# Find number of moles of HCl to change pH of Histidine buffer.

1. Dec 4, 2011

### Sprinkles

1. The problem statement, all variables and given/known data

There is 400 mL of 0.2 M histidine buffer- pH 11. Find the moles of HCl that are needed to change the pH to 3.2.

2. Relevant equations

pH= pKa+log/[a]

3. The attempt at a solution

First, this is how you would go from acid to base (acid-->base):

1. There is 400 mL of 0.2 M histidine buffer- pH 3.2 Find the moles of OH that are needed to change the pH to 11.

H+2 <--> H+1 <-->H0 <-->H-1

1) Find a and b using H-H, and pKa values (2.1-pKa of COOH and 3.2 in one equation, and 9.6-pKa of NH3+ and 11 in the second equation).
a+b+1= total1) [This total is the total # of equivalents.]

2) Multiply .4 L by 0.2 M.=total2)

3) Multiple total1) and total2) together to get the final answer.

Now my question:

What if we had to go backwards- from base to acid?

There is 400 mL of 0.2 M histidine buffer- pH 11. Find the moles of HCl that are needed to change the pH to 3.2.

Would I subtract -1 now? (Please see 1).) I am just confused on how to go backwards (base to acid). How should the number change? When going from a-->b you need moles of OH- to increase the pH, but from b-->a you would need moles of HCl to decrease the pH, so the number of moles should be different, I assume. However, when calculating it, I get the same number of moles for both H+ and OH-.

Thank you!

Thank you!

Last edited: Dec 4, 2011
2. Dec 4, 2011

### epenguin

If I understand your question I don't understand your perplexity - the number of moles should be exactly the same both ways.

Also please check back the question. Histidine has another ionising group of pK about 6 you have not mentioned.

Offhand on a quick look you ought to realise that the answer has to be quite close to 2 moles added base per mole of histidine, probably more important than the exact number - is that what you got?

3. Dec 4, 2011

### Sprinkles

^Thank you so much epenguin!! I appreciate your time.

Let me explain my reasoning in a different way:

We're trying to find the number of moles of HCl that are required to change the pH from 11 to a pH of 3.2 (of this Histidine buffer solution).

We would still need to do two H-H equations (1st H-H: use pKa of COOH, 2.1 and pH, 3.2 / 2nd H-H: use pKa of NH3+, 9.6 and pH 11). Find "a" by using the 1st H-H ("a"=.07359), and find "b" using the 2nd H-H ("b"=.9693).

The total number of moles of Histidine= 0.4 L x 0.2 M= 0.08 moles of Histidine

[*Assuming that the addition of acid doesn't interfere w/volume of solution.*]

To find the total number of moles of H+ = (a=acid)(moles of histidine) + (b=base)(moles of histdine) (+/-) ( ______ )

(.07359)(0.08) + (.9693)(0.08) (+ or -) _______

I'm getting stuck with the last blank. Would I subtract or add 0.08 now?

To find the moles of OH- needed to change pH from 3.2 to 11, you would add another 0.08 moles of Histdine (in that last blank).
So, to find the moles of H+ needed to change pH from 11 to 3.2, does this mean that you need to subtract 0.08 moles of Histidine?

I am getting confused, because I get the same answer for both the number of moles of OH- needed to go up the curve, and for the number of moles
of H+ needed to go down the curve. My reasoning is because it's all along the same curve, and it's an equilibrium.

In terms of ionizable groups, this is the equation that I have:

His+2 <---> His+1 <---->His0 <----> His-1
(pKa of COOH 2.1)-->inbetween His +2 and His+1
(pKa of Histidine 6.0) --> inbetween His +1 and His 0
(pKa of NH3+ 9.6)-->inbetween His0 and His-1

Last edited: Dec 4, 2011
4. Dec 4, 2011

### epenguin

To avoid complication of inessentials I’ll take a case of 1l of 1M solutions.

Just take the case of a simple base RNH2. At pH 3.2 error is totally negligible in assuming it 100% RNH3+. To go to high pH you have to add 1 mole of NaOH. To go to only pH 11 you add instead a bit less, roughly 0.96 moles. So you’ll have in your solution about 0.96 M RNH2 AND 0.04 M RNH3+, exact amount calculate by HH equation.

