Chemical Forums
Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: Solerb on January 13, 2007, 12:28:08 PM
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Propanoic acid 0.74 g (MWt = 74) was dissolved in 250ml distilled water in a volumetric flask.
50 ml of this acid solution was titrated against 0.01M KOH
(a) Calculate the dissociation constant and the pKa of the acid if the pH of the acid solution was found to be 3.14
(b) What is the pH of the KOH solution ?
(c) determine the pH after 75ml of base have been added and the pH at equivalance
(d) what is the pH of the buffer solution prepared by mixing 300ml of the acid with 700ml of the base
thanks ppl
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A: http://www.ou.edu/provost/integrity/
B: http://www.chemicalforums.com/index.php?topic=289.0
C: http://dictionary.reference.com/search?q=attempt
D: http://en.wikipedia.org/wiki/Chemistry
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Hehe oki i will give my answers then i am not saying that my exam is on monday
but 1 week time. Come on I forgot this staff i need a brush up, these are basic a level i know but a year passed and forgot things hehe
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thanks for shoing me the rule let me give u till what i am up to i found out the number of moles of propanoic whiich is 0.002 moles
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now the equation is
CH3CH2COOH +h2o ->(equilbrium) H+ + CH3CH2COO-
Ka = [H+] + [CH3CH2COO-]
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[CH3CH2COOH]
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pKa = -log Ka
pH= -log[H+]
3.14= -log[H+]
H+= 0.497 ?
I think i am doing something wrong
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pKa = -log Ka
pH= -log[H+]
3.14= -log[H+]
H+= 0.497 ?
I think i am doing something wrong
You are.
pH = -log(H+)
H+ = 10-pH
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0.000724? isn't that a small number i know we are dealing with propanoic weak acid
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0.000724? isn't that a small number i know we are dealing with propanoic weak acid
That is correct. Sure it is a small number. But look at what you have.
0.74 g of a weak acid with MW 74g. You only have 0.01mol of the acid. You then mix it in 0.250L. This means you have a 0.04M, a fairly "weak" solution of a weak acid.
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wait i think i got it
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H+ = square root of Ka X c
7.24x10 -4 = kax 0.04
= 1.31x10-5
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pka=-log1.31x10-5
= 4.88
b) OH- = square root of kw x c
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ka
1x10-14 x 0.01
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1.31x10-5
=2.76x10-6
pOH = -log[OH-]
=5.56
=14-5.56 = 8.44
Am I right?
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My answers came pls can somebody check
a 4.88
b 8.44
c 8.54
d 5.95
Are these figures right ?
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First lets start with A
What grade class is this?
In highschool, or for a quick estimate I would accept the (Ka * x)1/2 method.
But for real work I would expect people to use then
Ka = [H+] * [A-]
[HA]
*note* If you have not covered this before because you are in highschool, you can ignore it, but read it to see the % difference.
In your case that becomes:
( 7.224*10-4)2
0.04 - 7.224x10-4
= 1.336*10-5
Which is only 1.5% difference from what you got (before -log), but is still a better and more precise method of calculating.
And then B)
I am not sure what you are doing? This is a strong base.
You should know that:
Kw = [H+][OH-] = 1.0x10-14
You know Kw, and OH- because this is a strong base; one of the assumptions we are allowed to make is that a strong base (or acid) dissociates completely.
[H+] = Kw = 1.0x10-14
[OH-] 0.01
If you are in high school, please repost c and d in the high school forum, as in this forum people are expecting you to be in a college level analytical class. If you need more explanation for a and b just ask, but because you did not show what you tried for c and d I feel no desire to check if you are correct. It is more important to understand what/why then to just get the correct answer. Btw, question C you gave one answer, but from your question it should have two answers.
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In highschool, or for a quick estimate I would accept the (Ka * x)1/2 method.
I have no idea about American HS, but I doubt at any level you are asked to remember just a formula without understanding anything (unless American HS has gone so low ;) ).
Whether this approach gives correct results or not depends on whether some approximation holds - and it can be easily checked if it holds.
http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base
(Shane: I am more then sure that you know 5% rule, I just can't agree with the approach of accepting the method in HS).
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I AM A University student (malta) B Pharmacy 1st year . we use to do them a year ago... i do not know the system u r dealing with
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In highschool, or for a quick estimate I would accept the (Ka * x)1/2 method.
I have no idea about American HS, but I doubt at any level you are asked to remember just a formula without understanding anything (unless American HS has gone so low ;) ).
Whether this approach gives correct results or not depends on whether some approximation holds - and it can be easily checked if it holds.
http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base
(Shane: I am more then sure that you know 5% rule, I just can't agree with the approach of accepting the method in HS).
The problem is, most American high schools have gone that low. :(
In fact, the majority of undergrad colleges now seem to be going that route. :(
In high school I would focus more on concepts and theories and less on computational aspects of chemistry my self. Plus, most Americans suck big time at math, unfortunately, I am not sure if the average American high school student knows the quadratic formula.
That is not to say that all are stupid, many come out of high school with enough knowledge to skip entire freshman level college chemistry.
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Nucular potatoe you mean...
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Not to stray off topic but a lot of the concepts that we should have learned in HS are being learned now. Plus all the in-depth stuff and that is why a lot of students have difficulty with it. The basics were never completely understood.