always equal at the half way point and H^+ = Ka so pH = pKa. NOW, with that in mind, using activities may change that slightly but this should give you a clearer picture of where to start.
I'm just more confused..but after gathering my thoughts together and thingking about what you said..I came up with this.
since the buffer eqzn you said that the acid remaining is equal to the salt formed..
would that mean that since it is 1/2 the eq point that 1/2 the moles have been used up of the base. then would there be 1/2 the molesof salt formed from the total?
(0.05) for the [NaA] and [HA]? and that would go into the eqzn for finding the activity?
But why is H^{+}= to Ka??
and pH= pKa? I think I need a visualization of why that is but I think that it's b/c since the acid and conj base are = they would cancel out?
But if that is then what's the point of the activity coeficents?
Now I've gotten myself in a confused mode again...
good grief _;;
For the moment, forget about activity and activity coefficients. We will simply say activity and molarity are the same.
Yes, you are right. Whatever HA we begin with for the titration, we will be halfway to the equivalence point, so 1/2 of the HA has been used, no excess NaOH is present, and 1/2 of the HA remains. Of course, 1/2 of the HA has been converted to the salt; therefore, HA = A.
HA ==> H^+ + A^
Ka = (H^+)(A^)/(HA). Solve for (H^+).
(H^+) = Ka*[(HA)/(A^)].
Since (HA) = (A^), then (H^+)= Ka
Taking the negative log of each side gives us
log(H^+) = log Ka
But log(H^+) = pH
and log Ka = pKa so
pH = pKa half way to the equivalence point.
You can visualize this, perhaps, by starting with say 100 mL of 0.1 HA and titrate with 0.1 NaOH.
So we start with 0.01 mols HA. The equivalence point will be at 100 mL NaOH, so 50.0 mL will be the half way point. At 50.0 mL of 0.1 M NaOH, we have added 0.005 mols NaOH. That uses us 0.005 mols HA, leaving 0.005 mols HA remaining and forms 0.005 mols NaA. So you see, halfway to the equivalence point, HA really is equal to A^. No matter what numbers you choose, halfway to the equivalence point they will ALWAYS be equal (in concentration, anyway). I didn't convert to M by dividing mols by volume because the volume is the same for both acid remaining and salt formed and the volume ALWAYS cancels, also. Neat, huh?
Now you take the (H^+) = Ka*[(HA)/(A^)] and plug in activity corrections by using the activity coefficients. I hope this clears up any confusion.
Hm...this is interesting
this is what I did after reading what you said and If I got it correctly my answer was correct, just that the ratio numbers were wrong, correct?
assuming it was using 1L
.5ml x (0.1mol NaOH/L)= 0.05mol NaOH
aNaA= [0.05][0.854]=0.0427
aHA= [0.05][1.00]= 0.05
[H30+]=Ka (cHA/cNaA)
2.40e5=Ka*1.1709
Ka=2.04e5
pKa= 4.69 ===> the exact same thing!?!
is this correct?
so if that IS correct then the only wrong thing was the ratio numbers right?
And that is kinda neat that the volume cancels but I'll admit I just put it into there for other volume questions just for the fun of it..XD weird eh?
I also have another Question, for a lab report when doing the header is it just the # of the page on the top or it includes the title of the lab report or your last name +pg #?
My lab report title is kind of long
Determination of % of tartaric acid (C2H4O2(COOH)2) in wine by titration with standardized sodium hydroxide (NaOH)
Yes, you get the same answer because cHA and cNaA cancels; therefore, no matter what numbers you place for those two values, it has no effect on the answer. I suggest, however, that if you want to do it right, you might put in 0.0333. Two or three examples will show you why. Suppose we start with 50 mL of 0.1 M HA and titrate with 0.1 M NaOH. At the halfway point, we have
5 millimols HA to start  2.5 millimols NaOH added leaving 2.5 millimols HA and forming 2.5 millimols NaA.
(HA) = 2.5/75 = 0.0333 M
(NaA) = 2.5/75 = 0.0333 M.
Start with 100 mL 0.1 M HA.
We have 10 millimols to start  5 millimols NaOH added leaving 5 millimols HA and forming 5 millimols NaA.
(HA) = 5/150 = 0.0333 M
(NaA) 5/150 = 0.0333 M
Start with 200 mL of 0.1 M HA.
20 millimols HA to start  10 millimols NaOH leaving 10 millimols HA and forming 10 millimols NaA.
(HA) = 10/300 = 0.0333 M
(NaA) = 10/300 = 0.0333 M
I don't see anything wrong with the way you have worked the problem. We are assuming, I suppose, that the pH of 4.62 measures the ACTIVITY of the H+.
As for the lab report, I make several suggestions.
1. I would write out percent and not use the symbol.
2. I would not use the formula of tartaric acid in the title.
3. I would not use the titration with standardized sodium hydroxide (NaOH).
Something like,
"Determination of Tartaric Acid in Wine" is what I would use for the title. You don't want to write the repoert in the title which is what using specifics does. If you want to differentiate (since there are many ways to determine tartaric acid, perhaps you might want to add the word
"Volumetric Determination of Tartaric......."
4. I would repost under the title English, report writing, or something to catch the attention of the English profs on this board. They can tell you exactly how to do citations and report writing including what goes at the top of each page.
I get it now...seriously I hope I get all this by Monday night which is of course the day before my final exam on Tuesday for this class... O.O
Hm..The funny thing is that I only get my #'s if I calculate the concentration using 1000ml or 1L for the moles as the total volume so that the 1/2 vol would be 500ml...
