Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: sallyhansen on April 22, 2013, 10:09:24 PM
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Alright so I have been posting these questions up on a whole bunch of forums and no one seems to be answering any of the questions I have posted up, so it would be much appreciated if someone where to help me out with the following questions, because I feel they are easier then they seem but I just over analyze things. Any help on the following questions will be appreciated. Thanks in advance
1. Use the initial pH of the acetic
acid solution to find the initial [H3O+] and initial [CH3COO]-(2)
Initial pH = 2.58
2. Assume that the amount of CH3COOH
that ionizes is small compared with the initial concentration of the
acid. Calculate Ka for the acid. (2)
HC2H3O2(aq) + H2O(l) ---> H3O+(aq)+ C2H3O2-(aq)
Ka = [H3O+(aq)][ C2H3O2-(aq)]
[HC2H3O2(aq)]
3. Refer to the volume of NaOH on
your graph from question 2. Calculate half this volume and on your graph,
find the pH when the solution was half neutralized. (1)
half of it would make it 6 mL
4. Calculate Ka when
the acetic acid was half-neutralized. How does this value compare with
your Ka value for acetic acid? (2)
5. Calculate the percent difference
between your value for Ka (from the calculations and the graph)
and the accepted value. (3)
6. Do the values you calculated for
[H3O+] and [CH3COOH] prove that CH3COOH
is a weak acid? Explain. (2)
You see that a majority of the questions are only worth 2 marks, how can that be when there is so much that should be done for them?
{MOD Edit: add useful title}
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You must show you have attempted the question, this is a Forum Rule (http://www.chemicalforums.com/index.php?topic=65859.0).
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I really suck at chemistry and this is my second time taking it and I cant even understand the simplest thing in chemistry. Please *delete me*
1. Use the initial pH of the acetic acid solution to find the initial [H3O+] and initial [CH3COO]-(2)
Initial pH = 2.58
My answer is as follows
pH = -log[H+]=2.58
= [H+] = 10-2.58
= 2.63*10-3 m
Is this correct
2. Assume that the amount of CH3COOH that ionizes is small compared with the initial concentration of the
acid. Calculate Ka for the acid. (2)
The formula would be below
HC2H3O2(aq) + H2O(l) ---> H3O+(aq)+ C2H3O2-(aq)
and my Ka constant is this, but I do not understand what I am supposed to do after this.
Ka = [H3O+(aq)][ C2H3O2-(aq)]
[HC2H3O2(aq)]
The procedure is below to give out some information to those that will help me with these questions.
Procedure:
1. Record the molar concentration of the NaOH solution.
0.1 mol/L
2. Produce a table to record your data. It should have one column for volume of NaOH added and one column for pH.
3. Obtain 50 mL of acetic acid and place it into a beaker.
4. Place 50.0 mL of NaOH into the burette.
5. Pipet 25.0 mL of acetic acid into the Erlenmeyer flask. Add two drops of phenolphthalein to the acid.
6. Record the initial pH of the solution.
7. Add 1.00 mL of NaOH from the burette to the Erlenmeyer until the pH reaches 5.00. Record the volume to two decimal places. Measure the pH of the solution each time you add NaOH.
8. Above pH=5.00, add NaOH in 0.10 or 0.20 mL portions. Record the volume at which the phenolphthalein turns pink. phenolphthalein turned pink at 12.0 mL
9. Continue to add NaOH until the pH reaches 11.00. Above pH=11.00, add 0.10 mL portions until the pH reaches 12.00.
Note: All you actually have to do is start the burette running, the values will record as you progress through the lab.
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sallyhansen: I hope you don't mind my merging your two posts. There's no need for multiple threads on the same topic. You've done some work, can you see what needs to be done? Can you write out that question? And can you do it or attempt to do that missing answer? You haven't really asked us anything, except why this work has such low marks -- that's not a question we can answer.
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Assume that the amount of CH3COOH that ionizes is small compared with the initial concentration of the
acid. Calculate Ka for the acid. (2)
What I have done is below, but I do not understand why you would assume CH3COOH that was ionizes is small compared to the intital concentration of the acid. So what I have done below is just put the generic answers that I think is needed to answer the question, but don't know how to apply them.
