[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 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}

[Dan]

You must show you have attempted the question, this is a Forum Rule.

[sallyhansen]

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.
 

[Arkcon]

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.

[sallyhansen]

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)]

[sallyhansen]

okay if the concentration is 0.1 mol/L I would just then do x^2/Co and calculate it that way?

[sallyhansen]

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!

[sallyhansen]

whoops accidentally put 0.2 it is 0.1

[Borek]

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.

[sallyhansen]

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 ...

[sallyhansen]

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?

[Borek]

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.

[sallyhansen]

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?

[sallyhansen]

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.

[Borek]

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?