But if I use any other volume it comes out the way your #'s do with the 0.0333M...why is that ?..
Does this mean that next time if I encounter something like this where they don't give the exact vol used I should arbitrarily go and not use 1L as the total vol but rather something less?
(I can't believe I used 15 min to analyze this...sad...but of course sleep is elusive when finals are around the corner)
When I was typing the title I had a feeling that I shouldn't have used the % sign.
If I'm not supposed to use the tartaric acid formula in the title, Is It fine to use it in the report itself?
And I guess this is redundant but I also assume that everytime I put the words tartaric acid that I shouldn't go and type out the whole formula out..
luckily I didn't print my paper yet though...
"Determination of Tartaric Acid in Wine" is what I would use for the title. You don't want to write the repoert in the title which is what using specifics does. If you want to differentiate (since there are many ways to determine tartaric acid, perhaps you might want to add the word
"Volumetric Determination of Tartaric......."
Just great, so there ARE many other ways to go and find the % of tartaric acid in wine...I didn't write background info on that and how there are supposed to be benefits from determining the % tartaric acid by titration.I guess cost would be a possitive thing, but I don't know the other ways people determine the % tartaric acid in wine so I probably shouldn't say that..basically...no time for that research but If it was a 25 point report instead of a 5 point one...and IF they gave it about 3 weeks before instead of last week, I probably would have included that information. So for now it shall remain backgroundless and citation less as too since all I used was the text
But why wouldn't specifics be in the title of a lab report?
And I will post that question about what goes at the top of each page in a report
Thank You Very Much Dr.Bob for all the help AND also for the advice on the title of my lab report. =D
1. Let's talk about using 1 L to start. I think it still comes out to be 0.0333M.
1000 mL x 0.l M HA = 100 mmols to start.
half way to eq pt of 0.1 M NaOH is 500 mL. 500 mL x 0.1 M NaOH = 50 mmols NaOH.
100 HA  50 NaOH = 50 mmols HA remaining, 50 mmols NaA formed, and total volume is 1000 mL + 500 mL = 1500 mL.
(HA) = 50 mmols/1500 mL = 0.0333 M
(A^) = 50 mmols/1500 mL  0.0333 M
Using the formula for tartaric acid is fine for the first time it is used in the body of the report. After that refer to it only as the name (unless of course you are writing an equation.) Most specifics are omitted from the title because they belong in the body of the report. (You can't write the report in the title.)The title tells the reader what the article is about, the abstract (which usually follows the title and is just before the body) goes into a little of the specifics and a quick summary of the results to tell the reader if there is a need to read the entire report. This is followed by the history (other methods, similar procedures, etc).You don't have time to do the history now. I don't KNOW that there are other methods of doing tartaric acid in wine but I'm willing to bet there are several. Using "Volumetric" to start the title is a sure way of telling the reader how it is being done without using the stinger in your original title of "with standardized sodium hydroxide (NaOH)."
Hm..It was the total volume that got me..=d
So the title is just a generality.
Okay, I did include all that for the components of the report. But there are probable a whole bunch of things wrong with it but I'm not going to post my whole report b/c that would be crazy...so all I need to do is fix the title and click print, and get back to studying of course..
THANKS again Dr.Bob =)
Respond to this Question
Similar Questions

man alive! can someone check for me?
I am so bad when it comes to mathI I think I am slow! okay so my question is : a recrusive sequence is defined by tn= 2tn1, where t1= 5 define the first five terms of this sequnce: so would I go like this to figure them out: 2tn1, … 
math!
george entered a function into his calculator andfound the following partail sums s1=0.0016 s2= 0.0096 s3= 0.0496 s4= 0.2496 s5= 1.2496 determine the genral term of the corresponding sequence how would I approach this question? 
check!
Write a recursive formula that generates the terms of the following: . . .1, 3, 9, 27 my answer: t1=1, n>2 > tn= 3tn1 is this correct? 
chemistry
The definition of pH is: pH = log 10[H^+] what is the defintion of the [OH^]... is it pOH = log 10[OH^]? 
chemistry
I was way off on the other problem. But I see how you got the answer. Last one...thanks i'm learning more here thn class What is the sum of the coefficients (including “1”) of the following reaction? 
halflife
If 250mg of a radioactive element decays to 200mg in 48 hours, find the halflife of the element. ln(N o /N) = kt N o =250 mg N = 200 mg t = 48 hours. Solve for k, then put k into the following equation and solve for t 1/2 . k = 0.693/t … 
alg2 (sequences)
how do i find the pattern for: 4,8,12,16 if the last number is negative, multiply by 1 then add 4 if the last number is positive, add 4, then multiply by 1 or how about term n+1 = (abs(term n ))*(1)^n, where term 1 = 4, n>0 … 
Math
If there is a recursively defined sequence such that a 1 = sqrt(2) a n + 1 = sqrt(2 + a n ) Show that a n < 2 for all n ≥ 1 
Statistics
Consider pairs <x,y>,...,<xn,yn> for i=1, ...,n. zi=cxi+d Express b sub(z*x) in terms of b sub(y*x) and a sub(z*x) in terms of a sub(y*x). I have been stumped on this one for hours! Sub (y*x) means its a subscript. Thank … 
Calculus
A baseball, hit 3 feet above the ground, leaves the bat at an angle of 45' and is caught by an outfielder 3 feet above the ground and 300 feet from home plate. What is the intitial speed of the ball, and how high does it rise?