The formula would be below
HC2H3O2(aq) + H2O(l) ---> H3O+(aq)+ C2H3O2-(aq)
and my Ka constant is this, but I do not understand what I am supposed to do after this.
Ka = [H3O+(aq)][ C2H3O2-(aq)]
[HC2H3O2(aq)]
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okay if the concentration is 0.1 mol/L I would just then do x^2/Co and calculate it that way?
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just realized I need to use an ICE table and this was the steps I took to get the answer
HC2H3O2 H3O+ C2H3O2-
I (0.2) 0 0
C -x +x +x
E (0.2-x) x x
Ka = [0.1]
=1.8*10^-5(0.1) = x^2
=1.3*10^-3
Hopefully I did this right if not please clarify where it is I need to make changes Thanks!
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whoops accidentally put 0.2 it is 0.1
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OK for concentrations of H+ and CH3COO-.
I don't see how to calculate Ka without knowing initial concentration, and you never explained whether 0.1M was given to you, or is it just some random value out of nothing. But if it was given then you start in the right direction (your ICE table is correct), but I don't understand what you did later.
Your previous approach (x2/C) was closer, and the result you get with the ICE table should be very similar.
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the 0.1 mol/L was calculated like so
4. Use the ratio in which the acid and base react, determined from the chemical equation. Calculate the molar concentration of the acetic acid. (2)
CH3COOH (aq) + NaOH (aq) NaCH3COO(aq) + H2O(l)
From the equation, there was 1 mole of NaOH neutralized with 1 mole of acetic acid; so therefore, there are 0.0015 moles of acid.
n=CV
0.0015= C(0.015)
C= 0.1 mol/L
So it was not something I pulled out from thin air ...
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Okay I am completely confused I am not given the Ka value but I am supposed to calculate it, how would I go about doing this?
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It is perfectly correct to use 0.1 that you determined experimentally, and if that's where the 0.1M comes from it is OK - you just didn't told us where it comes from, and without this information it was not clear what you are doing.
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So the answer to the question is correct? Because what I do not get is that I have to find the Ka value of the acetic acid and I plugged in the value of acetic acid to find the x value, so how would that answer the question?
I have a few more questions
8. Refer to the volume of NaOH on your graph from question 2. Calculate half this volume and on your graph, find the pH when the solution was half neutralized. (1)
Original volume of NaOH = 12
Half the volume of NaOH = 6
pH for 12 mL of NaOH = 10.66
pH for 6 mL of NaOH = 4.58
9. Calculate Ka when the acetic acid was half-neutralized. How does this value compare with your Ka value for acetic acid? (2)
So I am not sure about this one would I make the 0.1 mol/L halved? Making it 0.05 mol/L
or would I do the question like this
0.1 mol/L of NaOH and 6 mL of NaOH 0.006L
(0.006L)(0.1mol/L)
=0.0006 moles
CH3COOH (aq) + NaOH (aq) NaCH3COO(aq) + H2O(l)
From the equation, there was 1 mole of NaOH neutralized with 1 mole of acetic acid; so therefore, there are 0.0006 moles of acid.
n=CV
0.0006 = C(0.006)
C= 0.1 mol/L
And if that were correct would I not be completing the same ice table twice?
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Suppose you are titrating a weak acid, HA, with NaOH.
HA <===> H+ + A-
Ka = [H+][A-] / [HA]
The term "half titration" simply means that half as much NaOH is added as it would take to reach the end point. You normally would not do this type of titration, but it is useful to determine the experimental value of a Ka.
At the half-titration point, the amount of HA left is equal to the amount of A- formed. At this point, the Ka = [H+].
Ka = [H+][A-] / [HA] = [H+] (when A- equals HA)
If the pH of the solution is measured at this point, it equals the pKa.
If Ka = [H+], the pKa = pH.
Thus, a half titration can be used to determine an experimental Ka value. Of course, you have to do a complete titration first, then redo the titration but add only have as much NaOH to reach the half titration point.
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Because what I do not get is that I have to find the Ka value of the acetic acid and I plugged in the value of acetic acid to find the x value, so how would that answer the question?
