Titration Lab questions help needed!?

[sallyhansen]

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 K_a for the acid. (2)

HC₂H₃O_2(aq) + H₂O_{(l) --->} H₃O⁺_(aq)+ C₂H₃O₂^-_(aq)

K_a = [H₃O⁺_(aq)][ C₂H₃O₂^-_(aq)]____

[HC₂H₃O_2(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 K_a when
the acetic acid was half-neutralized. How does this value compare with
your K_a value for acetic acid? (2)

5. Calculate the percent difference
between your value for K_a (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?

[Murka]

Sorry, in the middle of a lecture, check this out see if this helps and we'll continue

[sallyhansen]

omg this is excellent. So the part after the pH is the second question onwards correct? Also do you mind continuing you are a life saver.

I am not really sure what the amount of ionization will be but I will put the procedure of the lab below.

 

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.

Edited April 23, 2013 by sallyhansen

[Murka]

I don't understand what would you want me to do, question 2 depends on the concentration of the acetic acid, and all the other answered are gained experimentally from your experiment's results i.e. half titration of the acetic acid from the naoh etc'.

Also pay attention that from the Henderson-Hasselbalch equation (http://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation) we see that when the concentrations of Acid and Base are equal the pH=pKa, thus the Ka of the acid can be calculated very easily.

[sallyhansen]

No but for this question how would I go about answering this

 

2. Assume that the amount of CH₃COOH that ionizes is small compared with the initial concentration of the acid. Calculate K_a for the acid. (2)

 

How do I calculate Ka?

HC₂H₃O_2(aq) + H₂O_{(l) --->} H₃O⁺_(aq)+ C₂H₃O₂^-_(aq)

K_a = [H₃O⁺_(aq)][ C₂H₃O₂^-_(aq)]____

[HC₂H₃O_2(aq)]

[Murka]

Read my scan, I've explained it, but it seems like you have missing data since you need the concentration of the unionized acid (CH3COOH) to calculate it.

[sallyhansen]

so once I find the concentration I would use the equation that was on the scan

[sallyhansen]

okay the concentration is 0.1 mol/L

[sallyhansen]

Okay I have finally answered this question (I know I am slow ...) but here is it please tell me if I have completed it correctly

 

1. Assume that the amount of CH₃COOH
that ionizes is small compared with the initial concentration of the
acid. Calculate K_a for the acid. (2)

HC₂H₃O_2(aq) + H₂O_(l) à 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

K_a = [H₃O⁺][C₂H₃O₂^-]

HC₂H₃O₂

K_a= [x][x]

[0.1]

(1.8 x 10^-5)(0.1)= x²

1.8 x 10^-6= x²

(square root both sides)

x = 1.342 x10^-3

[Murka]

You're coming at this from the wrong angle. 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.

[sallyhansen]

So I am not understanding how I am supposed to approach the question anymore

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?

[Murka]

You didn't use the Ka of the acetic acid to find X, you used the pH=-log[H+] of the acid to find x=[H+]=[A-], then you use this X and [HA]=0.1M (from your data) to find Ka.

Then you're being asked for the experimental data. your acid was neutralized at 12 ml of NaOH, right? This means that it was half-neutralized at 6.

You know the pH at 6, what do you know about the half-titration point?

What is special about it?

leading you there...

 

So, assuming 8 is correct, what do you know about 9?

[sallyhansen]

The Ka value is 1.8 x 10^{-5 do I plugged that into the Ka value part to find the answer for X?}

So I would assume that I am 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.

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?

whoops the answer is 6.0 x 10^-9

[sallyhansen]

not a very helpful forum. If anyone decides they need to get some better explanation please visit this forum http://www.chemicalforums.com/index.php?topic=67705.30#top To get the questions that I have asked in a reasonable amount of time.

 

Thanks for the help I got, but not as useful as I thought it would be! :@