[Solerb]

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

[enahs]

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

[Solerb]

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

[Solerb]

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

[Solerb]

now the equation is

CH3CH2COOH +h2o ->(equilbrium) H+ + CH3CH2COO-

Ka = [H+] + [CH3CH2COO-]
        ------------------------------
         [CH3CH2COOH]

[Solerb]

pKa = -log Ka
pH= -log[H+]

3.14= -log[H+]
H+= 0.497 ?

I think i am doing something wrong

[enahs]

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

[Solerb]

0.000724? isn't that a small number i know we are dealing with propanoic weak acid

[enahs]

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.

[Solerb]

wait i think i got it

[Solerb]

H+ =  square root of Ka X c

7.24x10 -4 = kax 0.04
= 1.31x10-5

[Solerb]

pka=-log1.31x10-5
= 4.88

b) OH- = square root of kw x c
                                --------
                                  ka

1x10-14 x 0.01
-----------
1.31x10-5

=2.76x10-6

pOH = -log[OH-]
=5.56
=14-5.56 = 8.44



Am I right?

[Solerb]

My answers came pls can somebody check

a 4.88
b 8.44
c 8.54
d 5.95

Are these figures right ?

[enahs]

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.

[Borek]

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).