Before we get to your other questions - I have no idea what you are asking about. You are asked to calculate Ka. You can write Ka definition - it contains concentrations of undissociated acetic acid, acetate anion and H+. [H+] you know from pH measurement, concentration of acetate anion is identical to concentration of H+, concentration of undissociated acetc acid you know from the determination of the total concentration and ICE table (or from the approximation, that they don't change by much). Plug, multiply, divide, done, you have Ka value.
Which part is confusing?
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Of course, you have to do a complete titration first, then redo the titration but add only have as much NaOH to reach the half titration point.
Not exactly - it is enough that you do titration once, registering whole titration curve. Half titration point is already there then, you just have to find it.
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Okay so for the part of the first question I find confusing is that from question #1 you are given pH and from there you can calculate [H+] and [CH3COO-] which are both X.
Then, notice that in question #2 you're not given the Ka, but expected to calculate it with the X from question #1.
So the first wuestion I had was this
1. Use the initial pH of the acetic acid solution to find the initial [H3O+] and initial [CH3COO]-(2)
Initial pH = 2.58
pH = -log[H+]=2.58
= [H+] = 10^-2.58
= 2.63*10^-3 m
so What I would do is (2.63 x 10^-3)/0.1 = 0.0263 or 2.63 x 10^-2
Would that be the correct answer then?
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Okay so for the part of the first question I find confusing is that from question #1 you are given pH and from there you can calculate [H+] and [CH3COO-] which are both X.
Then, notice that in question #2 you're not given the Ka, but expected to calculate it with the X from question #1.
What does the fact that something is marked with X has to do with the calculations? Something is an unknown, once the problem is solved and you know its value, it becomes a known, no matter what symbol is/was/will be used.
So the first wuestion I had was this
1. Use the initial pH of the acetic acid solution to find the initial [H3O+] and initial [CH3COO]-(2)
Initial pH = 2.58
pH = -log[H+]=2.58
= [H+] = 10^-2.58
= 2.63*10^-3 m
so What I would do is (2.63 x 10^-3)/0.1 = 0.0263 or 2.63 x 10^-2
Would that be the correct answer then?
Close, but you should check your math. You were right up to 2.63×10-3 M.
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whoops forgot to square it the answer is 6.0 x 10-9
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okay another question I would like to attempt
10. Calculate the percent difference between your value for Ka (from the calculations and the graph) and the accepted value.
Would I use the initial Ka value which was 6.916 x 10-6 or the half neutralized one which was 6.0 x 10-9?
And I am not sure what it means by the calculation of the graph. Also is there a formula that is needed to calculate it?
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Would I use the initial Ka value which was 6.916 x 10-6 or the half neutralized one which was 6.0 x 10-9?
Both, each separately.
And I am not sure what it means by the calculation of the graph.
Are we reading the same question? Not "calculations of the graph" but "calculations AND the graph". First refers to the number you just calculated, the other to the number you are expected to read from the titration curve.
Also is there a formula that is needed to calculate it?
(Kaexp/Katables-1)*100%
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so the ka expected is the 6.0 x 10-9 and the other one and they are each done separately and the Ka table is the value I would get on the table which would be 1.8 x 10^-5?
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so the ka expected is the 6.0 x 10-9
Please pay attention to what you write, you make so many small mistakes it is pretty difficult o know what you are talking about.
exp=experimental
Basically it is about calculating ratio of the value you got and the value taken from tables. Trick is, what you will get is something like 120% (for example). You are asked about the difference, so it is 20%. That was already reflected by the formula I have listed.
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Also you wrote Katable-1 would I subtract 1.8 x 10^-5 by 1?
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But basically this is what I got
% difference = Ka Experimental/ Ka Given * 100
= 6.916 x 10^-6/ 1.8 x 10^-5 * 100
= 38.4%
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And for the second calculation I got
Calculation #2
% difference = Ka Experimental/ Ka Given * 100
= 6.0 x 10-9/ 1.8 x 10-5 * 100
= 0.033%
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Where did you got 6.0×10-9 from.
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But basically this is what I got
% difference = Ka Experimental/ Ka Given * 100
= 6.916 x 10^-6/ 1.8 x 10^-5 * 100
= 38.4%
If one is 38.4% of the other, difference is NOT 38.4%.
