[mspike6]

Hello,
I have been trying to get help on this question (ionline),  but i couldn't find anyone who can explain it to me, instead of just giving me the answer.
I really appreciate it if you are willing to give me the answer right away, but i really prefer it if someone explained to me how to solve it myself, especially that i am struggling in this unite.

without further ado, here is the question:

"Buffers that match the pH of blood and the cytoplasm are commonly required in scientific experiments. If you were studying the effects of lactic acid on muscles and needed to design a buffer that would have a pH of 7.2, what chemical components would you select and in what proportions? Include relevant calculations in your answer."

Please, if someone could explain it to me step by step, i would REALLY REALLY appreciate it !

Thanks guys !!

[Jorriss]

So, from what I recall, a buffer is an acid and its conjugate base used to maintain the pH of a solution. And a buffer works best if the pKa of the acid is near the desired pH of solution.

To design a buffer at pH 7.2 we go to the Henderson-Hasselbalch equation:

pH=pKa +log (A-/HA)


pH=7.2 - as given in the question.

pKa is upto us, we choose the acid. But we want something in the range of the pH. Hypochlorous acid's pKa is 7.53 if I recall, and it's fairly close, so let's use that.

pKa = 7.53



Now, plug those in and we solve for A-/HA, which will give us the ratio of conjugate base: acid needed for our buffer solution.

Hope that helps (and that it's right )

[Borek]

You are wrong and right at the same time. Right - that's the correct approach to buffer calculation/preparation. Wrong - hypochlorous acid is a strong oxidant, and as such it is toxic, so of no use for biology experiments.

Think phosphoric acid or carbonic acid, or something like Tris, BES or MOPS.

[mspike6]

Am confused again

Can someone Simplify it to me ?

So far i understand that :

1) we need the pH=pK = 7.2 (or as near to that number as possible )
 2) From the "Relative Strengths of Acids and Bases" Table, the Dihydrogen Phosphate Ion has Ka - 6.2 * 10^-8 , Since pK= -Log [Ka] the H₂PO₄^- Has pK = 7.21
3) So i guess we can use the H₂PO₄^- as our weak acid.

That's what i got, can someone explain to me from here on ? and tell me if what i understand is correct ?

Thanks guys, really appreciate the help !

[Jorriss]

You are wrong and right at the same time. Right - that's the correct approach to buffer calculation/preparation. Wrong - hypochlorous acid is a strong oxidant, and as such it is toxic, so of no use for biology experiments.

Ahhh true, I was just thinking what I knew of with the right pKa, roughly. Good catch

[Borek]

Now you have to identify acid and conjugate base.

[mspike6]

Now you have to identify acid and conjugate base.

Isn't the
H₂PO₄^- is the Acid, and HPO₄^2- it's conjuget base ?

and what do i do next ?

Thanks

[Jorriss]

Now you have to identify acid and conjugate base.

Isn't the
H₂PO₄^- is the Acid, and HPO₄^2- it's conjuget base ?

and what do i do next ?

Thanks

Yeah.

Check this out, http://www.scifun.org/chemweek/BioBuff/BioBuffers.html

[mspike6]

[/quote]
Yeah.

Check this out, http://www.scifun.org/chemweek/BioBuff/BioBuffers.html
[/quote]
Awesome link, really clarified alot of things to me .

So here is my Solution that i would write if such a question came in am exam.

Solution


It is best to find an Acid which has a pKa equal to (or as close as possible) to the pH we need.
From the "Relative Strengths of Acids and Bases" Table we can see that dihydrogen Phosphate ion has a Ka = 6.2 * 10^-8. Since pKa = -Log[Ka], The pKa for the dihydrogen Phosphate is 7.21, which is perfect for our need.

H₂PO₄^-  H⁺ + HPO₄^2-


We need the pH to be 7.2, and since pH = -Log[H⁺] , the Concentration of Hydrogen ions should be 10^-7.2 = 6.2 * 10^-8 mol/L

Since the ration of H⁺ to HPO₄^2- is 1:1, than the Concentration of  H⁺ is equal to the concentration of  HPO₄^2- equal to 6.2 * 10^-8 mol/L

Now we need to know the concentration of the dihydrogen Phosphate ion (Acid) that we gonna use.

Ka = [H⁺]*[HPO₄^2-] / H₂PO₄^-

H₂PO₄^- = [H⁺]*[HPO₄^2-] / Ka
H₂PO₄^- = [6.2 * 10^-8]² / 6.2 * 10^-8
H₂PO₄^- =6.2 * 10^-8


-----------------------------------------------------------------------

Wow, i must be doing something wrong...everything is coming out to be  6.2 * 10^-8..

Can someone tell me what is wrong in my steps ?
and if, it turned out to be right ...how do i get the concentration of the Conjugate base that i need to use ?

Thanks 

[Jorriss]

You need the henderson-hasselbalch equation now.

pH=pKa+logA-/HA, where HA is your acid - dihydrogen phosphate, and A- is your conjugate base, hydrogen phosphate.

[mspike6]

Am i correct so far?

Cause if i used pH=pKa+logA-/HA
The concentration of the baser will also be 6.2 * 10^-8 which seems like a universal number in this question

[Jorriss]

I think you are misunderstanding aspects of a buffer.

A buffer solution contains both the conjugate base AND the acid, but it's not a numerical value - like 6.98 x 10 ^-8 or whatever you get, but a ratio between the conjugate base and acid, such as one part acid for every one part base.

Now, let's look at the Henderson-Hasselbalch equation.

pH=pKa + log (A-/HA)

That log(A-/HA) term, that's the business end. It gives us the ratio between acid and conjugate base we're looking for. And conveniently, we have the two other terms - pH and pKa - that we need to find it!

Let's plug our pH and pKa into the equation.

The desired pH = 7.2 and the pKa of dihydrogen phosphate is 7.198 or about 7.2 also.


7.2=7.198 + log(A-/HA)

log(A-/HA)=0

A-/HA=1

so the ratio wanted between hydrogen phosphate and dihydrogen phosphate is one! Which is very easy



...Unless I'm wrong again...

[mspike6]

I think you are misunderstanding aspects of a buffer.

A buffer solution contains both the conjugate base AND the acid, but it's not a numerical value - like 6.98 x 10 ^-8 or whatever you get, but a ratio between the conjugate base and acid, such as one part acid for every one part base.

Now, let's look at the Henderson-Hasselbalch equation.

pH=pKa + log (A-/HA)

That log(A-/HA) term, that's the business end. It gives us the ratio between acid and conjugate base we're looking for. And conveniently, we have the two other terms - pH and pKa - that we need to find it!

Let's plug our pH and pKa into the equation.

The desired pH = 7.2 and the pKa of dihydrogen phosphate is 7.198 or about 7.2 also.


7.2=7.198 + log(A-/HA)

log(A-/HA)=0

A-/HA=1

so the ratio wanted between hydrogen phosphate and dihydrogen phosphate is one! Which is very easy



...Unless I'm wrong again...

Thanks Once again !!
it is way simpler and easier than what was in my mind.