To get back to pH 3.2 you’ll have to add the same amount, 0.96 moles say, of HCl! What you’re really doing is reacting your RNH2 with H+ to reprotonate them. But along with the H+ you have to add Cl-.

(I guess the confusing thing about these calculations is that the H+ which is the concentration usually of interest is such a small fraction of the ions present that for some parts of the calculation you ignore it, you ignore the thing that is of interest!)

This principle is still the same however many ionisations there are, however complicated the scheme. The moles of strong base you add to increase the pH from x to y = amount of strong acid to decrease from y to x.

In your case you have to add 1 mole in going from pH 3.2 to 11 to get your imidazole group of histidine from totally protonated to totally unprotonated (well below to well above the pK). You seem on second reading to be perhaps aware of this.

There is additionally a bit less than 1 mole of base to be added to nearly deprotonate the RNH3+ as explained above.

On the other hand at pH 3.2, there is a small fraction of the carboxyl group still protonated. I think this will be about 0.08 M. So you will have to add a corresponding extra amount of base to deprotonate this.

So yes the answer to your question is the two effects are opposite, there will be a subtraction, not addition.

I hope this is not additionally confusing (I see now I have done the inverse problem by should be OK) and that it helps you think your way through.

The scheme is HOOC-Im+-NH3+ $\Leftrightarrow$ -OOC-Im+-NH3+ $\Leftrightarrow$ OOC-Im-NH3+ $\Leftrightarrow$ -OOC-Im-NH2
(left out the loose protons).

5. Dec 4, 2011

### Sprinkles

Thank you so much for that explanation! I'm not really advanced in pH/buffer questions, so it's a tad confusing, but I still appreciate your reply.

So in 1) you're saying that 0.08 moles of histidine will always be the case for both when you're going up the pH and down?

and by 2) you mean that it would look like this:

The total number of moles of HCl that are required to change the pH from 11 to a pH of 3.2 (of this Histidine buffer solution)=

(a=acid)(moles of histidine) + (b=base)(moles of histidine) - (moles of histidine)

(.07359)(0.08) + (.9693)(0.08) - 0.08
At pH of 3.2, histidine has a net charge of +1.

----

versus

There is 400 mL of 0.2 M histidine buffer- pH 3.2 Find the moles of OH- that are needed to change the pH to 11.
Where this would be: (.07359)(0.08) + (.9693)(0.08) + 0.08 <----This is the correct answer when going from
acid to base (pH 3.2 to pH 11).
At pH 11, histidine has a net charge of 0.

Last edited: Dec 4, 2011
6. Dec 5, 2011

### epenguin

The equilibria written better:

HOOC-ImH+-NH3+ $\Leftrightarrow$ -OOC-ImH+-NH3+ $\Leftrightarrow$ -OOC-Im-NH3+ $\Leftrightarrow$ -OOC-Im-NH2

Net charges: +2, +1, 0, -1

If I understand the question, yes, I confirm.

You can also think of it like this. If I add a number of moles NaOH upping the pH then the same number moles HCl going down I get to the pH I started. If these numbers are equal, it is like I added NaCl - which is 'neutral' and doesn't change pH. Conversely if they are not equal it is like I added some acid or alkali which does change the pH

More useful almost than exact calculations are rough ones that help you see where you are and what is happening or should be.

So remember that if you are 1 pH unit away from the pK you are getting out out of the buffering zone, the unfavoured form is about 10% of the total, 2 units away it is near 1%, 3 units, 0.1% and so on. From that you could have guessed roughly about how much the fractions mentioned were. Also from that you can see that at pH 3.2 there is so little of the 0 and -1 forms you can ignore them, while at pH 11 you can ignore the +1 and +2 forms.

So, we maybe saying the same thing in different ways, to get from pH 11 to 3.2 (1l 1M total histidine) you will be adding a bit less of than a mole, 0.96 moles, of HCl to protonate the amino group of the not quite total -1 form that there is, 1 mole to protonate the imidzole group that brings you to the +1 form and then 0.038 mole to get the small fraction of +2 that there is at pH 3.2, I make it just over 2: 2.036. I expected it very close to 2.

At end just multiply by your 0.08 to change it to your volume and concentration

Last edited: Dec 5, 2011
7. Dec 5, 2011

### Sprinkles

^Thank you so much for the reply, epenguin!!!! :)
I appreciate the time that you spent, and the help that you have provided.

Happy Holidays!