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Where did you got 6.0×10-9 from.
This was the second calculation I got when acetic acid is half neutralized
9. Calculate Ka when the acetic acid was half-neutralized. How does this value compare with your Ka value for acetic acid? (2)
pH for 6 mL = 4.58
pH = -log[H+]=4.58
= [H+] = 10^-4.58
= 2.63 x 10^-5 mol/L
HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2-(aq)
HC2H3O2 H3O+ C2H3O2-
Initial 0.1 0 0
Change -x +x +x
Equilibrium 0.1-x (the x is relatively small compared to 0.1) x x
Ka = [H3O+][C2H3O2-]
HC2H3O2
Ka = [0.1]
Ka= [2.63 x 10^-5]^2
[0.1]
Ka = 6.0 x 10^-9
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But basically this is what I got
% difference = Ka Experimental/ Ka Given * 100
= 6.916 x 10^-6/ 1.8 x 10^-5 * 100
= 38.4%
If one is 38.4% of the other, difference is NOT 38.4%.
so the difference would be 100 - 38.4 = 61.6?
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so the difference would be 100 - 38.4 = 61.6?
That would be my approach.
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This was the second calculation I got when acetic acid is half neutralized
9. Calculate Ka when the acetic acid was half-neutralized. How does this value compare with your Ka value for acetic acid? (2)
pH for 6 mL = 4.58
(some wrong calculations not quoted)
You already wrote earlier:
At the half-titration point, the amount of HA left is equal to the amount of A- formed. At this point, the Ka = [H+].
(...)
If Ka = [H+], the pKa = pH.
Why don't you follow what you already wrote, instead of trying to reinvent the wheel?
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so for my other calculation I ended up getting 0.033%, which would then be 100 - 0.033 = 99.67%. Does that not seem to high? And also I do not understand that concept of Ka = [H+] would that mean it is 0.1 mol/L. Totally confused now.
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Okay so for the part of the first question I find confusing is that from question #1 you are given pH and from there you can calculate [H+] and [CH3COO-] which are both X.
Then, notice that in question #2 you're not given the Ka, but expected to calculate it with the X from question #1.
What does the fact that something is marked with X has to do with the calculations? Something is an unknown, once the problem is solved and you know its value, it becomes a known, no matter what symbol is/was/will be used.
So the first wuestion I had was this
1. Use the initial pH of the acetic acid solution to find the initial [H3O+] and initial [CH3COO]-(2)
Initial pH = 2.58
pH = -log[H+]=2.58
= [H+] = 10^-2.58
= 2.63*10^-3 m
so What I would do is (2.63 x 10^-3)/0.1 = 0.0263 or 2.63 x 10^-2
Would that be the correct answer then?
Close, but you should check your math. You were right up to 2.63×10-3 M.
You quoted me being correct earlier but I did some mathematical error and forgot to square it, so the basis of my question is correct here right?
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I think I just had a eureka moment
Ka = [H3O+][C2H3O2-]
HC2H3O2
Ka = [0.1]
But since Ka = [H+] at the half way point the answer would be
Ka = [H+]
= 2.63 x 10^-5
Calculation #2
% difference = Ka Experimental/ Ka Given * 100
= 2.63 x 10-5/ 1.8 x 10-5 * 100
= 1.461
100 – 1.461 = 98.54%
*Please be right*
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I do not understand that concept of Ka = [H+] would that mean it is 0.1 mol/L. Totally confused now.
No, it wouldn't mean 0.1 mol. You are mistaking concentration of H+ (given by pH) with the analytical (total) concentration of the acetic acid.
since Ka = [H+] at the half way point the answer would be
Ka = [H+]
= 2.63 x 10^-5
Yes, you got it correctly now (assuming that's the correct concentration of H+).
% difference = Ka Experimental/ Ka Given * 100
= 2.63 x 10-5/ 1.8 x 10-5 * 100
OK so far.
= 1.461
Nope. This is without multiplying by 100%.
100 – 1.461 = 98.54%
1.461 was already wrong, but here even the subtraction is not correct - both in terms of logic and in terms of the math.
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oo goodness I am making silly mistakes thanks for notifying me
Calculation #2
% difference = Ka Experimental/ Ka Given * 100
= 2.63 x 10-5/ 1.8 x 10-5 * 100
= 146.11% (what does this mean if it is over 100%)
100 – 146.11 = 46.11%
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So just a summary of what I did and my answers to them
1. Write the balanced chemical equation for the neutralization reaction you observed showing state symbols. (2)
CH3COOH (aq) + NaOH (aq) CH3COONa(aq) + H2O(l)
Acetic acid + sodium hydroxide → sodium acetate + water
2. Plot a graph of your data, with pH on the vertical axis and volume of NaOH on the horizontal axis. Your graph should show a steep rise in pH as the volume of NaOH becomes enough to neutralize the acetic acid. Take the midpoint of this steep rise and read off the volume of NaOH. This is the volume of NaOH that was needed to neutralize all the acetic acid. Compare the volume on your graph with the volume you recorded when the phenolphthalein indicator first turned pink. (5)
The midpoint of the steep rise was 12.0 mL of NaOH that was needed to neutralize all the acetic acid.
The recorded value of phenolphthalein when it first turned pink was also at 12.0 mL of NaOH.
When an indicator is used in a titration, the color change occurs at what is called the endpoint. If the indicator has been properly selected, this point will be the same as the equivalence point. When a pH meter is used, the pH of the solution is recorded as the titrant is added. The pH versus the volume of titrant added can be plotted on what is called a titration curve. In this case the equivalence point occurs at the point where very small additions of titrant cause a very rapid rise in the pH.
3. Determine the amount of NaOH added (in mols).(1)
0.1 mol/L of NaOH and 15 mL of NaOH 0.015L
(0.015L)(0.1mol/L)
=0.0015 moles
4. Use the ratio in which the acid and base react, determined from the chemical equation. Calculate the molar concentration of the acetic acid. (2)
CH3COOH (aq) + NaOH (aq) CH3COONa(aq) + H2O(l)
From the equation, there was 1 mole of NaOH neutralized with 1 mole of acetic acid; so therefore, there are 0.0015 moles of acid.
n=CV
0.0015= C(0.015)
C= 0.1 mol/L
5. Write the expression for Ka for the ionization of acetic acid in water. (2)
HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2-(aq)
Ka = [H3O+(aq)][ C2H3O2-(aq)]_____
[HC2H3O2(aq)]
6. Use the initial pH of the acetic acid solution to find the initial [H3O+] and initial [CH3COO]-(2)
Initial pH = 2.58
pH = -log[H+]=2.58
= [H+] = 10-2.58
= 2.63 x 10-3 mol/L for both [H3O+] and [CH3COO]-
7. Assume that the amount of CH3COOH that ionizes is small compared with the initial concentration of the acid. Calculate Ka for the acid. (2)
HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2-(aq)
HC2H3O2 H3O+ C2H3O2-
Initial 0.1 0 0
Change -x +x +x
Equilibrium 0.1-x (the x is relatively small compared to 0.1) x x
Ka = [H3O+][C2H3O2-]
HC2H3O2
Ka = [0.1]
Ka= [2.63 x 10-3]2
[0.1]
Ka = 6.916 x 10-6
8. Refer to the volume of NaOH on your graph from question 2. Calculate half this volume and on your graph, find the pH when the solution was half neutralized. (1)
Original volume of NaOH = 12
Half the volume of NaOH = 6
pH for 12 mL of NaOH = 10.66
pH for 6 mL of NaOH = 4.58
9. Calculate Ka when the acetic acid was half-neutralized. How does this value compare with your Ka value for acetic acid? (2)
pH for 6 mL = 4.58
pH = -log[H+]=4.58
= [H+] = 10-4.58
= 2.63 x 10-5 mol/L
HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2-(aq)
HC2H3O2 H3O+ C2H3O2-
Initial 0.1 0 0
Change -x +x +x
Equilibrium 0.1-x (the x is relatively small compared to 0.1) x x
Ka = [H3O+][C2H3O2-]
HC2H3O2
Ka = [0.1]
But since Ka = [H+] at the half way point the answer would be
Ka = [H+]
= 2.63 x 10-5
10. Calculate the percent difference between your value for Ka (from the calculations and the graph) and the accepted value. (3)
Calculation #1
% difference = Ka Experimental/ Ka Given * 100
= 6.916 x 10-6/ 1.8 x 10-5 * 100
= 38.4%
100 – 38.4 = 61.6%
Calculation #2
% difference = Ka Experimental/ Ka Given * 100
= 2.63 x 10-5/ 1.8 x 10-5 * 100
= 146.11%
100 – 146.11 = 46.11%
11. Do the values you calculated for [H3O+] and [CH3COOH] prove that CH3COOH is a weak acid? Explain. (2)
The values calculated for [H3O+] and [CH3COOH], which is 2.63 x 10-3 mol/L and 2.63 x 10-5 mol/L for when it was half-neutralized, show that it has not been completely ionized. So the stronger the acid the larger their Ka value is, but the weaker the acid the smaller their Ka value is. A strong acid is a better proton donor, resulting in more products. Since the concentration of the products is in the numerator of the Ka expression, the stronger the acid, the larger the Ka. The two values for the acetic acid are Ka = 6.916 x 10-6 and Ka = 2.63 x 10-5. The exponents show that the value contains numerous zeros before it not having a large number for the Ka shows that acetic acid is a weak acid.
AND there is an attachment of how my graph looks like
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The midpoint of the steep rise was 12.0 mL of NaOH
The recorded value of phenolphthalein when it first turned pink was also at 12.0 mL of NaOH.
0.1 mol/L of NaOH and 15 mL of NaOH looks like
So 12, or 15?
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The total amount is 15 but at the midpoint it was 12 so half of that would be 6
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So why do you use 15 mL to calculate amount of acid?
And 12 mL was not a midpoint, but endpoint. Midpoint was 6 mL.
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So why do you use 15 mL to calculate amount of acid?
And 12 mL was not a midpoint, but endpoint. Midpoint was 6 mL.
okay if the midpoint is 6 then that is when it was neutralized meaning that the half neutralized is 3? And I use 15 mL because as you can see on the graph that is where the volume and pH correlation ends ... ?
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Midpoint is the point in the middle, half way between the start and equivalence point - that's where the acid was half neutralized. So if the equivalence point was at 12 mL, midpoint was at 6 mL.
To calculate amount of titrant (and stoichiometric equivalent of titrated substance) you should use volume required to arrive at the end point (so 12 mL), not some random point after that.
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Okay so changing everything else up since the pH calculation would have changed since the half neutralized point would now be 3 and the pH for it is 4.17 these are my answers again
1. Write the balanced chemical equation for the neutralization reaction you observed showing state symbols. (2)
CH3COOH (aq) + NaOH (aq) CH3COONa(aq) + H2O(l)
Acetic acid + sodium hydroxide → sodium acetate + water
2. Plot a graph of your data, with pH on the vertical axis and volume of NaOH on the horizontal axis. Your graph should show a steep rise in pH as the volume of NaOH becomes enough to neutralize the acetic acid. Take the midpoint of this steep rise and read off the volume of NaOH. This is the volume of NaOH that was needed to neutralize all the acetic acid. Compare the volume on your graph with the volume you recorded when the phenolphthalein indicator first turned pink. (5)
The midpoint of the steep rise was 6.0 mL of NaOH that was needed to neutralize all the acetic acid.
The recorded value of phenolphthalein when it first turned pink was at 12.0 mL of NaOH.
When an indicator is used in a titration, the color change occurs at what is called the endpoint. If the indicator has been properly selected, this point will be the same as the equivalence point. When a pH meter is used, the pH of the solution is recorded as the titrant is added. The pH versus the volume of titrant added can be plotted on what is called a titration curve. In this case the equivalence point occurs at the point where very small additions of titrant cause a very rapid rise in the pH.
3. Determine the amount of NaOH added (in mols).(1)
0.1 mol/L of NaOH and 12 mL of NaOH 0.012L
(0.012L)(0.1mol/L)
=0.0012 moles
4. Use the ratio in which the acid and base react, determined from the chemical equation. Calculate the molar concentration of the acetic acid. (2)
CH3COOH (aq) + NaOH (aq) CH3COONa(aq) + H2O(l)
From the equation, there was 1 mole of NaOH neutralized with 1 mole of acetic acid; so therefore, there are 0.0012 moles of acid.
n=CV
0.0012= C(0.012)
C= 0.1 mol/L
5. Write the expression for Ka for the ionization of acetic acid in water. (2)
HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2-(aq)
Ka = [H3O+(aq)][ C2H3O2-(aq)]_____
[HC2H3O2(aq)]
6. Use the initial pH of the acetic acid solution to find the initial [H3O+] and initial [CH3COO]-(2)
Initial pH = 2.58
pH = -log[H+]=2.58
= [H+] = 10-2.58
= 2.63 x 10-3 mol/L for both [H3O+] and [CH3COO]-
7. Assume that the amount of CH3COOH that ionizes is small compared with the initial concentration of the acid. Calculate Ka for the acid. (2)
HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2-(aq)
HC2H3O2 H3O+ C2H3O2-
Initial 0.1 0 0
Change -x +x +x
Equilibrium 0.1-x (the x is relatively small compared to 0.1) x x
Ka = [H3O+][C2H3O2-]
HC2H3O2
Ka = [0.1]
Ka= [2.63 x 10-3]2
[0.1]
Ka = 6.916 x 10-6
8. Refer to the volume of NaOH on your graph from question 2. Calculate half this volume and on your graph, find the pH when the solution was half neutralized. (1)
Original volume of NaOH = 6
Half the volume of NaOH = 3
pH for 6 mL of NaOH = 4.58
pH for 3 mL of NaOH = 4.12
9. Calculate Ka when the acetic acid was half-neutralized. How does this value compare with your Ka value for acetic acid? (2)
pH for 3 mL = 4.12
pH = -log[H+]=4.12
= [H+] = 10-4.12
= 7.6 x 10-5 mol/L
HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2-(aq)
HC2H3O2 H3O+ C2H3O2-
Initial 0.1 0 0
Change -x +x +x
Equilibrium 0.1-x (the x is relatively small compared to 0.1) x x
Ka = [H3O+][C2H3O2-]
HC2H3O2
Ka = [0.1]
But since Ka = [H+] at the half way point the answer would be
Ka = [H+]
= 7.6 x 10-5
10. Calculate the percent difference between your value for Ka (from the calculations and the graph) and the accepted value. (3)
Calculation #1
% difference = Ka Experimental/ Ka Given * 100
= 6.916 x 10-6/ 1.8 x 10-5 * 100
= 38.4%
100 – 38.4 = 61.6%
Calculation #2
% difference = Ka Experimental/ Ka Given * 100
= 7.6 x 10-5 / 1.8 x 10-5 * 100
= 422.22%
100 – 422.22= 322.22%
11. Do the values you calculated for [H3O+] and [CH3COOH] prove that CH3COOH is a weak acid? Explain. (2)
The values calculated for [H3O+] and [CH3COOH], which is 2.63 x 10-3 mol/L and 7.6 x 10-5 mol/L for when it was half-neutralized, show that it has not been completely ionized. So the stronger the acid the larger their Ka value is, but the weaker the acid the smaller their Ka value is. A strong acid is a better proton donor, resulting in more products. Since the concentration of the products is in the numerator of the Ka expression, the stronger the acid, the larger the Ka. The two values for the acetic acid are Ka = 6.916 x 10-6 and Ka = 7.6 x 10-5. The exponents show that the value contains numerous zeros before it not having a large number for the Ka shows that acetic acid is a weak acid.
For the part that I bolded the percent difference is a very large number, how may that be?
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Okay so changing everything else up since the pH calculation would have changed since the half neutralized point would now be 3
No, it won't be. Midpoint and the half neutralization is the same, as explained in my previous post.
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Okay so changing everything else up since the pH calculation would have changed since the half neutralized point would now be 3
No, it won't be. Midpoint and the half neutralization is the same, as explained in my previous post.
okay so everything that I did before was correct except the whole 12 and 15 volume thing. And midpoint and half neutralization mean the same thing. Got it thanks so much, you have been a great *delete me*